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cdots","display":"\\cdots","kind":"use","occurrences":[657]},{"id":"symbol-942faac63e3ce23f","tex":"\\longrightarrow","aliases":"⟶ longrightarrow","display":"\\longrightarrow","kind":"use","occurrences":[1549,1922,1947]},{"id":"symbol-22449108dc31ebd8","tex":"\\longmapsto","aliases":"⟼ longmapsto","display":"\\longmapsto","kind":"use","occurrences":[1484]}],"contexts":["\\[\nR(n)=\\sum_{d^2\\mid n}\\mu(d)\\chi_5(d)d^{10}\\tau(n/d^2).\n\\]","\\section{The coefficient and its local parameters}\nThe normalized form is\n\\[\n\\Delta(q)=q\\prod_{m\\ge1}(1-q^m)^{24}\n         =\\sum_{n\\ge1}\\tau(n)q^n,\\qquad \\tau(1)=1.\n\\]\nAll coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of $\\Delta/q$. For the latter coefficients, $f_m=\\tau(m+1)$ and $f_0=1$.","oefficient and its local parameters}\nThe normalized form is\n\\[\n\\Delta(q)=q\\prod_{m\\ge1}(1-q^m)^{24}\n         =\\sum_{n\\ge1}\\tau(n)q^n,\\qquad \\tau(1)=1.\n\\]\nAll coefficient indices in this book are positive integers. A stored value at index zero is a programming sentinel unless a displayed formula explicitly introduces the coefficients of $\\Delta/q$. For the latter coefficients, $f_m=\\tau(m+1)$ and $f_0=1$.","At a prime $p$, the default local parameters are $a=\\tau(p)$ and $b=p^{11}$. The scalar recurrence has $T_0=1$, $T_1=a$ and $T_r=aT_{r-1}-bT_{r-2}$ for $r\\ge2$. Thus $T_r=\\tau(p^r)$. A construction that permits an arbitrary integer $a$ states that change explicitly. Such a recurrence need not be the local sequence of $\\Delta$.","The symbol $T_k(p)$ in the trace-formula chapter denotes a Hecke operator on the weight-$k$ cusp-form space. It is an operator, whereas $T_r$ in a scalar recurrence is an integer or an integer polynomial. The weight $k$ and the exponent $r$ serve different roles.","The valuation $v_\\ell(x)$ is the exponent of $\\ell$ in a nonzero rational number $x$, with $v_\\ell(0)=\\infty$. The quadratic character used in the radial extraction is\n\\[\n\\chifive(d)=\n\\begin{cases}\n0&5\\mid d,","The valuation $v_\\ell(x)$ is the exponent of $\\ell$ in a nonzero rational number $x$, with $v_\\ell(0)=\\infty$. The quadratic character used in the radial extraction is\n\\[\n\\chifive(d)=\n\\begin{cases}\n0&5\\mid d,\\\\\n1&d\\equiv1,4\\pmod5,\\\\\n-1&d\\equiv2,3\\pmod5.\n\\end{cases}\n\\]\nThe character value and the tenth power in the scaling formula are both essential.","\\section{Packets, content and the coefficient shift}\nCoordinates have indices $0,1,2,3,4$. A shell is\n\\[\n\\Vn=\\left\\{v\\in\\Z^5:\nv_i\\equiv i\\pmod5,\\quad\n\\sum_i v_i=0,\\quad\n\\sum_i v_i^2=10n\\right\\}.\n\\]\nThe standard orientation is\n\\[\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=D(v)/288.\n\\]\nThe identity is $\\tau(n)=\\sum_{v\\in\\Vn}W(v)$. No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of $W$ is proved separately from the core identity.","dinates have indices $0,1,2,3,4$. A shell is\n\\[\n\\Vn=\\left\\{v\\in\\Z^5:\nv_i\\equiv i\\pmod5,\\quad\n\\sum_i v_i=0,\\quad\n\\sum_i v_i^2=10n\\right\\}.\n\\]\nThe standard orientation is\n\\[\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=D(v)/288.\n\\]\nThe identity is $\\tau(n)=\\sum_{v\\in\\Vn}W(v)$. No absolute values occur in that sum. Distinct residue classes ensure that no Vandermonde factor is zero. Integrality of $W$ is proved separately from the core identity.","The corresponding five-core partition has size $n-1$. In particular, the coefficient at $n=1$ corresponds to the empty partition. The involution is $J(v)_i=-v_{-i\\bmod5}$. A $J$-orbit may contain one or two packets, and its mass is the sum of their weights.","The content of a packet is the positive gcd of its five coordinates. Content-one packets are primitive. Multiplication followed by the prescribed residue reindexing gives\n\\[\n\\tau(n)=\\sum_{g^2\\mid n}\\chifive(g)g^{10}R(n/g^2),\n\\qquad\nR(n)=\\sum_{d^2\\mid n}\\mu(d)\\chifive(d)d^{10}\\tau(n/d^2).\n\\]\nThe boundary values are $R(1)=1$ and $R(p)=\\tau(p)$. At a prime shell the only possible content is one, so the extraction has not created a new prime coefficient.","\\section{Three polynomial operations}\nThe ordinary hyperplane Laplacian is $\\Delta_H$. The conjugated operator $\\mathcal L$ is characterized by $\\Delta_H(DQ)=D\\mathcal LQ$. Harmonicity of $DQ$ means $\\mathcal LQ=0$. It does not mean that $Q$ itself is harmonic.","A universal null is a polynomial $Q$ for which\n\\[\n\\sum_{v\\in\\Vn}W(v)Q(v)=0\n\\]\nfor every positive $n$. A shell-null polynomial satisfies that equation only at a specified shell. A polynomial interpolating prescribed values on a shell need satisfy neither condition. The difference between these three operations is responsible for several of the obstruction results.","For a universal null, replacing $W(v)$ by $W(v)(1+Q(v))$ preserves the shell total. A sign conclusion requires real corrected weights and at least one strict nonzero contribution. There is no order relation on complex weights. The all-polynomial content theorem assumes a finite symmetric polynomial and makes no claim for arbitrary infinite series or nonsymmetric functions.","\\section{Local and global conventions}\nFor $\\ell\\ne p$, the arithmetic-Frobenius convention in the integral $\\ell$-adic representation gives the characteristic polynomial\n\\[\nX^2-\\tau(p)X+p^{11}.\n\\]\nThe cohomological conventions in the filtered-module discussion are stated there separately. An abstract trace-zero filtered module and a representation realizing it are local objects. A comparison identifying its normalized Frobenius trace with the actual coefficient $\\tau(p)$ is additional mathematical data.","ic-Frobenius convention in the integral $\\ell$-adic representation gives the characteristic polynomial\n\\[\nX^2-\\tau(p)X+p^{11}.\n\\]\nThe cohomological conventions in the filtered-module discussion are stated there separately. An abstract trace-zero filtered module and a representation realizing it are local objects. A comparison identifying its normalized Frobenius trace with the actual coefficient $\\tau(p)$ is additional mathematical data.","The fixed-modulus survivor theorem concerns the actual joint finite image, including the cyclotomic factor. It does not assume that its constituent images are independent. Its quantifiers are\n\\[\n\\text{for every fixed }M,\\quad \\text{there exist suitable actual primes }p.\n\\]\nThe prime may change with $M$. The theorem does not supply one prime for every modulus.","\\paragraph{S06: Deligne's bound and normalization.}\nThe modular representation and purity inputs give $|\\tau(p)|\\le2p^{11/2}$ \\cite{DeligneModular,DeligneWeil}. The modular paper's representation construction and the Weil theorem supply different parts of this implication. Their Frobenius conventions require translation. This upper bound includes zero and is never used as a positive lower bound.","\\paragraph{S07: classical congruences.}\nSerre's 1967/68 exposition supplies the checked modulo-$691$ identity and the modulo-$23$ prime classification \\cite{SerreCongruences}. The latter excludes $p=23$. Representation existence is still conjectural in the relevant passage of that early exposition. The existence input in this book therefore uses the later established representation statement.","\\paragraph{S10: a cited historical computation.}\nDerickx, van Hoeij and Zeng establish the strict historical range\n\\[\nn<816212624008487344127999\n\\]\nin their checked 2014 version, Corollary~1.2 \\cite{DHZ}. The endpoint is not asserted to be a zero. This source-certified range was not rerun for the present edition and is not presented as the latest record.","\\paragraph{S16: readable classical leaves.}\nBalakrishnan, Craig and Ono provide the modulo-$5$ congruence at equation~(1.2), the multiplicativity and recurrence statements at Theorem~2.2, and the direct odd-square parity argument in Section~3 of the checked 2020 version \\cite{BCO}. These are the exact leaves used here. This edition has not audited the paper's separate integral-point or omitted-value arguments.","\\section{Convolutions and trace moments}\n\\paragraph{S12: the Hurwitz moment normalization.}\nLygeros and Rozier's 2010 primary paper gives the trace moments and Catalan inversion used in Chapter~5 \\cite{LRTrace}. The class-number function includes imprimitive forms with its specified weights. The $t=0$ term has a half-weight in the zeroth moment. The integral normalization is $6H$. Removing any of these qualifications would change the rows of the inverse system.","\\paragraph{S15: differential equations.}\nFonseca records the normalized Ramanujan differential relations used in the polynomial computation \\cite{Fonseca}. The project-specific $S_4$ cancellation is verified algebraically in the main text and by an exact symbolic remainder. Modularity and the level-one dimension input for the cusp identity are separate contracts \\cite{Stein}. A finite output check is not substituted for the universal algebraic identity.","\\section{Cores, harmonic theta series and valence}\n\\paragraph{S02: Han's packet identity.}\nHan's Definition~1.1 and Theorems~1.1--1.2 in the checked first arXiv version provide the five-core specialization and Vandermonde product \\cite{Han}. The specialization $t=5$, $z=25$ gives the shell norm $10n$ after the coefficient shift. With the orientation used here, the normalization is $D/288$. The identity permits cancellation and does not imply that its terms have one sign.","\\paragraph{S13: packet cardinality.}\nGarvan, Kim and Stanton's Theorem~4 and equation~(5.3) give the five-core count \\cite{GKS}. In the present conventions this reads\n\\[\na_5(n-1)=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\]\nFor a prime different from $5$, the shell cardinality is $p+\\chifive(p)$. This is the number of packets, not their weighted sum.","\\paragraph{S03: harmonic theta modularity.}\nIbukiyama and Williams, Section~3.4 of the checked version, give the vector-valued harmonic theta theorem with weight $\\nu+r/2$ \\cite{IW}. Here the rank is four and the harmonic polynomial is alternating of degree $10+d$. The scalar full-level transformation follows from the separate sign-line descent in Chapter~8. It is not silently imported as part of the general theorem.","ocal realization}\n\\paragraph{S05: the actual compatible representations.}\nSerre's 1973 exposition gives the established integral rank-two representations and the arithmetic-Frobenius polynomial \\cite{SerreBourbaki}. The fixed-modulus proof uses their actual joint finite quotient. It does not require an image classification, surjectivity onto an ambient matrix group, or independence across primes $\\ell$.","\\paragraph{S09: weak admissibility and admissibility.}\nThe Colmez--Fontaine author manuscript supplies the implication from weak admissibility to admissibility and the $N=0$ crystalline realization in the specified setting \\cite{CF}. The constructed rank-two module has equal Hodge and Newton totals and no proper Frobenius-stable rational line. Those checks concern a local object. They do not identify it with the global representation of $\\Delta$.","issibility.}\nThe Colmez--Fontaine author manuscript supplies the implication from weak admissibility to admissibility and the $N=0$ crystalline realization in the specified setting \\cite{CF}. The constructed rank-two module has equal Hodge and Newton totals and no proper Frobenius-stable rational line. Those checks concern a local object. They do not identify it with the global representation of $\\Delta$.","The most direct coefficient check covers every $1\\le n\\le512$. The triangular recurrence, the defining binomial product and an independent logarithmic-derivative divisor-sum engine agree exactly. All three use integers. This range is distinct from the historical final-arithmetic program's product check through $128$, and from the independent product check in the packet-tree program through $512$.","The arithmetic survivor search uses $\\tau$ through $10000$ and twenty-four selected moduli. Twenty-two moduli have a matching prime in that bounded search, supplying $22\\times32=704$ local recurrence checks. Moduli $11$ and $19$ have no observed witness in that search. This is a search outcome, not a contradiction of the density theorem. Modulus one is included as a trivial case.","The synthetic recurrence suite has $256$ moduli, three parameter choices and thirty-two positive exponents, for $24576$ checks. Its value $p=5M-1$ need not be prime. Separately, the $1229$ prime-indexed companion matrices through $10000$ are constructed local models. They are not sampled global Frobenius matrices.","\\section{Packet and polynomial checks}\nThe five-core suite checks $405$ shell sums and $326585$ packet polynomials, with $308$ prime shell counts. Twenty-five coefficients also agree with an independent hook-partition calculation. The tree suite enumerates all shells through $512$ and the two additional shells $737$ and $2411$, giving $514$ shell sums and $95937$ packets. It checks $3084$ sector partitions and explicit sharp-boundary witnesses at sixty-four depths.","The shell at $737$ is a sparse additional witness. The replay does not enumerate every shell from $513$ through $736$. The universal threshold formula rests on the separate lower-bound proof and the symbolic witness families.","The retained universal-null suite checks fourteen exact rational sign certificates. Its degree-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.","sal-null suite checks fourteen exact rational sign certificates. Its degree-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.","ull suite checks fourteen exact rational sign certificates. Its degree-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.","checks fourteen exact rational sign certificates. Its degree-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.",". Its degree-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.","-at-most-sixteen dual at $31$ concerns the seven stored corrections, not all possible corrections. The primitive transport suite checks thirty-six symbolic harmonic certificates, $128$ Mobius identities, $144$ profiles and the stored rational transfer. The independent $31$ calculation obtains coarse, fine and detail ranks $5$, $15$ and $10$ over the rationals and the two prime fields $65537$ and $1000003$, with structural upper bounds.","The content obstruction is the exact shell $121$. Its masses are\n\\[\n-25439104668+25937424601=498319933.\n\\]\nThe computation establishes these integers. The theorem that every finite symmetric universal null conserves each content is a separate proof. Likewise, the interpolation suite's finite orbit checks and two explicit failures do not prove an all-degree fullness theorem.","\\section{Large modular coefficients}\nThe valuation suite regenerates coefficients at all $13848$ primes through $150000$. Its proposed valuation equalities are tested only on their stated candidate residue classes. Independent Eisenstein convolutions verify the selected counterexample coefficients. The dyadic obstruction uses twelve selected primes through $1032191$, with nineteen fresh modular comparisons.","The CRT/NTT computation returns residues modulo $2^{19}$. It is not a reconstruction of those large integer tau coefficients. Within each convolution, however, the integer coefficient is reconstructed before reduction. With\n\\[\nL=1032191,\\qquad M=524288,\n\\]\nthe nonnegative representative bound is\n\\[\nL(M-1)^2=283725419316756479\n <998244353\\cdot1004535809\n =1002772198720536577.\n\\]\nThe common transform length $2^{21}$ divides both finite-field multiplicative periods. This bound, primitive-root checks and integer-type limits are part of the computation contract. Modular agreement alone would not justify CRT reconstruction without them.","T computation returns residues modulo $2^{19}$. It is not a reconstruction of those large integer tau coefficients. Within each convolution, however, the integer coefficient is reconstructed before reduction. With\n\\[\nL=1032191,\\qquad M=524288,\n\\]\nthe nonnegative representative bound is\n\\[\nL(M-1)^2=283725419316756479\n <998244353\\cdot1004535809\n =1002772198720536577.\n\\]\nThe common transform length $2^{21}$ divides both finite-field multiplicative periods. This bound, primitive-root checks and integer-type limits are part of the computation contract. Modular agreement alone would not justify CRT reconstruction without them.","The full replay uses a $600$-second wall-clock bound per suite, a $570$-second per-process CPU cap, inherited one-core affinity and a two-million-byte combined log limit. Numerical-library thread caps are cooperative. No address-space cap is claimed. Exact arithmetic programs can consume substantial time or memory outside the recorded ranges.","The universal null construction depends on an exact differential operator. This appendix derives that operator from the Euclidean metric, rather than treating its power-sum formulas as a black box. It also explains the Newton reductions used by the implementation and gives two low-degree harmonic checks. All polynomial identities are on the hyperplane $H=\\{v\\in\\R^5:\\sum_i v_i=0\\}$, with its induced Euclidean metric. The modularity and valence contracts remain the ones stated in Chapters~\\ref{ch:theta} and~\\ref{ch:nulls}.","Write $s_m=\\sum_{i=0}^4v_i^m$, with $s_0=5$ and $s_1=0$, and let $D=\\prod_{i<j}(v_i-v_j)$. We distinguish the ordinary Laplacian $\\Delta_H$ from the conjugated operator $\\mathcal L$ defined by $\\Delta_H(DQ)=D\\mathcal LQ$. The factor $D$ changes radial coefficients substantially. Confusing the two operators would invalidate the harmonicity test.","\\section{Two descriptions of the same metric}\nThe orthogonal projection from $\\R^5$ to $H$ is\n\\begin{equation}\\label{appop:projection}\nP=I_5-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nwhere $\\mathbf1=(1,1,1,1,1)^T$. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies $P^2=P$, and has image $H$. For an ambient polynomial $F$, its gradient along $H$ is $P\\nabla F$. Its Laplacian along $H$ is\n\\begin{equation}\\label{appop:ambientlap}\n\\Delta_H(F|_H)\n=\\left(\\sum_{i=0}^4\\partial_i^2F\n-\\frac15\\left(\\sum_{i=0}^4\\partial_i\\right)^2F\\right)\\Big|_H.\n\\end{equation}\nOnly tangential derivatives enter. Thus this expression is independent of the polynomial extension away from $H$.","on{Two descriptions of the same metric}\nThe orthogonal projection from $\\R^5$ to $H$ is\n\\begin{equation}\\label{appop:projection}\nP=I_5-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nwhere $\\mathbf1=(1,1,1,1,1)^T$. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies $P^2=P$, and has image $H$. For an ambient polynomial $F$, its gradient along $H$ is $P\\nabla F$. Its Laplacian along $H$ is\n\\begin{equation}\\label{appop:ambientlap}\n\\Delta_H(F|_H)\n=\\left(\\sum_{i=0}^4\\partial_i^2F\n-\\frac15\\left(\\sum_{i=0}^4\\partial_i\\right)^2F\\right)\\Big|_H.\n\\end{equation}\nOnly tangential derivatives enter. Thus this expression is independent of the polynomial extension away from $H$.","etric}\nThe orthogonal projection from $\\R^5$ to $H$ is\n\\begin{equation}\\label{appop:projection}\nP=I_5-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nwhere $\\mathbf1=(1,1,1,1,1)^T$. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies $P^2=P$, and has image $H$. For an ambient polynomial $F$, its gradient along $H$ is $P\\nabla F$. Its Laplacian along $H$ is\n\\begin{equation}\\label{appop:ambientlap}\n\\Delta_H(F|_H)\n=\\left(\\sum_{i=0}^4\\partial_i^2F\n-\\frac15\\left(\\sum_{i=0}^4\\partial_i\\right)^2F\\right)\\Big|_H.\n\\end{equation}\nOnly tangential derivatives enter. Thus this expression is independent of the polynomial extension away from $H$.","projection from $\\R^5$ to $H$ is\n\\begin{equation}\\label{appop:projection}\nP=I_5-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nwhere $\\mathbf1=(1,1,1,1,1)^T$. Indeed, subtracting the coordinate mean leaves a zero coordinate sum. The matrix is symmetric, satisfies $P^2=P$, and has image $H$. For an ambient polynomial $F$, its gradient along $H$ is $P\\nabla F$. Its Laplacian along $H$ is\n\\begin{equation}\\label{appop:ambientlap}\n\\Delta_H(F|_H)\n=\\left(\\sum_{i=0}^4\\partial_i^2F\n-\\frac15\\left(\\sum_{i=0}^4\\partial_i\\right)^2F\\right)\\Big|_H.\n\\end{equation}\nOnly tangential derivatives enter. Thus this expression is independent of the polynomial extension away from $H$.","P^2=P$, and has image $H$. For an ambient polynomial $F$, its gradient along $H$ is $P\\nabla F$. Its Laplacian along $H$ is\n\\begin{equation}\\label{appop:ambientlap}\n\\Delta_H(F|_H)\n=\\left(\\sum_{i=0}^4\\partial_i^2F\n-\\frac15\\left(\\sum_{i=0}^4\\partial_i\\right)^2F\\right)\\Big|_H.\n\\end{equation}\nOnly tangential derivatives enter. Thus this expression is independent of the polynomial extension away from $H$.","For direct symbolic computation, use four coordinates $x_0,\\ldots,x_3$ and substitute\n\\[\nv=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3).\n\\]\nThe coordinate tangent vectors are $b_a=e_a-e_4$. Their Gram matrix and its inverse are\n\\begin{equation}\\label{appop:chartmetric}\nG=I_4+\\mathbf1\\mathbf1^T,\n\\qquad G^{-1}=I_4-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nnow with a four-entry vector $\\mathbf1$. Matrix multiplication verifies the inverse. The factor $1/5$ remains because the missing fifth coordinate is determined by the four chosen coordinates.","oordinates $x_0,\\ldots,x_3$ and substitute\n\\[\nv=(x_0,x_1,x_2,x_3,-x_0-x_1-x_2-x_3).\n\\]\nThe coordinate tangent vectors are $b_a=e_a-e_4$. Their Gram matrix and its inverse are\n\\begin{equation}\\label{appop:chartmetric}\nG=I_4+\\mathbf1\\mathbf1^T,\n\\qquad G^{-1}=I_4-\\frac15\\mathbf1\\mathbf1^T,\n\\end{equation}\nnow with a four-entry vector $\\mathbf1$. Matrix multiplication verifies the inverse. The factor $1/5$ remains because the missing fifth coordinate is determined by the four chosen coordinates.","The chart Laplacian is consequently\n\\begin{equation}\\label{appop:chartlap}\n\\Delta_H=\\sum_{a,b=0}^3\n\\left(\\delta_{ab}-\\frac15\\right)\\frac{\\partial^2}{\\partial x_a\\partial x_b}.\n\\end{equation}\nThe coordinates $x_a$ are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, $s_2=x^TGx$, so its Hessian is $2G$ and\n\\[\n\\Delta_Hs_2=\\operatorname{tr}(G^{-1}2G)=8.\n\\]\nThe rank of the Euclidean space is four, giving twice that rank. Later we obtain $\\mathcal Ls_2=48$, an intentionally different number.","ta_{ab}-\\frac15\\right)\\frac{\\partial^2}{\\partial x_a\\partial x_b}.\n\\end{equation}\nThe coordinates $x_a$ are not orthonormal. Using only the four diagonal second derivatives would use a different metric. As a normalization check, $s_2=x^TGx$, so its Hessian is $2G$ and\n\\[\n\\Delta_Hs_2=\\operatorname{tr}(G^{-1}2G)=8.\n\\]\nThe rank of the Euclidean space is four, giving twice that rank. Later we obtain $\\mathcal Ls_2=48$, an intentionally different number.","\\section{Conjugating by the Vandermonde}\nThe Vandermonde is alternating under coordinate permutations. Since $\\Delta_H$ commutes with those permutations, $\\Delta_HD$ is alternating of degree eight. Every alternating polynomial vanishes on each hyperplane $v_i=v_j$. In the polynomial ring of $H$, the ten corresponding linear factors are distinct and divide it. Their product has degree ten. Therefore $\\Delta_HD=0$.","ction{Conjugating by the Vandermonde}\nThe Vandermonde is alternating under coordinate permutations. Since $\\Delta_H$ commutes with those permutations, $\\Delta_HD$ is alternating of degree eight. Every alternating polynomial vanishes on each hyperplane $v_i=v_j$. In the polynomial ring of $H$, the ten corresponding linear factors are distinct and divide it. Their product has degree ten. Therefore $\\Delta_HD=0$.","On the open set where $D\\ne0$, the product rule gives\n\\begin{equation}\\label{appop:conjugation}\n\\frac{\\Delta_H(DQ)}{D}\n=\\Delta_HQ+2\\nabla_H\\log D\\cdot\\nabla_HQ.\n\\end{equation}\nHere $\\nabla\\log D$ denotes the rational logarithmic derivative $\\nabla D/D$. It does not require choosing a complex logarithm or assuming that $D$ is positive. For symmetric $Q$, the apparent poles cancel and the right side extends to a polynomial on all of $H$.","roduct rule gives\n\\begin{equation}\\label{appop:conjugation}\n\\frac{\\Delta_H(DQ)}{D}\n=\\Delta_HQ+2\\nabla_H\\log D\\cdot\\nabla_HQ.\n\\end{equation}\nHere $\\nabla\\log D$ denotes the rational logarithmic derivative $\\nabla D/D$. It does not require choosing a complex logarithm or assuming that $D$ is positive. For symmetric $Q$, the apparent poles cancel and the right side extends to a polynomial on all of $H$.","There is a second explanation of the extension. The polynomial $\\Delta_H(DQ)$ is alternating, so $D$ divides it. Its quotient is symmetric. Thus the operator $\\mathcal L$ is well defined on symmetric polynomials even at coordinate collisions. The packet shell itself has no collisions because its residues are distinct. The polynomial extension is nevertheless needed when constructing and comparing symbolic identities.","Translation invariance of $D$ gives\n\\[\n\\sum_i\\frac{\\partial_iD}{D}=0,\n\\qquad\n\\frac{\\partial_iD}{D}=\\sum_{j\\ne i}\\frac1{v_i-v_j}.\n\\]\nIts ambient logarithmic gradient is already tangential. This removes a potential extra projection term in the derivation below.","\\section{Deriving the power-sum operator}\nFor $m\\ge2$, differentiation and projection give\n\\begin{align}\n(\\nabla_Hs_m)_i\n &=m\\left(v_i^{m-1}-\\frac15s_{m-1}\\right),\\label{appop:gradsm}","\\section{Deriving the power-sum operator}\nFor $m\\ge2$, differentiation and projection give\n\\begin{align}\n(\\nabla_Hs_m)_i\n &=m\\left(v_i^{m-1}-\\frac15s_{m-1}\\right),\\label{appop:gradsm}\\\\\n\\Delta_Hs_m\n &=\\frac45m(m-1)s_{m-2}.\\label{appop:lapsm}\n\\end{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqre","\\Delta_Hs_m\n &=\\frac45m(m-1)s_{m-2}.\\label{appop:lapsm}\n\\end{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis f","\\Delta_Hs_m\n &=\\frac45m(m-1)s_{m-2}.\\label{appop:lapsm}\n\\end{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is","\\Delta_Hs_m\n &=\\frac45m(m-1)s_{m-2}.\\label{appop:lapsm}\n\\end{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$,","\\Delta_Hs_m\n &=\\frac45m(m-1)s_{m-2}.\\label{appop:lapsm}\n\\end{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is …","d{align}\nIn the logarithmic term, the coordinate-mean subtraction vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","vanishes because the logarithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","ithmic gradient has zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","zero sum. Pair the ordered indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","ed indices $i,j$ and $j,i$ to obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","in\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","log D\\cdot\\nabla_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","_Hs_m\n=2m\\sum_{i<j}\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}.\n\\]\nThe divided difference is the finite polynomial\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","1}}{v_i-v_j}\n=\\sum_{a=0}^{m-2}v_i^av_j^{m-2-a}.\n\\]\nThe sum over $a$ is symmetric in the two coordinates. Hence twice the sum over $i<j$ equals the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","ls the sum over $i\\ne j$. For each $a$, that ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","ordered sum is $s_as_{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","{m-2-a}-s_{m-2}$. There are $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","re $m-1$ values of $a$. We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.",". We obtain\n\\[\n2\\nabla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","abla_H\\log D\\cdot\\nabla_Hs_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-m(m-1)s_{m-2}.\n\\]\nAdding \\eqref{appop:lapsm} proves\n\\begin{equation}\\label{appop:lsm}\n\\boxed{\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2}.}\n\\end{equation}\nThis formula is asserted for $m\\ge2$. The boundary values are separately $\\mathcal Ls_0=0$ and $\\mathcal Ls_1=0$, since $s_0$ is constant and $s_1$ vanishes on $H$. No negative-index power sum is introduced.","The first values are\n\\begin{equation}\\label{appop:firstoperator}\n\\mathcal Ls_2=48,\\qquad\n\\mathcal Ls_3=0,\\qquad\n\\mathcal Ls_4=\\frac{188}{5}s_2,\\qquad\n\\mathcal Ls_5=46s_3.\n\\end{equation}\nFor example, at $m=2$ the first term is $2s_0^2=50$, while the subtraction is $2s_0/5=2$. The additional $40$ compared with $\\Delta_Hs_2=8$ comes from differentiating the degree-ten factor $D$.","\\section{The chain-rule metric and polynomial assembly}\nThe same projected gradients give\n\\begin{equation}\\label{appop:chainmetric}\n\\Gamma_{mn}:=\\nabla_Hs_m\\cdot\\nabla_Hs_n\n=mn\\left(s_{m+n-2}-\\frac15s_{m-1}s_{n-1}\\right).\n\\end{equation}\nExpanding the product in \\eqref{appop:gradsm} gives the first power sum once and the coordinate-mean product with coefficient $-1/5$. This identity includes $m,n\\ge1$, with the evident zero tangential gradient for $s_1$.","ly}\nThe same projected gradients give\n\\begin{equation}\\label{appop:chainmetric}\n\\Gamma_{mn}:=\\nabla_Hs_m\\cdot\\nabla_Hs_n\n=mn\\left(s_{m+n-2}-\\frac15s_{m-1}s_{n-1}\\right).\n\\end{equation}\nExpanding the product in \\eqref{appop:gradsm} gives the first power sum once and the coordinate-mean product with coefficient $-1/5$. This identity includes $m,n\\ge1$, with the evident zero tangential gradient for $s_1$.","If $Q=F(s_2,s_3,s_4,s_5)$, the ordinary second-order chain rule and \\eqref{appop:conjugation} yield\n\\begin{equation}\\label{appop:chainrule}\n\\mathcal LQ\n=\\sum_{m=2}^5\\frac{\\partial F}{\\partial s_m}\\mathcal Ls_m\n+\\sum_{m,n=2}^5\n\\frac{\\partial^2F}{\\partial s_m\\partial s_n}\\Gamma_{mn}.\n\\end{equation}\nThe second sum runs over ordered pairs. Thus a mixed derivative contributes twice when $m\\ne n$. If one instead sums over $m\\le n$, the off-diagonal terms require a factor two. These are two equivalent conventions, but their factors must not be mixed.","_5)$, the ordinary second-order chain rule and \\eqref{appop:conjugation} yield\n\\begin{equation}\\label{appop:chainrule}\n\\mathcal LQ\n=\\sum_{m=2}^5\\frac{\\partial F}{\\partial s_m}\\mathcal Ls_m\n+\\sum_{m,n=2}^5\n\\frac{\\partial^2F}{\\partial s_m\\partial s_n}\\Gamma_{mn}.\n\\end{equation}\nThe second sum runs over ordered pairs. Thus a mixed derivative contributes twice when $m\\ne n$. If one instead sums over $m\\le n$, the off-diagonal terms require a factor two. These are two equivalent conventions, but their factors must not be mixed.","Useful instances are\n\\begin{align}\n\\mathcal L(s_2^2)&=104s_2,\\label{appop:lradialtwo}\\\\\n\\mathcal L(s_2^3)&=168s_2^2,\\label{appop:lradialthree}\\\\\n\\mathcal L(s_2s_4)&=64s_4+\\frac{188}{5}s_2^2,\\label{appop:ls2s4}\\\\\n\\mathcal L(s_3^2)&=18s_4-\\frac{18}{5}s_2^2.\\label{appop:ls3square}\n\\end{align}\nFor the first identity, $2s_2\\mathcal Ls_2+2\\Gamma_{22}=96s_2+8s_2$. For the mixed product, $s_4\\mathcal Ls_2+s_2\\mathcal Ls_4+2\\Gamma_{24}$ gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.","\\mathcal L(s_3^2)&=18s_4-\\frac{18}{5}s_2^2.\\label{appop:ls3square}\n\\end{align}\nFor the first identity, $2s_2\\mathcal Ls_2+2\\Gamma_{22}=96s_2+8s_2$. For the mixed product, $s_4\\mathcal Ls_2+s_2\\mathcal Ls_4+2\\Gamma_{24}$ gives the displayed coefficients. Each calculation is a check of the ordered-pair convention.","The metric formula may produce $s_6,s_7,s_8$. They are not new independent variables. The invariant ring on $H$ is generated by $s_2,s_3,s_4,s_5$, and Newton's identities reduce higher indices exactly.","\\section{Newton reductions with zero first power sum}\nLet $e_j$ be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is\n\\[\n\\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,\n\\]\nbecause $e_1=s_1=0$. The first Newton identities give\n\\begin{align}\ne_2&=-\\frac12s_2,& e_3&=\\frac13s_3,\\label{appop:e23}","\\section{Newton reductions with zero first power sum}\nLet $e_j$ be the elementary symmetric polynomials in the five coordinates. Their characteristic polynomial is\n\\[\n\\prod_i(T-v_i)=T^5+e_2T^3-e_3T^2+e_4T-e_5,\n\\]\nbecause $e_1=s_1=0$. The first Newton identities give\n\\begin{align}\ne_2&=-\\frac12s_2,& e_3&=\\frac13s_3,\\label{appop:e23}\\\\\ne_4&=\\frac18s_2^2-\\frac14s_4,&\ne_5&=\\frac15s_5-\\frac16s_2s_3.\\label{appop:e45}\n\\end{align}\nFor instance, the degree-five identity is $s_5+e_2s_3-e_3s_2-5e_5=0$. The two middle terms combine to $-5s_2s_3/6$, fixing the sign in $e_5$.","e_4&=\\frac18s_2^2-\\frac14s_4,&\ne_5&=\\frac15s_5-\\frac16s_2s_3.\\label{appop:e45}\n\\end{align}\nFor instance, the degree-five identity is $s_5+e_2s_3-e_3s_2-5e_5=0$. The two middle terms combine to $-5s_2s_3/6$, fixing the sign in $e_5$.","Every coordinate satisfies its characteristic polynomial. Multiplying that equality by $v_i^{k-5}$ and summing, for $k\\ge6$, gives\n\\begin{equation}\\label{appop:newtonrec}\ns_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.\n\\end{equation}\nThis recurrence uses only lower indices. Substitution at $k=6,7$ yields\n\\begin{align}\ns_6&=\\frac34s_2s_4+\\frac13s_3^2-\\frac18s_2^3,\\label{appop:s6}","Every coordinate satisfies its characteristic polynomial. Multiplying that equality by $v_i^{k-5}$ and summing, for $k\\ge6$, gives\n\\begin{equation}\\label{appop:newtonrec}\ns_k=-e_2s_{k-2}+e_3s_{k-3}-e_4s_{k-4}+e_5s_{k-5}.\n\\end{equation}\nThis recurrence uses only lower indices. Substitution at $k=6,7$ yields\n\\begin{align}\ns_6&=\\frac34s_2s_4+\\frac13s_3^2-\\frac18s_2^3,\\label{appop:s6}\\\\\ns_7&=\\frac7{10}s_2s_5+\\frac7{12}s_3s_4\n-\\frac7{24}s_2^2s_3.\\label{appop:s7}\n\\end{align}\nThe same rule gives\n\\begin{equation}\\label{appop:s8}\ns_8=\\frac14s_4^2+\\frac14s_2^2s_4\n-\\frac1{16}s_2^4+\\frac8{15}s_3s_5.\n\\end{equation}\nEach term has the required invariant degree. This supplies a useful local check before substitution into a larger operator expression.","s_7&=\\frac7{10}s_2s_5+\\frac7{12}s_3s_4\n-\\frac7{24}s_2^2s_3.\\label{appop:s7}\n\\end{align}\nThe same rule gives\n\\begin{equation}\\label{appop:s8}\ns_8=\\frac14s_4^2+\\frac14s_2^2s_4\n-\\frac1{16}s_2^4+\\frac8{15}s_3s_5.\n\\end{equation}\nEach term has the required invariant degree. This supplies a useful local check before substitution into a larger operator expression.","At the first packet $v=(0,1,2,-2,-1)$, the first power sums are $s_2=10$, $s_3=0$, $s_4=34$, $s_5=0$. Formula \\eqref{appop:s6} gives $255-125=130$, matching the direct sum of sixth powers. Formula \\eqref{appop:s7} gives zero, matching cancellation of the opposite coordinates. Formula \\eqref{appop:s8} gives $514$. These examples fix normalizations. The polynomial recurrences, rather than these examples, establish the identities on every vector in $H$.","wer sums are $s_2=10$, $s_3=0$, $s_4=34$, $s_5=0$. Formula \\eqref{appop:s6} gives $255-125=130$, matching the direct sum of sixth powers. Formula \\eqref{appop:s7} gives zero, matching cancellation of the opposite coordinates. Formula \\eqref{appop:s8} gives $514$. These examples fix normalizations. The polynomial recurrences, rather than these examples, establish the identities on every vector in $H$.","\\section{Radial factors and harmonic projection}\nLet $h_\\ell$ be homogeneous harmonic of degree $\\ell$ on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","nd harmonic projection}\nLet $h_\\ell$ be homogeneous harmonic of degree $\\ell$ on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","ion}\nLet $h_\\ell$ be homogeneous harmonic of degree $\\ell$ on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","be homogeneous harmonic of degree $\\ell$ on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","nic of degree $\\ell$ on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","on the rank-four space $H$. Euler's identity is $v\\cdot\\nabla_Hh_\\ell=\\ell h_\\ell$. The product rule for the radial factor $s_2^j$ gives\n\\begin{equation}\\label{appop:radial}\n\\Delta_H(s_2^jh_\\ell)\n=4j(\\ell+j+1)s_2^{j-1}h_\\ell,\\qquad j\\ge1.\n\\end{equation}\nAt $j=0$ the left side is zero. At $j=1,\\ell=0$ the coefficient is $8$, agreeing with $\\Delta_Hs_2$. At $j=1,\\ell=10$ it is $48$, agreeing with $\\Delta_H(s_2D)=48D$.","For a homogeneous polynomial $P$ of degree $M$, a concrete top-degree harmonic projection in dimension four is\n\\begin{equation}\\label{appop:hprojection}\n\\mathcal H_M(P)=\\sum_{j=0}^{\\lfloor M/2\\rfloor}\n\\frac{(-1)^j(M-j)!}{4^j j!M!}\\,s_2^j\\Delta_H^jP.\n\\end{equation}\nFor $M=0$ or $1$ the sum contains only $j=0$. In general, if $a_j$ denotes its coefficient, the product rule gives the cancellation condition\n\\[\na_j+4(j+1)(M-j)a_{j+1}=0.\n\\]\nThe factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its difference from $P$ is divisible by $s_2$.","ogeneous polynomial $P$ of degree $M$, a concrete top-degree harmonic projection in dimension four is\n\\begin{equation}\\label{appop:hprojection}\n\\mathcal H_M(P)=\\sum_{j=0}^{\\lfloor M/2\\rfloor}\n\\frac{(-1)^j(M-j)!}{4^j j!M!}\\,s_2^j\\Delta_H^jP.\n\\end{equation}\nFor $M=0$ or $1$ the sum contains only $j=0$. In general, if $a_j$ denotes its coefficient, the product rule gives the cancellation condition\n\\[\na_j+4(j+1)(M-j)a_{j+1}=0.\n\\]\nThe factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its difference from $P$ is divisible by $s_2$.","j!M!}\\,s_2^j\\Delta_H^jP.\n\\end{equation}\nFor $M=0$ or $1$ the sum contains only $j=0$. In general, if $a_j$ denotes its coefficient, the product rule gives the cancellation condition\n\\[\na_j+4(j+1)(M-j)a_{j+1}=0.\n\\]\nThe factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its difference from $P$ is divisible by $s_2$.","H^jP.\n\\end{equation}\nFor $M=0$ or $1$ the sum contains only $j=0$. In general, if $a_j$ denotes its coefficient, the product rule gives the cancellation condition\n\\[\na_j+4(j+1)(M-j)a_{j+1}=0.\n\\]\nThe factorial coefficients satisfy this condition. The last possible iterated Laplacian has no further nonzero derivative, so the resulting polynomial is harmonic. Its difference from $P$ is divisible by $s_2$.","Repeating the construction on that lower-degree quotient produces the harmonic decomposition\n\\begin{equation}\\label{appop:decomposition}\nP=\\sum_{j=0}^{\\lfloor M/2\\rfloor}s_2^jh_{M-2j}.\n\\end{equation}\nIt is unique. If such a sum were zero, apply the Laplacian as many times as the largest occupied radial exponent. All smaller-exponent terms vanish. Formula \\eqref{appop:radial} leaves a nonzero scalar times the last harmonic polynomial, so that polynomial must be zero. Descending removes every term.","When $P=DQ$ is alternating, each projected harmonic component is alternating because the Laplacian and multiplication by $s_2$ commute with permutations. Each nonzero component is divisible by $D$ and has degree at least ten. Thus for a correction of homogeneous degree $d$, only $j\\le d/2$ can occur. This is the degree bound used in the quasimodular argument for nonhomogeneous content conservation.","Harmonic projection is not preservation of shell values. The elementary counterexample is $P=s_2D$, of degree $12$. Since $\\Delta_HP=48D$ and the next Laplacian is zero,\n\\[\n\\mathcal H_{12}(s_2D)=s_2D-\\frac1{48}s_2(48D)=0.\n\\]\nYet on $\\Vn$ the original polynomial equals $10nD$, which is nonzero at every packet. The operation removes radial information. It cannot silently turn a shell interpolant into a universally null polynomial with the same values.","\\section{Two explicit harmonic moment checks}\nThe degree-four correction\n\\[\nK_4=130s_4-47s_2^2\n\\]\nhas\n\\[\n\\mathcal LK_4\n=130\\frac{188}{5}s_2-47(104s_2)=0.\n\\]\nThus $DK_4$ is harmonic. At the first packet, $W=1$ and $K_4=130\\cdot34-47\\cdot100=-280$. Its theta form has weight $16$, whose cusp space is spanned by $\\Delta E_4$ with first coefficient one. Consequently the theta form is $-280\\Delta E_4$, giving the moment identity already stated in Chapter~\\ref{ch:nulls}.","Likewise, set\n\\[\nK_6=90s_2s_4-27s_2^3-320s_3^2.\n\\]\nUsing \\eqref{appop:lradialthree}--\\eqref{appop:ls3square}, its operator image is\n\\begin{align*}\n\\mathcal LK_6={}&(5760-5760)s_4","Likewise, set\n\\[\nK_6=90s_2s_4-27s_2^3-320s_3^2.\n\\]\nUsing \\eqref{appop:lradialthree}--\\eqref{appop:ls3square}, its operator image is\n\\begin{align*}\n\\mathcal LK_6={}&(5760-5760)s_4\\\\\n&+(3384-4536+1152)s_2^2=0.\n\\end{align*}\nAt the first packet its value is $90\\cdot10\\cdot34-27\\cdot1000=3600$. The weight-$18$ cusp space is spanned by $\\Delta E_6$, again normalized with first coefficient one. Its theta form is therefore $3600\\Delta E_6$.","&+(3384-4536+1152)s_2^2=0.\n\\end{align*}\nAt the first packet its value is $90\\cdot10\\cdot34-27\\cdot1000=3600$. The weight-$18$ cusp space is spanned by $\\Delta E_6$, again normalized with first coefficient one. Its theta form is therefore $3600\\Delta E_6$.","The retained programs organize these same operations using sparse rational polynomials. A direct four-coordinate implementation can instead substitute $v_4=-\\sum_{a=0}^3x_a$ into $DQ$ and apply \\eqref{appop:chartlap}. Agreement between these organizations checks the symbolic construction through different algebraic representations. The universal justification remains the metric calculation, the exact polynomial identities and the stated harmonic theta and valence contracts. None of those statements alone excludes the unresolved zero prime parameter of $\\Delta$.","um_{a=0}^3x_a$ into $DQ$ and apply \\eqref{appop:chartlap}. Agreement between these organizations checks the symbolic construction through different algebraic representations. The universal justification remains the metric calculation, the exact polynomial identities and the stated harmonic theta and valence contracts. None of those statements alone excludes the unresolved zero prime parameter of $\\Delta$.","This appendix gives calculations that a reader can reproduce without trusting a status flag. The objects are the normalized coefficients of $\\Delta$, the auxiliary shell polynomials, and one rational transfer certificate at $31$. Each calculation ends at its actual mathematical boundary. Agreement of finite outputs does not supply the missing all-prime nonvanishing argument.","The complete integer inputs and outputs are in the companion evidence archive. Paths below are relative to its extracted project root. A JSON integer remains an integer even when it has many digits. An encoded rational pair $[a,b]$ means $a/b$, with a nonzero denominator. Converting these objects to floating point before checking an equality changes the contract.","Write\n\\[\n F(q)=\\frac{\\Delta(q)}q=\\prod_{k\\ge1}(1-q^k)^{24}\n      =\\sum_{m\\ge0}f_mq^m.\n\\]\nThen $f_0=1$ and $\\tau(n)=f_{n-1}$. This shift is essential. A program that stores $\\tau(0)=0$ as a sentinel must not confuse that entry with the constant coefficient $f_0$.","For direct product expansion, terms of degree at most $m$ depend only on factors with $k\\le m$. The exact binomial expansion\n\\[\n (1-q^k)^{24}=\\sum_{j=0}^{24}(-1)^j\\binom{24}{j}q^{kj}\n\\]\ntherefore supplies a finite algorithm for each desired coefficient. Three short calculations are\n\\begin{align*}\n f_1&=-24,","For direct product expansion, terms of degree at most $m$ depend only on factors with $k\\le m$. The exact binomial expansion\n\\[\n (1-q^k)^{24}=\\sum_{j=0}^{24}(-1)^j\\binom{24}{j}q^{kj}\n\\]\ntherefore supplies a finite algorithm for each desired coefficient. Three short calculations are\n\\begin{align*}\n f_1&=-24,\\\\\n f_2&=\\binom{24}{2}-24=276-24=252,\\\\\n f_3&=-\\binom{24}{3}+(-24)(-24)-24\\\\\n    &=-2024+576-24=-1472.\n\\end{align*}\nThe last line comes respectively from the $q^3$ term in the first factor, the product of its $q$ term with the second factor's $q^2$ term, and the third factor's $q^3$ term. Thus it computes $\\tau(4)$, not $\\tau(3)$.","&=-2024+576-24=-1472.\n\\end{align*}\nThe last line comes respectively from the $q^3$ term in the first factor, the product of its $q$ term with the second factor's $q^2$ term, and the third factor's $q^3$ term. Thus it computes $\\tau(4)$, not $\\tau(3)$.","The logarithmic derivative provides a different exact engine. With $\\sigma_1(j)=\\sum_{d\\mid j}d$, comparison of coefficients in\n\\[\n q\\frac{F'(q)}{F(q)}=-24\\sum_{j\\ge1}\\sigma_1(j)q^j\n\\]\ngives\n\\begin{equation}\n m f_m=-24\\sum_{j=1}^{m}\\sigma_1(j)f_{m-j}\n \\qquad(m\\ge1).\n \\label{eq:workbook-divisor-recurrence}\n\\end{equation}\nFor $m=3$, the divisor sums are $1,3,4$. Hence\n\\[\n f_3=-8\\bigl(252+3(-24)+4\\bigr)=-8\\cdot184=-1472.\n\\]\nAn implementation should check that the numerator in \\eqref{eq:workbook-divisor-recurrence} is divisible by $m$, rather than rounding the quotient.","frac{F'(q)}{F(q)}=-24\\sum_{j\\ge1}\\sigma_1(j)q^j\n\\]\ngives\n\\begin{equation}\n m f_m=-24\\sum_{j=1}^{m}\\sigma_1(j)f_{m-j}\n \\qquad(m\\ge1).\n \\label{eq:workbook-divisor-recurrence}\n\\end{equation}\nFor $m=3$, the divisor sums are $1,3,4$. Hence\n\\[\n f_3=-8\\bigl(252+3(-24)+4\\bigr)=-8\\cdot184=-1472.\n\\]\nAn implementation should check that the numerator in \\eqref{eq:workbook-divisor-recurrence} is divisible by $m$, rather than rounding the quotient.","Jacobi gives a third engine. Put $T_j=j(j+1)/2$ and $a_j=(-1)^j(2j+1)$. The series\n\\[\n A(q)=\\prod_{k\\ge1}(1-q^k)^3=1+\\sum_{j\\ge1}a_jq^{T_j}\n\\]\nsatisfies $F=A^8$. The identity $AqF'=8FqA'$ yields\n\\begin{equation}\n m f_m=\\sum_{\\substack{j\\ge1","Jacobi gives a third engine. Put $T_j=j(j+1)/2$ and $a_j=(-1)^j(2j+1)$. The series\n\\[\n A(q)=\\prod_{k\\ge1}(1-q^k)^3=1+\\sum_{j\\ge1}a_jq^{T_j}\n\\]\nsatisfies $F=A^8$. The identity $AqF'=8FqA'$ yields\n\\begin{equation}\n m f_m=\\sum_{\\substack{j\\ge1\\\\T_j\\le m}}\n          a_j(9T_j-m)f_{m-T_j}.\n \\label{eq:workbook-triangular-recurrence}\n\\end{equation}\nAt $m=3$, only $T_1=1$, $T_2=3$ contribute. Its numerator is\n\\[\n (-3)(9-3)(252)+5(27-3)(1)=-4536+120=-4416.\n\\]\nDivision by $3$ again gives $-1472$. The direct product, divisor recurrence and triangular recurrence reach the same integer through different algebraic operations.","T_j\\le m}}\n          a_j(9T_j-m)f_{m-T_j}.\n \\label{eq:workbook-triangular-recurrence}\n\\end{equation}\nAt $m=3$, only $T_1=1$, $T_2=3$ contribute. Its numerator is\n\\[\n (-3)(9-3)(252)+5(27-3)(1)=-4536+120=-4416.\n\\]\nDivision by $3$ again gives $-1472$. The direct product, divisor recurrence and triangular recurrence reach the same integer through different algebraic operations.","\\Needspace{11\\baselineskip}\n\\begin{center}\n\\begin{tabular}{r r r r}\n\\toprule\n$m$ & $f_m=\\tau(m+1)$ & $m$ & $f_m=\\tau(m+1)$","\\midrule\n0 & $1$ & 5 & $-6048$","1 & $-24$ & 6 & $-16744$","2 & $252$ & 7 & $84480$","3 & $-1472$ & 8 & $-113643$","4 & $4830$ & 9 & $-115920$","The current independent comparison covers every $\\tau(n)$ for $1\\le n\\le512$. The historical final-arithmetic run's direct-product comparison stopped at $128$, while the packet-tree run compared through $512$. These are different documented ranges. The new three-engine table supplies its own bound and does not infer one from an older summary.","For a packet $v$, set $c_i=(v_i-i)/5$ and form\n\\[\n P_v(T)=\\prod_{0\\le i<j\\le4}\\bigl((i-j)+T(c_i-c_j)\\bigr).\n\\]\nIts physical value is $P_v(5)=288W(v)$. Summing the full shell gives\n\\[\n P_n(T)=\\sum_{v\\in\\Vn}P_v(T)=\\sum_{j=0}^{10}B_jT^j,\n \\qquad P_n(5)=288\\tau(n).\n\\]\nThe polynomial has a different variable from the prime index. A root in $T$ or in a rescaled variable does not represent a nearby prime coefficient.","For a nonzero $P_n$, use the convention $v_5(0)=+\\infty$ and define\n\\[\n \\mu_n=\\min_j\\bigl(j+v_5(B_j)\\bigr),\\qquad\n G_n(U)=5^{-\\mu_n}P_n(5U)=\\sum_{j=0}^{10}C_jU^j.\n\\]\nEvery $C_j=B_j5^{j-\\mu_n}$ is an integer. The power of $5$ should be removed only after verifying divisibility for each coefficient. Since $288$ is a $5$-adic unit,\n\\begin{equation}\n v_5(\\tau(n))=\\mu_n+v_5(G_n(1)).\n \\label{eq:workbook-normalized-valuation}\n\\end{equation}\nAll coefficients in the next two tables are exact integers from the freshly reproduced shell data. They allow a reader to check the normalization and every displayed residue without re-enumerating a large shell.","At $n=499$ the shell has $500$ packets. Of their weights, $262$ are positive and $238$ negative. Their sum is\n\\[\n \\tau(499)=-108877719272500.\n\\]\nThe minimum layer valuation is $\\mu_{499}=3$.","\\Needspace{17\\baselineskip}\n\\begin{center}\n\\begingroup\\small\n\\setlength{\\tabcolsep}{5pt}\n\\begin{tabular}{r r r r}\n\\toprule\n$j$ & $B_j$ in $P_{499}$ & $C_j$ in $G_{499}$ & $j+v_5(B_j)$","\\midrule\n0 & $144000$ & $1152$ & 3","1 & $-360000$ & $-14400$ & 5","2 & $1063040$ & $212608$ & 3","3 & $28418240$ & $28418240$ & 4","4 & $-148872808$ & $-744364040$ & 4","5 & $209474504$ & $5236862600$ & 5","6 & $2422451240$ & $302806405000$ & 7","7 & $16445000$ & $10278125000$ & 11","8 & $8107402848$ & $25335633900000$ & 8","9 & $-11893316304$ & $-185833067250000$ & 9","10 & $-1160632800$ & $-90674437500000$ & 12","Summing the second coefficient column and using the normalization checks gives\n\\begin{align*}\n P_{499}(5)&=-31356783150480000=288\\tau(499),\\\\\n G_{499}(1)&=-250854265203840.\n\\end{align*}\nReducing the normalized coefficients gives\n\\begin{align*}\n G_{499}(U)&\\equiv2+3U^2\\pmod5,","G_{499}(1)&=-250854265203840.\n\\end{align*}\nReducing the normalized coefficients gives\n\\begin{align*}\n G_{499}(U)&\\equiv2+3U^2\\pmod5,\\\\\n G_{499}(U)&\\equiv2+8U^2+15U^3+10U^4\\pmod{25}.\n\\end{align*}\nThe first polynomial has the simple roots $1$ and $4$ modulo $5$. At $U=1$, the exact sum is $10$ modulo $25$ and its derivative is $1$ modulo $5$. Consequently\n\\[\n G_{499}(1+5h)\\equiv10+5h\\pmod{25}.\n\\]\nThe five possibilities are $10,15,20,0,5$ as $h$ runs from $0$ to $4$. The root in the branch $U\\equiv1\\pmod5$ therefore lifts to $U=16\\pmod{25}$.","G_{499}(U)&\\equiv2+8U^2+15U^3+10U^4\\pmod{25}.\n\\end{align*}\nThe first polynomial has the simple roots $1$ and $4$ modulo $5$. At $U=1$, the exact sum is $10$ modulo $25$ and its derivative is $1$ modulo $5$. Consequently\n\\[\n G_{499}(1+5h)\\equiv10+5h\\pmod{25}.\n\\]\nThe five possibilities are $10,15,20,0,5$ as $h$ runs from $0$ to $4$. The root in the branch $U\\equiv1\\pmod5$ therefore lifts to $U=16\\pmod{25}$.","This lifted root is not the physical target. The target remains $U=1$, where $v_5(G_{499}(1))=1$. Formula \\eqref{eq:workbook-normalized-valuation} gives $v_5(\\tau(499))=3+1=4$.","The following exact root history records both simple branches. Each entry lies between $0$ and $5^k-1$ and evaluates to zero modulo $5^k$. Consecutive entries agree modulo the preceding power. These conditions are stronger than displaying an approximate root.","\\Needspace{18\\baselineskip}\n\\begin{center}\n\\begin{tabular}{r r r r}\n\\toprule\n$k$ & $5^k$ & Root reducing to $1$ & Root reducing to $4$","The fixed degree $10$ does not impose a universal ceiling on the valuation of polynomial values at varying inputs. A simple $5$-adic root supplies arbitrarily accurate integer approximations. That observation concerns the auxiliary variable. It neither changes the value at $1$ nor supplies interpolation of the prime-indexed family.","At $16249$ the shell contains $16250$ packets, with $8284$ positive and $7966$ negative weights. The exact coefficient is\n\\[\n \\tau(16249)=133116215274958169206250.\n\\]\nHere $\\mu_{16249}=4$. The full coefficient table is as follows.","\\Needspace{17\\baselineskip}\n\\begin{center}\n\\begingroup\\small\n\\setlength{\\tabcolsep}{4pt}\n\\begin{tabular}{r r r r}\n\\toprule\n$j$ & $B_j$ in $P_{16249}$ & $C_j$ in $G_{16249}$ & $j+v_5(B_j)$","\\midrule\n0 & $4680000$ & $7488$ & 4","1 & $-13320000$ & $-106560$ & 5","2 & $-199541600$ & $-7981664$ & 4","3 & $-26375464320$ & $-5275092864$ & 4","4 & $737398074300$ & $737398074300$ & 6","5 & $-39838758304992$ & $-199193791524960$ & 5","6 & $-746041848478320$ & $-18651046211958000$ & 7","7 & $-6990990388478280$ & $-873873798559785000$ & 8","8 & $21389931029028020$ & $13368706893142512500$ & 9","9 & $-64003552771660728$ & $-200011102411439775000$ & 9","10 & $3937759175521630080$ & $61527487117525470000000$ & 11","The exact sums are\n\\begin{align*}\n P_{16249}(5)&=38337469999187952731400000,\\\\\n G_{16249}(1)&=61339951998700724370240.\n\\end{align*}\nThe first equals $288\\tau(16249)$, and the second equals the first divided by $625$. Modulo $5$, the normalized polynomial is\n\\[\n 3+U^2+U^3=(U-1)^2(U-2).\n\\]\nModulo $25$, the table gives\n\\[\n G_{16249}(U)\\equiv13+15U+11U^2+11U^3+15U^5.\n\\]\nThus $G_{16249}(1)\\equiv15\\pmod{25}$ while $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $","G_{16249}(1)&=61339951998700724370240.\n\\end{align*}\nThe first equals $288\\tau(16249)$, and the second equals the first divided by $625$. Modulo $5$, the normalized polynomial is\n\\[\n 3+U^2+U^3=(U-1)^2(U-2).\n\\]\nModulo $25$, the table gives\n\\[\n G_{16249}(U)\\equiv13+15U+11U^2+11U^3+15U^5.\n\\]\nThus $G_{16249}(1)\\equiv15\\pmod{25}$ while $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","_{16249}(1)&=61339951998700724370240.\n\\end{align*}\nThe first equals $288\\tau(16249)$, and the second equals the first divided by $625$. Modulo $5$, the normalized polynomial is\n\\[\n 3+U^2+U^3=(U-1)^2(U-2).\n\\]\nModulo $25$, the table gives\n\\[\n G_{16249}(U)\\equiv13+15U+11U^2+11U^3+15U^5.\n\\]\nThus $G_{16249}(1)\\equiv15\\pmod{25}$ while $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","nd{align*}\nThe first equals $288\\tau(16249)$, and the second equals the first divided by $625$. Modulo $5$, the normalized polynomial is\n\\[\n 3+U^2+U^3=(U-1)^2(U-2).\n\\]\nModulo $25$, the table gives\n\\[\n G_{16249}(U)\\equiv13+15U+11U^2+11U^3+15U^5.\n\\]\nThus $G_{16249}(1)\\equiv15\\pmod{25}$ while $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","[\n 3+U^2+U^3=(U-1)^2(U-2).\n\\]\nModulo $25$, the table gives\n\\[\n G_{16249}(U)\\equiv13+15U+11U^2+11U^3+15U^5.\n\\]\nThus $G_{16249}(1)\\equiv15\\pmod{25}$ while $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","ile $G'_{16249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","249}(1)\\equiv0\\pmod5$. For every integer or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","ger or $5$-adic integer $h$, polynomial Taylor expansion gives\n\\[\n G_{16249}(1+5h)\\equiv G_{16249}(1)+5hG'_{16249}(1)\n                    \\equiv15\\pmod{25}.\n\\]\nTerms of order at least two have a factor $25$. This is a uniform obstruction throughout that residue branch, not merely the observation that five trial inputs fail. Every input congruent to $1$ modulo $5$ has normalized valuation exactly $1$.","The other residual root should not be discarded. At $U=2$, the normalized value is $5$ modulo $25$ and its derivative is $1$ modulo $5$. That simple branch lifts to $22$ modulo $25$. The conclusion is that the double branch at $1$ does not lift, rather than that the polynomial has no $5$-adic roots anywhere.","At the physical input $1$, the coefficient valuation is $4+1=5$. The two examples have different root geometry but both settle their actual target value by an exact modulo-$25$ calculation.","The tested auxiliary equality was\n\\[\n v_\\ell(\\tau(p))=v_\\ell(p+1),\n \\qquad p\\ne\\ell,\\quad\\ell\\mid(p+1),\\quad p\\text{ prime}.\n\\]\nThe complete search through $150000$ contains $13848$ primes. The following counts refer to the indicated candidate sets, not to every prime without a residue condition.","\\Needspace{11\\baselineskip}\n\\begin{center}\n\\begingroup\\small\n\\begin{tabular}{r r r r r r}\n\\toprule\n$\\ell$ & First $p$ & $v_\\ell(\\tau(p))$ & $v_\\ell(p+1)$ & Candidates & Discrepancies","The exact decisive values, including the discrepancy in the opposite $5$-adic direction, are\n\\begin{align*}\n \\tau(499)&=-108877719272500,","The exact decisive values, including the discrepancy in the opposite $5$-adic direction, are\n\\begin{align*}\n \\tau(499)&=-108877719272500,\\\\\n \\tau(1249)&=27757672778617250,\\\\\n \\tau(8747)&=348766170858826454196,\\\\\n \\tau(81919)&=-904107824789465028682055680.\n\\end{align*}\nFor example, $1250=2\\cdot5^4$, while the second integer has $5$-adic valuation $3$. Hence a proposed one-sided lift $v_5(\\tau(p))\\ge v_5(p+1)$ also fails at this actual prime.","\\tau(81919)&=-904107824789465028682055680.\n\\end{align*}\nFor example, $1250=2\\cdot5^4$, while the second integer has $5$-adic valuation $3$. Hence a proposed one-sided lift $v_5(\\tau(p))\\ge v_5(p+1)$ also fails at this actual prime.","These four coefficients were checked independently by the Eisenstein convolution. If $\\sigma_r(n)=\\sum_{d\\mid n}d^r$, the coefficient of $(E_4E_8-E_6^2)/1728$ is obtained from\n\\begin{align*}\n1728\\tau(n)={}&240\\sigma_3(n)+480\\sigma_7(n)+1008\\sigma_5(n)","These four coefficients were checked independently by the Eisenstein convolution. If $\\sigma_r(n)=\\sum_{d\\mid n}d^r$, the coefficient of $(E_4E_8-E_6^2)/1728$ is obtained from\n\\begin{align*}\n1728\\tau(n)={}&240\\sigma_3(n)+480\\sigma_7(n)+1008\\sigma_5(n)\\\\\n &+115200\\sum_{k=1}^{n-1}\\sigma_3(k)\\sigma_7(n-k)\\\\\n &-254016\\sum_{k=1}^{n-1}\\sigma_5(k)\\sigma_5(n-k).\n\\end{align*}\nThis computes the selected integers without the triangular recurrence. The numerator is divided exactly by $1728$. The replay checks this independent formula for $1\\le n\\le128$ and at the four displayed primes, giving $132$ checks.","&-254016\\sum_{k=1}^{n-1}\\sigma_5(k)\\sigma_5(n-k).\n\\end{align*}\nThis computes the selected integers without the triangular recurrence. The numerator is divided exactly by $1728$. The replay checks this independent formula for $1\\le n\\le128$ and at the four displayed primes, giving $132$ checks.","The phrase ``first failure'' has two limits. It concerns the ordered search in its declared candidate set, and its exhaustive range ends at $150000$. A larger prime may supply another failure. The displayed nonzero integers refute an auxiliary valuation identity, not Lehmer's conjecture.","The degree-at-most-$22$ certificate has six coarse sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let\n\\[\n A\\in\\Z^{6\\times16},\\quad\n M_c\\in\\Z^{6\\times18},\\quad M_f\\in\\Z^{16\\times18}.\n\\]\nThe fine-to-coarse parent map completely specifies $A$: its column $j$ has a $1$ in the parent row and zero elsewhere. The original fine masses are given below. Let $s_f$ denote their sixteen-entry column vector. Indices start at $0$.","sectors and sixteen fine sectors. Both correction matrices have eighteen columns. Let\n\\[\n A\\in\\Z^{6\\times16},\\quad\n M_c\\in\\Z^{6\\times18},\\quad M_f\\in\\Z^{16\\times18}.\n\\]\nThe fine-to-coarse parent map completely specifies $A$: its column $j$ has a $1$ in the parent row and zero elsewhere. The original fine masses are given below. Let $s_f$ denote their sixteen-entry column vector. Indices start at $0$.","\\Needspace{15\\baselineskip}\n\\begin{center}\n\\begin{tabular}{r r r @{\\qquad} r r r}\n\\toprule\n$j$ & Parent & Fine mass & $j$ & Parent & Fine mass","\\midrule\n0 & 0 & $2354352$ & 8 & 0 & $-11844448$","1 & 2 & $-19467448$ & 9 & 1 & $-7162848$","2 & 4 & $-832048$ & 10 & 0 & $-31156048$","3 & 4 & $15074752$ & 11 & 2 & $4356352$","4 & 4 & $1881152$ & 12 & 2 & $32821152$","5 & 5 & $4248552$ & 13 & 2 & $1262352$","6 & 5 & $7759752$ & 14 & 0 & $-7455448$","7 & 4 & $-31808448$ & 15 & 3 & $-12874848$","For instance, row $0$ aggregates fine indices $0,8,10,14$:\n\\[\n 2354352-11844448-31156048-7455448=-48101592.\n\\]\nThe six coarse masses are therefore\n\\[\n s_c=(-48101592,-7162848,18972408,\n      -12874848,-15684592,12008304)^t.\n\\]\nTheir sum, and the sum of the fine masses, is $-52843168$. These original masses differ in order from the lexicographic orbit bars because the certificate orders the residue-sector keys.","The full machine certificate is\n\\begin{center}\\small\n\\path{evidence/computation/replays/transport/p31_transport_certificate.json}.\n\\end{center}\nIt supplies an $18\\times13$ rational matrix $N$ whose columns form a basis of $\\ker M_c$, and the detail matrix $B=M_fN$ of size $16\\times13$. The basic exact checks are\n\\[\n AM_f=M_c,\\quad As_f=s_c,\\quad M_cN=0,\n \\quad B=M_fN,\\quad AB=0.\n\\]\nThe last equation shows that a detail correction changes no coarse mass.","The rational row-reduction results are\n\\[\n \\rank M_c=5,\\qquad\\rank M_f=15,\\qquad\\rank B=10.\n\\]\nAn independent check obtains the same lower bounds over both prime fields $\\mathbb F_{65537}$ and $\\mathbb F_{1000003}$. Every denominator used in a reduction is invertible there. A nonzero minor in either field proves a nonzero rational minor after clearing denominators.","The corresponding upper bounds have direct structural explanations. Each correction matrix has column sum zero, giving bounds $5$ and $15$. Every row of $A$ has a nonempty, disjoint support, so $\\rank A=6$ and $\\dim\\ker A=10$. Since $AB=0$, the detail rank is at most $10$. The upper and lower bounds establish the displayed ranks exactly.","The coarse coefficient vector $a\\in\\Q^{18}$ makes\n\\[\n x=s_f+M_fa,\\qquad Ax=(0,0,0,0,0,-52843168)^t.\n\\]\nThe certificate then chooses a coefficient vector $d\\in\\ker M_c$. Its final fine vector $y=x+M_fd$ is zero except at index $6$, where it equals $-52843168$.","One parent gives a concrete view of this second stage. Parent $5$ has fine children $5,6$. Just after the coarse stage their values are\n\\begin{align*}\n x_5&=-\\frac{161598180629860124657534386744}\n                 {23300823161076013875625},","One parent gives a concrete view of this second stage. Parent $5$ has fine children $5,6$. Just after the coarse stage their values are\n\\begin{align*}\n x_5&=-\\frac{161598180629860124657534386744}\n                 {23300823161076013875625},\\\\\n x_6&=-\\frac{1069691132209170737342448593256}\n                 {23300823161076013875625}.\n\\end{align*}\nTheir exact sum is $-52843168$. The detail stage adds $-x_5$ to child $5$ and $x_5$ to child $6$, giving $(0,-52843168)$ in that parent. All other parents retain their zero totals. This is a rational conservation calculation, not a numerical near-zero test.","x_6&=-\\frac{1069691132209170737342448593256}\n                 {23300823161076013875625}.\n\\end{align*}\nTheir exact sum is $-52843168$. The detail stage adds $-x_5$ to child $5$ and $x_5$ to child $6$, giving $(0,-52843168)$ in that parent. All other parents retain their zero totals. This is a rational conservation calculation, not a numerical near-zero test.","To verify the whole certificate, parse the rational pairs, check the matrix dimensions, verify all five identities, and compare both stored stages entry by entry. Then check $M_cd=0$ and the single surviving final entry. The script \\path{code/check_transport_certificate.py} performs the independent equality and prime-field checks.","This example does not assume a uniform rank theorem. Nor does full rank prove a uniformly nonzero survivor. At a shell whose original total were zero, complete zero-total transfer could still leave every final mass zero. At $31$ the separately computed exact total is nonzero, so the displayed survivor is authenticated for this one shell.","A version $2$ manifest records the actual argument vector, exact arithmetic domain, semantic bounds, software versions and executed limits. Its \\texttt{input\\_artifacts} contain input checksums before and after execution. Its \\texttt{declared\\_results} name the scientific files the child must create. Output checksums cover those files and the execution logs.","The count $704$ means $22$ selected modulus examples times $32$ exponents, not $704$ distinct primes. Of $24$ selected moduli, $11$ and $19$ had no observed matching prime through $10000$. That finite absence does not contradict positive density. The $24576$ synthetic cases mean $256$ moduli times three trace parameters times $32$ exponents. Their parameter $p=5M-1$ need not be prime, and their trace is not asserted to equal $\\tau(p)$.","ected modulus examples times $32$ exponents, not $704$ distinct primes. Of $24$ selected moduli, $11$ and $19$ had no observed matching prime through $10000$. That finite absence does not contradict positive density. The $24576$ synthetic cases mean $256$ moduli times three trace parameters times $32$ exponents. Their parameter $p=5M-1$ need not be prime, and their trace is not asserted to equal $\\tau(p)$.","\\bibitem{Lehmer}\nD. H. Lehmer, \\emph{The vanishing of Ramanujan's function $\\tau(n)$},\nDuke Mathematical Journal \\textbf{14} (1947), 429--433.\n\\href{https://doi.org/10.1215/S0012-7094-47-01436-1}{DOI: 10.1215/S0012-7094-47-01436-1}.\n\\href{https://oeis.org/A000594/a000594.pdf}{Original scan}.\nThe scan was authenticated, but its original proof was not readable through the source-check route. The prime reduction is a classical attributed input and is proved separately in this reader.","\\bibitem{DHZ}\nMaarten Derickx, Mark van Hoeij, and Jinxiang Zeng,\n\\emph{Computing Galois representations and equations for modular curves $X_H(\\ell)$},\n\\href{https://arxiv.org/pdf/1312.6819v2}{arXiv:1312.6819v2}, 18 March 2014.\nCorollary~1.2, p.~2; its computational proof in \\S4, p.~11.\nThe cited nonvanishing range has the strict endpoint\n$n<816212624008487344127999$. This is a historical source result, not an edition replay or a verified latest record.","\\bibitem{DeligneModular}\nPierre Deligne, \\emph{Formes modulaires et repr\\'esentations $\\ell$-adiques},\nS\\'eminaire Bourbaki 1968/69, expos\\'e~355, published 1971, 139--172.\n\\href{https://www.numdam.org/item/SB_1968-1969__11__139_0.pdf}{NUMDAM original}.\nTheorem~5.6 supplies the implication from the Weil conjectures to the Ramanujan bound.\nThe author's \\href{https://publications.ias.edu/sites/default/files/5_FormesModulaires.pdf}{English translation}, Proposition~4.8 and equation~(5.7), was also checked.\nIts cohomological $F$ is inverse arithmetic Frobenius; the reader uses the rank-two arithmetic normalization specified in~\\cite{SerreBourbaki}.","/69, expos\\'e~355, published 1971, 139--172.\n\\href{https://www.numdam.org/item/SB_1968-1969__11__139_0.pdf}{NUMDAM original}.\nTheorem~5.6 supplies the implication from the Weil conjectures to the Ramanujan bound.\nThe author's \\href{https://publications.ias.edu/sites/default/files/5_FormesModulaires.pdf}{English translation}, Proposition~4.8 and equation~(5.7), was also checked.\nIts cohomological $F$ is inverse arithmetic Frobenius; the reader uses the rank-two arithmetic normalization specified in~\\cite{SerreBourbaki}.","\\bibitem{DeligneWeil}\nPierre Deligne, \\emph{La conjecture de Weil: I},\nPublications Math\\'ematiques de l'IH\\'ES \\textbf{43} (1974), 273--307.\n\\href{https://doi.org/10.1007/BF02684373}{DOI: 10.1007/BF02684373}.\n\\href{https://www.numdam.org/article/PMIHES_1974__43__273_0.pdf}{Theorem~1.6, p.~276}.\nFor the modular specialization giving $|\\tau(p)|\\le2p^{11/2}$, this purity theorem is used with the implication in~\\cite{DeligneModular}.","\\bibitem{SerreBourbaki}\nJean-Pierre Serre, \\emph{Congruences et formes modulaires (d'apr\\`es H. P. F. Swinnerton-Dyer)},\nS\\'eminaire Bourbaki 1971/72, expos\\'e~416, Lecture Notes in Mathematics \\textbf{317} (1973), 319--338.\n\\href{https://www.numdam.org/item/SB_1971-1972__14__319_0.pdf}{NUMDAM PDF}.\n\\S3.1, pp.~331--332, supplies the established integral $\\ell$-adic representations and their unramified Frobenius polynomials. Theorem~10 and \\S3.3, p.~335, concern image exceptions. These statements do not assert joint surjectivity for an arbitrary product of reductions.","\\bibitem{SerreCongruences}\nJean-Pierre Serre, \\emph{Une interpr\\'etation des congruences relatives \\`a la fonction $\\tau$ de Ramanujan},\nS\\'eminaire Delange--Pisot--Poitou \\textbf{9} (1967/68), no.~1, expos\\'e~14, 1--17.\n\\href{https://www.numdam.org/item/SDPP_1967-1968__9_1_A13_0.pdf}{NUMDAM PDF}.\n\\S2.2, equation~(21), p.~4, for modulo~691; \\S2.1, equation~(16), p.~3, and \\S4.5, p.~11, for the stated prime congruences modulo~23 and~$23^2$.\nThe representation existence is still conjectural in this 1968 text; its established version is cited from~\\cite{SerreBourbaki}.","e Serre, \\emph{Une interpr\\'etation des congruences relatives \\`a la fonction $\\tau$ de Ramanujan},\nS\\'eminaire Delange--Pisot--Poitou \\textbf{9} (1967/68), no.~1, expos\\'e~14, 1--17.\n\\href{https://www.numdam.org/item/SDPP_1967-1968__9_1_A13_0.pdf}{NUMDAM PDF}.\n\\S2.2, equation~(21), p.~4, for modulo~691; \\S2.1, equation~(16), p.~3, and \\S4.5, p.~11, for the stated prime congruences modulo~23 and~$23^2$.\nThe representation existence is still conjectural in this 1968 text; its established version is cited from~\\cite{SerreBourbaki}.","o-Niu Han, \\emph{The Nekrasov--Okounkov hook length formula: refinement, elementary proof, extension and applications},\n\\href{https://arxiv.org/pdf/0805.1398}{arXiv:0805.1398v1}, 9 May 2008,\nDefinition~1.1 and Theorems~1.1--1.2, p.~3.\nPublished in Annales de l'Institut Fourier \\textbf{60} (2010), 1--29,\n\\href{https://doi.org/10.5802/aif.2515}{DOI: 10.5802/aif.2515}.\nThe checked specialization is $t=5$, $z=25$: shell $\\sum v_i^2=10n$ and Vandermonde weight $D(v)/288$. Integrality and the sign consequences are separate project arguments.","an, \\emph{The Nekrasov--Okounkov hook length formula: refinement, elementary proof, extension and applications},\n\\href{https://arxiv.org/pdf/0805.1398}{arXiv:0805.1398v1}, 9 May 2008,\nDefinition~1.1 and Theorems~1.1--1.2, p.~3.\nPublished in Annales de l'Institut Fourier \\textbf{60} (2010), 1--29,\n\\href{https://doi.org/10.5802/aif.2515}{DOI: 10.5802/aif.2515}.\nThe checked specialization is $t=5$, $z=25$: shell $\\sum v_i^2=10n$ and Vandermonde weight $D(v)/288$. Integrality and the sign consequences are separate project arguments.","Nekrasov--Okounkov hook length formula: refinement, elementary proof, extension and applications},\n\\href{https://arxiv.org/pdf/0805.1398}{arXiv:0805.1398v1}, 9 May 2008,\nDefinition~1.1 and Theorems~1.1--1.2, p.~3.\nPublished in Annales de l'Institut Fourier \\textbf{60} (2010), 1--29,\n\\href{https://doi.org/10.5802/aif.2515}{DOI: 10.5802/aif.2515}.\nThe checked specialization is $t=5$, $z=25$: shell $\\sum v_i^2=10n$ and Vandermonde weight $D(v)/288$. Integrality and the sign consequences are separate project arguments.","refinement, elementary proof, extension and applications},\n\\href{https://arxiv.org/pdf/0805.1398}{arXiv:0805.1398v1}, 9 May 2008,\nDefinition~1.1 and Theorems~1.1--1.2, p.~3.\nPublished in Annales de l'Institut Fourier \\textbf{60} (2010), 1--29,\n\\href{https://doi.org/10.5802/aif.2515}{DOI: 10.5802/aif.2515}.\nThe checked specialization is $t=5$, $z=25$: shell $\\sum v_i^2=10n$ and Vandermonde weight $D(v)/288$. Integrality and the sign consequences are separate project arguments.","\\bibitem{GKS}\nFrank Garvan, Dongsu Kim, and Dennis Stanton, \\emph{Cranks and $t$-cores},\nInventiones Mathematicae \\textbf{101} (1990), 1--17.\n\\href{https://doi.org/10.1007/BF01231493}{DOI: 10.1007/BF01231493}.\nChecked \\href{https://www-users.cse.umn.edu/~stant001/PAPERS/cra2.pdf}{author manuscript},\n\\S2, Bijections~1--2; \\S5, Theorem~4 and equation~(5.3), printed p.~8.\nThe exact five-core count is translated to V-coded packets using Han's bijection; it is not a sign assertion about the weighted sum.","\\bibitem{IW}\nTomoyoshi Ibukiyama and Brandon Williams,\n\\emph{Hermitian modular forms and algebraic modular forms on $SO(6)$},\n\\href{https://arxiv.org/pdf/2505.23497}{arXiv:2505.23497v1},\nheader 29 May 2025, internal date 12 June 2025.\n\\S3.4, printed pp.~18--19, gives vector-valued harmonic theta modularity and weight $\\nu+n/2$.\nThe scalar full-level sign-line descent is a project argument. The paper's conjectural Hermitian correspondences are not assumed.","ine chapter \\href{https://wstein.org/books/modform/modform/level_one.html}{\\emph{Modular Forms of Level 1}}, checked 9 October 2026.\nTheorem~2.11 for valence; Theorem~2.14, Corollaries~2.15--2.16, and Theorem~2.17 for the level-one structure and dimensions.\nThe chapter's Eisenstein series differ by nonzero scalars from the reader's constant-one normalization; its displayed normalized identity is $\\Delta=(E_4^3-E_6^2)/1728$.","\\bibitem{Niebur}\nDouglas Niebur, \\emph{A formula for Ramanujan's $\\tau$-function},\nIllinois Journal of Mathematics \\textbf{19} (1975), 448--449.\n\\href{https://doi.org/10.1215/ijm/1256050746}{DOI: 10.1215/ijm/1256050746}.\nThe original proof could not be accessed through the check route. The universal convolution identity was checked in the primary-paper restatement~\\cite{MoriyaSmyth}; the differential reconstruction in this reader uses~\\cite{Fonseca,Stein}.","\\bibitem{MoriyaSmyth}\nB. K. Moriya and C. J. Smyth,\n\\emph{Index-dependent divisors of coefficients of modular forms},\n\\href{https://webhomes.maths.ed.ac.uk/~chris/papers/Moriya_Smyth_final.pdf}{author manuscript},\ndated 1 May 2013, accepted 28 April 2013.\n\\S7, printed pp.~10--11, restates Niebur's identity for every $n\\ge1$ and gives the symmetric half-sum convention.\n\\S2.1 explicitly retains the alternative $a(p)=0$ in the size/divisibility hypothesis.\nThe locators refer to the inspected manuscript.","yaSmyth}\nB. K. Moriya and C. J. Smyth,\n\\emph{Index-dependent divisors of coefficients of modular forms},\n\\href{https://webhomes.maths.ed.ac.uk/~chris/papers/Moriya_Smyth_final.pdf}{author manuscript},\ndated 1 May 2013, accepted 28 April 2013.\n\\S7, printed pp.~10--11, restates Niebur's identity for every $n\\ge1$ and gives the symmetric half-sum convention.\n\\S2.1 explicitly retains the alternative $a(p)=0$ in the size/divisibility hypothesis.\nThe locators refer to the inspected manuscript.","\\bibitem{LRTrace}\nNik Lygeros and Olivier Rozier,\n\\emph{A New Solution to the Equation $\\tau(p)\\equiv0\\pmod p$},\nJournal of Integer Sequences \\textbf{13} (2010), article~10.7.4.\n\\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.pdf}{Journal PDF},\n\\S\\S4--6, printed pp.~3--6, for the trace moments, Catalan inverse, and reduced-form convention.\nThe Hurwitz count includes imprimitive forms and exceptional weights $1/2$ and $1/3$; $6H$ is the stored integer. The zeroth moment keeps half of the $t=0$ term.","and Olivier Rozier,\n\\emph{A New Solution to the Equation $\\tau(p)\\equiv0\\pmod p$},\nJournal of Integer Sequences \\textbf{13} (2010), article~10.7.4.\n\\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.pdf}{Journal PDF},\n\\S\\S4--6, printed pp.~3--6, for the trace moments, Catalan inverse, and reduced-form convention.\nThe Hurwitz count includes imprimitive forms and exceptional weights $1/2$ and $1/3$; $6H$ is the stored integer. The zeroth moment keeps half of the $t=0$ term.","r Rozier,\n\\emph{A New Solution to the Equation $\\tau(p)\\equiv0\\pmod p$},\nJournal of Integer Sequences \\textbf{13} (2010), article~10.7.4.\n\\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.pdf}{Journal PDF},\n\\S\\S4--6, printed pp.~3--6, for the trace moments, Catalan inverse, and reduced-form convention.\nThe Hurwitz count includes imprimitive forms and exceptional weights $1/2$ and $1/3$; $6H$ is the stored integer. The zeroth moment keeps half of the $t=0$ term.","r,\n\\emph{A New Solution to the Equation $\\tau(p)\\equiv0\\pmod p$},\nJournal of Integer Sequences \\textbf{13} (2010), article~10.7.4.\n\\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.pdf}{Journal PDF},\n\\S\\S4--6, printed pp.~3--6, for the trace moments, Catalan inverse, and reduced-form convention.\nThe Hurwitz count includes imprimitive forms and exceptional weights $1/2$ and $1/3$; $6H$ is the stored integer. The zeroth moment keeps half of the $t=0$ term.",",\nJournal of Integer Sequences \\textbf{13} (2010), article~10.7.4.\n\\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.pdf}{Journal PDF},\n\\S\\S4--6, printed pp.~3--6, for the trace moments, Catalan inverse, and reduced-form convention.\nThe Hurwitz count includes imprimitive forms and exceptional weights $1/2$ and $1/3$; $6H$ is the stored integer. The zeroth moment keeps half of the $t=0$ term.","textbf{18} (1996), no.~2, 26--37.\nChecked \\href{https://pub.math.leidenuniv.nl/~lenstrahw/PUBLICATIONS/1996d/art.pdf}{author scan},\np.~31 for the natural-density convention and p.~34 for the conjugacy-class theorem.\nThe historical project URL failed; this author-hosted copy was inspected successfully.\nThe theorem applies to one fixed finite quotient and a nonempty conjugacy class, giving density $|C|/|G|$.","\\bibitem{CF}\nPierre Colmez and Jean-Marc Fontaine,\n\\emph{Construction des repr\\'esentations $p$-adiques semi-stables},\nInventiones Mathematicae \\textbf{140} (2000), 1--43.\n\\href{https://doi.org/10.1007/s002220000042}{DOI: 10.1007/s002220000042}.\nChecked \\href{https://webusers.imj-prg.fr/~pierre.colmez/CF.pdf}{37-page author manuscript}:\nTheorem~A, p.~2; \\S3.4, pp.~21--22; \\S4.1 and Corollary~4.3, pp.~23--24; the period-ring identity on p.~11.\nThese are manuscript locators. Local realization does not identify the constructed object with the restriction of Deligne's global representation.","\\bibitem{Withdrawn2025}\nMinjia Shi, Lu Wang, and Patrick Sol\\'e,\n\\emph{Proof of the Lehmer conjecture on Ramanujan's $\\tau$ function},\n\\href{https://arxiv.org/abs/2503.23498}{arXiv:2503.23498v2}, 1 April 2025, withdrawn.\nOnly the explicit withdrawal metadata was checked. This title is not cited as an established proof or as a theorem input.","\\chapter{The exact obligation that remains}\\label{ch:closeout}\n\\section{The target after every valid reduction}\nThe constructions in this reader return to one fixed arithmetic question. They have not replaced $\\Delta$ by a more convenient sequence, and they have not changed nonvanishing into a statement about typical primes. Their target is still every actual prime coefficient.","\\begin{proposition}[The remaining family]\\label{prop:remainingfamily}\nLehmer's conjecture is equivalent to the assertion\n\\begin{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","remaining family]\\label{prop:remainingfamily}\nLehmer's conjecture is equivalent to the assertion\n\\begin{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","ly]\\label{prop:remainingfamily}\nLehmer's conjecture is equivalent to the assertion\n\\begin{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","ly}\nLehmer's conjecture is equivalent to the assertion\n\\begin{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","is equivalent to the assertion\n\\begin{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","in{equation}\\label{eq:remainingfamily}\nR(p)=\\tau(p)\\ne0\n\\qquad\\text{for every prime }p\\equiv4\\pmod5.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:primereduction} reduces the conjecture to prime indices. The congruence \\eqref{eq:primemod5} excludes a zero at primes congruent to $1,2,3$ modulo $5$, and the explicit value $\\tau(5)=4830$ handles $5$. The primitive transform equals $\\tau$ at a prime by \\eqref{eq:rprime}. Thus exactly the displayed unexcluded family remains.\n\\end{proof}","On each such prime shell, all packets have content one. The residual is therefore the full signed sum\n\\begin{equation}\\label{eq:primepacketobligation}\n\\tau(p)=\\sum_{v\\in\\mathcal V_p}W(v).\n\\end{equation}\nThe summands are nonzero integers, and for $p>20$ they are units at the index prime. Neither property forbids their sum from being zero. The unresolved statement concerns the exact balance of their signed masses.","\\midrule\n$R(s^2)\\ne0$ for all $s$&Every even primitive local block survives, including a hypothetical zero prime parameter.","$R(n)=0\\iff\\tau(n)=0$&Primitive extraction encodes the same zero problem, with $R(p)=\\tau(p)$.","The arithmetic survivor theorem also has an all-modulus quantifier, but it is a limitation theorem for a proposed method. For every fixed positive modulus $M$, a positive-density set of actual primes satisfies\n\\[\np\\equiv-1\\pmod{5M},\n\\qquad R(p^r)\\equiv0\\pmod M\\quad(r\\ge1).\n\\]\nIt applies to the actual joint finite image, with no independence assumption. This is stronger than observing survivors in a bounded search. It still says nothing about an integer zero at one fixed prime, because that prime may change with $M$.","od. For every fixed positive modulus $M$, a positive-density set of actual primes satisfies\n\\[\np\\equiv-1\\pmod{5M},\n\\qquad R(p^r)\\equiv0\\pmod M\\quad(r\\ge1).\n\\]\nIt applies to the actual joint finite image, with no independence assumption. This is stronger than observing survivors in a bounded search. It still says nothing about an integer zero at one fixed prime, because that prime may change with $M$.","\\section{The missing arrows cannot be supplied by repetition}\nSeveral attractive final steps have failed in precisely stated forms. The integral prime-index interpolation violates a necessary divided-difference condition. The aggregate auxiliary polynomial does not inherit its individual factors' root interval. The proposed lifting equalities at $2,3,5$ have exact prime counterexamples. Strict preservation of sectors with opposite masses cannot yield a common-sign sum. Interpolation on one shell does not give universal nullness, and top-degree harmonic projection does not preserve the interpolated values.","The local filtered Frobenius construction and the arithmetic sequence $f$ test different kinds of retained information. The former realizes compatible local zero-trace data. The latter satisfies a specified collection of arithmetic rules while having a prime zero. Neither is identified unconditionally with an actual zero coefficient of $\\Delta$. Their role is to show why those weaker premises alone cannot force nonvanishing.","\\section{A sufficient adaptive residue input}\nOne precise route would be a uniform construction of residue certificates. Suppose there are explicitly defined functions on prime indices, with integer values $m(p)\\ge2$ and $b(p)$, for which one proves\n\\begin{equation}\\label{eq:adaptiveresidueinput}\n\\tau(p)\\equiv b(p)\\pmod{m(p)},\n\\qquad b(p)\\not\\equiv0\\pmod{m(p)}\n\\end{equation}\nfor every prime in \\eqref{eq:remainingfamily}. These two assertions would settle that family and hence Lehmer. The modulus may depend on $p$, so the fixed-modulus survivor theorem does not refute this proposed form of input.","tes. Suppose there are explicitly defined functions on prime indices, with integer values $m(p)\\ge2$ and $b(p)$, for which one proves\n\\begin{equation}\\label{eq:adaptiveresidueinput}\n\\tau(p)\\equiv b(p)\\pmod{m(p)},\n\\qquad b(p)\\not\\equiv0\\pmod{m(p)}\n\\end{equation}\nfor every prime in \\eqref{eq:remainingfamily}. These two assertions would settle that family and hence Lehmer. The modulus may depend on $p$, so the fixed-modulus survivor theorem does not refute this proposed form of input.","The issue is the proof of the second assertion for the actual coefficient, together with the identity linking the construction to $\\tau$. Defining $b(p)$ to be the unknown residue and then assuming it is nonzero would only restate the target. A useful construction must give an independently justified nonzero residue, with all exceptional primes and all precision requirements covered.","The index-dependent criterion already available from Deligne's bound is\n\\[\n\\tau(p)=0\\iff p^6\\mid\\tau(p),\\qquad p\\ge5.\n\\]\nCombined with the divisor convolution, it becomes\n\\begin{equation}\\label{eq:closeoutconvolution}\n840S_4(p)\\not\\equiv p^4C(p)\\pmod{p^6},\n\\qquad p>7,\n\\end{equation}\nas the exact desired exclusion. This gives a correctly typed target for a convolution route. No uniform arithmetic property excluding that residue has been proved here. A stronger valuation bound may suffice, but its proof must permit the zero case as a possibility until the bound itself excludes it.","\\section{A sufficient sign input}\nA second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials $Q_{p,j}$ and rational coefficients $a_{p,j}$, and put\n\\[\nF_p(v)=1+\\sum_j a_{p,j}Q_{p,j}(v).\n\\]\nThe choice may depend on $p$. Universal nullness must hold after the choice, so its weighted sum is still the actual $\\tau(p)$.","input}\nA second precise route would construct permitted corrections with an independently proved sign. At each remaining prime, choose finitely many rational universal null polynomials $Q_{p,j}$ and rational coefficients $a_{p,j}$, and put\n\\[\nF_p(v)=1+\\sum_j a_{p,j}Q_{p,j}(v).\n\\]\nThe choice may depend on $p$. Universal nullness must hold after the choice, so its weighted sum is still the actual $\\tau(p)$.","Suppose the construction proves a sign $\\varepsilon_p\\in\\{-1,1\\}$ for which\n\\begin{equation}\\label{eq:uniformsigninput}\n\\varepsilon_p W(v)F_p(v)\\ge0\n\\quad\\text{for every }v\\in\\mathcal V_p,\n\\end{equation}\nand proves that at least one of these finitely many quantities is strictly positive. Then summing \\eqref{eq:uniformsigninput} gives $\\varepsilon_p\\tau(p)>0$. This would be a valid all-prime certificate theorem.","The strict clause cannot be discarded. If the total were zero, a correction making every term zero would satisfy the weak inequalities. Nor can the strict clause be obtained merely from full transfer rank. Equalizing corrected masses to $\\tau(p)/N$ already uses the total whose nonvanishing is at issue.","The finite certificates at $7$ and $31$ show that this form of argument can work at concrete shells. A uniform construction needs a proof covering all remaining primes, with coefficients and inequalities derived from actual packet structure. The current family and its finite successes do not provide such a theorem.","\\section{Where a global input would have to enter}\nThe actual form has more structure than an arbitrary multiplicative recurrence sequence. A new global theorem could impose a relation or constraint that is absent from the countermodel and that excludes a zero prime parameter. The constraint must be stated exactly and derived for $\\Delta$. Merely naming modularity or compatibility does not prove a particular trace is nonzero.","Local compatibility also has a direction. A weakly admissible trace-zero object establishes that the selected local conditions permit zero. It does not establish that this object occurs as the restriction of the actual global representation. Proving such an identification at a prime, together with a proved comparison identifying its normalized Frobenius trace with $\\tau(p)$, would give an actual zero and would disprove Lehmer. Proving a global obstruction to that identification could instead contribute to nonvanishing. Neither direction has been supplied by the local construction alone.","Similarly, showing that the synthetic series $\\sum_{n\\ge1}f(n)q^n$ is a holomorphic cusp form of level one and weight $12$ would be new information. Its normalization $f(1)=1$ and the one-dimensional cusp space would then identify it with $\\Delta$. The present construction does not establish that premise. Calling the sequence nonmodular without an independent argument would also go beyond what was proved, since it could coincide with $\\tau$ if the chosen prime were already an actual zero.","eries $\\sum_{n\\ge1}f(n)q^n$ is a holomorphic cusp form of level one and weight $12$ would be new information. Its normalization $f(1)=1$ and the one-dimensional cusp space would then identify it with $\\Delta$. The present construction does not establish that premise. Calling the sequence nonmodular without an independent argument would also go beyond what was proved, since it could coincide with $\\tau$ if the chosen prime were already an actual zero.","\\begin{openproblem}[One actual coefficient at every remaining prime]\nFor every prime $p\\equiv4\\pmod5$, prove that the actual signed prime-shell total in \\eqref{eq:primepacketobligation} is nonzero. Supply an independent nonzero residue, a legitimate sign certificate with a strict term, or another fully justified mechanism excluding exact cancellation. A verified actual prime zero would instead disprove the conjecture.\n\\end{openproblem}","\\chapter{A fixed modular form and an exact question}\\label{ch:target}\n\\section{The coefficient that must not disappear}\nFor $|q|<1$ the product defining $\\Delta$ converges. As a formal power series it is even simpler: the coefficient of $q^N$ depends only on the first $N$ product factors. Thus each $\\tau(n)$ is an unambiguously defined integer. The opening terms are\n\\begin{align}\n\\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\\notag","\\chapter{A fixed modular form and an exact question}\\label{ch:target}\n\\section{The coefficient that must not disappear}\nFor $|q|<1$ the product defining $\\Delta$ converges. As a formal power series it is even simpler: the coefficient of $q^N$ depends only on the first $N$ product factors. Thus each $\\tau(n)$ is an unambiguously defined integer. The opening terms are\n\\begin{align}\n\\Delta(q)={}&q-24q^2+252q^3-1472q^4+4830q^5-6048q^6\\notag\\\\\n&-16744q^7+84480q^8-113643q^9-115920q^{10}+\\cdots.\n\\end{align}\nThe signs already change repeatedly. Nonvanishing cannot be inferred from positivity of the defining product, since positivity of a function at a real argument does not require positivity, or nonvanishing, of its Taylor coefficients.","\\begin{conjecture}[Lehmer]\nFor every integer $n\\ge1$, one has $\\tau(n)\\ne0$.\n\\end{conjecture}","The conjecture concerns Ramanujan's tau function. It is distinct from the conjecture bearing Lehmer's name about Mahler measure and from questions about Euler's totient. Throughout this book there is one fixed weight-$12$ cusp form, one fixed coefficient sequence and one fixed universal target. Auxiliary sequences introduced later will be explicitly named and will not silently replace $\\tau$.","Lehmer's 1947 paper records the multiplicative and prime-power structure that makes the prime version of the problem decisive \\cite{Lehmer}. Modern verified ranges are extraordinarily large. For example, Derickx, van Hoeij and Zeng proved nonvanishing for\n\\[\n0<n<816212624008487344127999\n\\]\nin their 2014 source \\cite{DHZ}. This is a source-certified historical range. It is not a range recomputed for this edition, and it is not asserted here to be the latest record. Its logical role is still finite: no finite endpoint reaches every positive integer.","\\section{Upper bounds and lower bounds}\nDeligne's theorem gives\n\\begin{equation}\\label{eq:deligne}\n|\\tau(p)|\\le2p^{11/2}\n\\end{equation}\nat a prime $p$ \\cite{DeligneWeil}. The exponent $11/2$ comes from weight $12$: the associated Frobenius eigenvalues have weight $11$. An upper bound confines a possible coefficient to an interval. It does not remove zero from that interval.","Integrality makes a residue argument useful. If $x\\in\\Z$, $|x|<L$ and $L\\mid x$, then $x=0$. If instead a nonzero residue modulo $L$ is established, then $x\\ne0$ without any size estimate. At primes $p\\ge5$, \\eqref{eq:deligne} yields\n\\[\n2p^{11/2}<p^6.\n\\]\nTherefore\n\\begin{equation}\\label{eq:p6}\n\\tau(p)=0\\quad\\Longleftrightarrow\\quad p^6\\mid\\tau(p),\\qquad p\\ge5.\n\\end{equation}\nThis is an exact criterion. It replaces equality to zero by a divisibility condition with a modulus depending on $p$. It does not show that the divisibility condition fails.","If instead a nonzero residue modulo $L$ is established, then $x\\ne0$ without any size estimate. At primes $p\\ge5$, \\eqref{eq:deligne} yields\n\\[\n2p^{11/2}<p^6.\n\\]\nTherefore\n\\begin{equation}\\label{eq:p6}\n\\tau(p)=0\\quad\\Longleftrightarrow\\quad p^6\\mid\\tau(p),\\qquad p\\ge5.\n\\end{equation}\nThis is an exact criterion. It replaces equality to zero by a divisibility condition with a modulus depending on $p$. It does not show that the divisibility condition fails.","For a nonzero actual coefficient, the same comparison gives $v_p(\\tau(p))\\le5$. A hypothetical assertion $v_p(\\tau(p))\\le3$ would be stronger than needed, and hence sufficient, but requires its own proof. One must not obtain such a valuation bound merely by rewriting the coefficient as another exact sum.","rticular coefficient survives. An exact expansion can evaluate it. A congruence can show it is nonzero modulo a specified integer. A sign certificate can rewrite it as a sum of one weak sign with at least one nonzero term. A valuation certificate can identify a uniquely least-valued summand in a nonarchimedean field. Each method has two parts: an identity connecting the certificate to the actual $\\tau(p)$, and an independently proved property that forces the identity's value away from zero.","\\begin{openproblem}[The exact prime obligation]\nProve $\\tau(p)\\ne0$ for every prime $p$, using information about the actual fixed form $\\Delta$ that does not already assume this conclusion. A verified actual prime with $\\tau(p)=0$ would instead disprove the conjecture.\n\\end{openproblem}","\\chapter{Exact coefficient engines}\\label{ch:engines}\n\\section{The product and a logarithmic recurrence}\nLet $\\sigma_j(n)=\\sum_{d\\mid n}d^j$ and write $\\sigma=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","oefficient engines}\\label{ch:engines}\n\\section{The product and a logarithmic recurrence}\nLet $\\sigma_j(n)=\\sum_{d\\mid n}d^j$ and write $\\sigma=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","recurrence}\nLet $\\sigma_j(n)=\\sum_{d\\mid n}d^j$ and write $\\sigma=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","et $\\sigma_j(n)=\\sum_{d\\mid n}d^j$ and write $\\sigma=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","and write $\\sigma=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","a=\\sigma_1$. Logarithmic differentiation of the product gives\n\\begin{equation}\\label{eq:logdelta}\n\\DQ\\Delta=\\left(1-24\\sum_{m\\ge1}\\sigma(m)q^m\\right)\\Delta,\n\\qquad \\DQ=q\\frac{d}{dq}.\n\\end{equation}\nIndeed, the contribution from the factor $(1-q^r)^{24}$ is\n\\[\n-24\\frac{rq^r}{1-q^r}=-24\\sum_{k\\ge1}rq^{rk}.\n\\]\nCollecting the terms with $rk=m$ gives $-24\\sigma(m)$. The leading factor $q$ contributes $1$.","Comparing coefficients of $q^n$ in \\eqref{eq:logdelta}, for $n>1$, yields\n\\begin{equation}\\label{eq:logrec}\n(n-1)\\tau(n)=-24\\sum_{m=1}^{n-1}\\sigma(m)\\tau(n-m),\n\\qquad \\tau(1)=1.\n\\end{equation}\nThis recurrence uses only smaller indices. Division by $n-1$ is exact because the recurrence came from the integral product, although an implementation should still check divisibility rather than silently round a quotient. It is a convenient independent coefficient engine at a moderate range.","For example,\n\\[\n\\tau(2)=-24\\sigma(1)\\tau(1)=-24.\n\\]\nAt $n=3$, the right side is $-24(-24+3)=504$, and division by $2$ gives $252$. These tiny checks fix the normalization, especially the leading $q$ and the exponent $24$.","\\section{Jacobi's triangular series}\nJacobi's identity is\n\\begin{equation}\\label{eq:jacobi}\n\\prod_{m\\ge1}(1-q^m)^3\n=\\sum_{j\\ge0}(-1)^j(2j+1)q^{j(j+1)/2}.\n\\end{equation}\nRaising this series to the eighth power and multiplying by $q$ gives $\\Delta$. The sparse triangular exponents make it useful for coefficient calculations. The source check in \\cite{BCO} also uses this identity to establish the parity criterion discussed below. It is an exact formal series identity, not a numerical approximation.","The edition's independent replay compares a defining-product calculation and a logarithmic recurrence through $n=512$. The historical arithmetic branch used the triangular engine through $10000$ and a separate product check through $128$. The historical packet-tree branch carried its own product comparison through $512$. These are different bounded contracts, and their endpoints must not be merged into an invented universal verification.","\\section{Eisenstein coefficients}\nWith the standard constant-one normalization,\n\\begin{align}\nE_4(q)&=1+240\\sum_{n\\ge1}\\sigma_3(n)q^n,\\\\\nE_6(q)&=1-504\\sum_{n\\ge1}\\sigma_5(n)q^n,\n\\end{align}\nand\n\\begin{equation}\\label{eq:deltaeisen}\n\\Delta=\\frac{E_4^3-E_6^2}{1728}.\n\\end{equation}\nSince $E_8=E_4^2$, one may equivalently write $(E_4E_8-E_6^2)/1728$. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator $1728$ is exact, and an integer implementation can test the required divisibility before extracting a coefficient.","E_6(q)&=1-504\\sum_{n\\ge1}\\sigma_5(n)q^n,\n\\end{align}\nand\n\\begin{equation}\\label{eq:deltaeisen}\n\\Delta=\\frac{E_4^3-E_6^2}{1728}.\n\\end{equation}\nSince $E_8=E_4^2$, one may equivalently write $(E_4E_8-E_6^2)/1728$. This last form supplies an independent divisor-sum convolution for selected large coefficients in the valuation branch. The denominator $1728$ is exact, and an integer implementation can test the required divisibility before extracting a coefficient.","The first coefficient of $E_4^3$ is $720$, whereas that of $E_6^2$ is $-1008$. Their difference is $1728$, so the leading coefficient of the quotient is $1$. A constant-one Eisenstein series is essential here. Some modular-form texts use a scaled Eisenstein basis. The normalization must be translated before coefficient identities are used.","\\section{Parity from the actual product}\nModulo $2$, the Frobenius identity gives\n\\[\n\\prod_{m\\ge1}(1-q^m)^{24}\n\\equiv\\prod_{m\\ge1}(1-q^{8m})^3.\n\\]\nUsing \\eqref{eq:jacobi}, whose coefficients $2j+1$ are odd, gives\n\\[\n\\Delta(q)\\equiv\\sum_{j\\ge0}q^{1+4j(j+1)}\n=\\sum_{j\\ge0}q^{(2j+1)^2}\\pmod2.\n\\]\nDistinct nonnegative $j$ give distinct exponents. Therefore\n\\begin{equation}\\label{eq:parity}\n\\tau(n)\\text{ is odd}\\quad\\Longleftrightarrow\\quad n\\text{ is an odd square}.\n\\end{equation}\nIn particular, $\\tau(p)$ is even at every odd prime. This elementary global property will later make the primitive odd-block multiplier odd. It is useful arithmetic, but is consistent with a zero at an odd prime because zero is even.","1}(1-q^m)^{24}\n\\equiv\\prod_{m\\ge1}(1-q^{8m})^3.\n\\]\nUsing \\eqref{eq:jacobi}, whose coefficients $2j+1$ are odd, gives\n\\[\n\\Delta(q)\\equiv\\sum_{j\\ge0}q^{1+4j(j+1)}\n=\\sum_{j\\ge0}q^{(2j+1)^2}\\pmod2.\n\\]\nDistinct nonnegative $j$ give distinct exponents. Therefore\n\\begin{equation}\\label{eq:parity}\n\\tau(n)\\text{ is odd}\\quad\\Longleftrightarrow\\quad n\\text{ is an odd square}.\n\\end{equation}\nIn particular, $\\tau(p)$ is even at every odd prime. This elementary global property will later make the primitive odd-block multiplier odd. It is useful arithmetic, but is consistent with a zero at an odd prime because zero is even.","\\figplaceholder{coefficients512}{The exact coefficients through $512$ on a signed logarithmic scale. The plotted ordinate is $\\sgn(\\tau(n))\\log_{10}(1+|\\tau(n)|)$. Sign changes and large values are visible, while every plotted coefficient is nonzero. This finite plot does not establish the conjecture.}","\\chapter{Why the problem lives at primes}\\label{ch:prime}\n\\section{The Hecke contracts}\nThe normalized form $\\Delta$ is a Hecke eigenform. Its coefficients satisfy\n\\begin{align}\n\\tau(mn)&=\\tau(m)\\tau(n),&&\\gcd(m,n)=1,\\label{eq:mult}","\\chapter{Why the problem lives at primes}\\label{ch:prime}\n\\section{The Hecke contracts}\nThe normalized form $\\Delta$ is a Hecke eigenform. Its coefficients satisfy\n\\begin{align}\n\\tau(mn)&=\\tau(m)\\tau(n),&&\\gcd(m,n)=1,\\label{eq:mult}\\\\\n\\tau(p^{r+1})&=\\tau(p)\\tau(p^r)-p^{11}\\tau(p^{r-1}),&&r\\ge1.\\label{eq:hecke}\n\\end{align}\nThe exponent $11$ again records weight $12$. The constant term for the local recurrence is $\\tau(1)=1$, and the first parameter is $a=\\tau(p)$. These classical statements occur in Lehmer's original paper and in the modern source \\cite{Lehmer,BCO}.","\\tau(p^{r+1})&=\\tau(p)\\tau(p^r)-p^{11}\\tau(p^{r-1}),&&r\\ge1.\\label{eq:hecke}\n\\end{align}\nThe exponent $11$ again records weight $12$. The constant term for the local recurrence is $\\tau(1)=1$, and the first parameter is $a=\\tau(p)$. These classical statements occur in Lehmer's original paper and in the modern source \\cite{Lehmer,BCO}.","For a fixed prime put $T_r=\\tau(p^r)$ and $b=p^{11}$. The generating series is\n\\begin{equation}\\label{eq:heckegen}\n\\sum_{r\\ge0}T_rX^r=\\frac1{1-aX+bX^2}.\n\\end{equation}\nMultiplication by the denominator gives $T_0=1$, $T_1=a$ and the recurrence. Thus this is a formal identity over $\\Z[[X]]$. No analytic continuation in the prime index is involved.","The first local blocks are\n\\[\nT_2=a^2-b,\\qquad T_3=a^3-2ab,\\qquad\nT_4=a^4-3a^2b+b^2.\n\\]\nEven if $a\\ne0$, it is not formally obvious from these expressions that none vanishes. The reduction to primes needs a short arithmetic argument.","\\section{The two roots and a root of unity}\nLet $\\alpha,\\beta$ be the complex roots of\n\\[\nX^2-aX+p^{11}=0.\n\\]\nThey are nonzero because their product is $p^{11}$. They are distinct: equality would require $a^2=4p^{11}$, but the $p$-adic valuation of the right side is odd for odd $p$, and is $13$ for $p=2$.","Solving the recurrence gives\n\\begin{equation}\\label{eq:rootrec}\nT_r=\\frac{\\alpha^{r+1}-\\beta^{r+1}}{\\alpha-\\beta}\n=\\sum_{j=0}^r\\alpha^{r-j}\\beta^j.\n\\end{equation}\nIf $T_r=0$ for some $r\\ge1$, then $\\zeta=\\alpha/\\beta$ is a root of unity with $\\zeta^{r+1}=1$. Consequently\n\\begin{equation}\\label{eq:rootunity}\n\\frac{a^2}{p^{11}}\n=\\frac{(\\alpha+\\beta)^2}{\\alpha\\beta}\n=2+\\zeta+\\zeta^{-1}.\n\\end{equation}\nThe left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between $0$ and $4$. Thus\n\\[\n\\frac{a^2}{p^{11}}\\in\\{0,1,2,3,4\\}.\n\\]\nThis argument uses the root-of-unity consequence of a zero, rather than assuming that an arbitrary ratio of Hecke roots is a root of unity.","$, then $\\zeta=\\alpha/\\beta$ is a root of unity with $\\zeta^{r+1}=1$. Consequently\n\\begin{equation}\\label{eq:rootunity}\n\\frac{a^2}{p^{11}}\n=\\frac{(\\alpha+\\beta)^2}{\\alpha\\beta}\n=2+\\zeta+\\zeta^{-1}.\n\\end{equation}\nThe left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between $0$ and $4$. Thus\n\\[\n\\frac{a^2}{p^{11}}\\in\\{0,1,2,3,4\\}.\n\\]\nThis argument uses the root-of-unity consequence of a zero, rather than assuming that an arbitrary ratio of Hecke roots is a root of unity.","$\\zeta=\\alpha/\\beta$ is a root of unity with $\\zeta^{r+1}=1$. Consequently\n\\begin{equation}\\label{eq:rootunity}\n\\frac{a^2}{p^{11}}\n=\\frac{(\\alpha+\\beta)^2}{\\alpha\\beta}\n=2+\\zeta+\\zeta^{-1}.\n\\end{equation}\nThe left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between $0$ and $4$. Thus\n\\[\n\\frac{a^2}{p^{11}}\\in\\{0,1,2,3,4\\}.\n\\]\nThis argument uses the root-of-unity consequence of a zero, rather than assuming that an arbitrary ratio of Hecke roots is a root of unity.","ha/\\beta$ is a root of unity with $\\zeta^{r+1}=1$. Consequently\n\\begin{equation}\\label{eq:rootunity}\n\\frac{a^2}{p^{11}}\n=\\frac{(\\alpha+\\beta)^2}{\\alpha\\beta}\n=2+\\zeta+\\zeta^{-1}.\n\\end{equation}\nThe left side is rational. The right side is an algebraic integer, because roots of unity are algebraic integers. A rational algebraic integer is an integer. It is real and lies between $0$ and $4$. Thus\n\\[\n\\frac{a^2}{p^{11}}\\in\\{0,1,2,3,4\\}.\n\\]\nThis argument uses the root-of-unity consequence of a zero, rather than assuming that an arbitrary ratio of Hecke roots is a root of unity.","\\section{Excluding the nonzero integer cases}\nFor $p\\ge5$, every $c\\in\\{1,2,3,4\\}$ is prime to $p$. The equality $a^2=cp^{11}$ would give an odd $p$-adic valuation on a square. It is impossible. Hence the only possible value in \\eqref{eq:rootunity} is $0$, and $a=0$.","At $p=2,3$, the actual parameters are\n\\[\n\\tau(2)=-24,\\qquad\\tau(3)=252.\n\\]\nTheir quotients $a^2/p^{11}$ are not integers. Thus no prime power of $2$ or $3$ vanishes. We obtain the exact local criterion:\n\\begin{equation}\\label{eq:tzero}\n\\tau(p^r)=0\\quad\\Longleftrightarrow\\quad\n\\tau(p)=0\\text{ and }r\\text{ is odd}.\n\\end{equation}\nThe forward implication is what we just proved. For the reverse implication and the even case, set $a=0$ in the recurrence:\n\\[\nT_{2m}=(-p^{11})^m,\\qquad T_{2m+1}=0.\n\\]","ers are\n\\[\n\\tau(2)=-24,\\qquad\\tau(3)=252.\n\\]\nTheir quotients $a^2/p^{11}$ are not integers. Thus no prime power of $2$ or $3$ vanishes. We obtain the exact local criterion:\n\\begin{equation}\\label{eq:tzero}\n\\tau(p^r)=0\\quad\\Longleftrightarrow\\quad\n\\tau(p)=0\\text{ and }r\\text{ is odd}.\n\\end{equation}\nThe forward implication is what we just proved. For the reverse implication and the even case, set $a=0$ in the recurrence:\n\\[\nT_{2m}=(-p^{11})^m,\\qquad T_{2m+1}=0.\n\\]","uad\\tau(3)=252.\n\\]\nTheir quotients $a^2/p^{11}$ are not integers. Thus no prime power of $2$ or $3$ vanishes. We obtain the exact local criterion:\n\\begin{equation}\\label{eq:tzero}\n\\tau(p^r)=0\\quad\\Longleftrightarrow\\quad\n\\tau(p)=0\\text{ and }r\\text{ is odd}.\n\\end{equation}\nThe forward implication is what we just proved. For the reverse implication and the even case, set $a=0$ in the recurrence:\n\\[\nT_{2m}=(-p^{11})^m,\\qquad T_{2m+1}=0.\n\\]","\\begin{theorem}[Prime reduction]\\label{thm:primereduction}\nFor every positive integer $n$,\n\\[\n\\begin{aligned}\n\\tau(n)=0\\quad\\Longleftrightarrow\\quad &\n\\text{some prime }p\\text{ occurs to an odd exponent in }n","\\begin{theorem}[Prime reduction]\\label{thm:primereduction}\nFor every positive integer $n$,\n\\[\n\\begin{aligned}\n\\tau(n)=0\\quad\\Longleftrightarrow\\quad &\n\\text{some prime }p\\text{ occurs to an odd exponent in }n\\\\\n&\\text{and }\\tau(p)=0.\n\\end{aligned}\n\\]\nIn particular, Lehmer's conjecture is equivalent to nonvanishing at every prime.\n\\end{theorem}\n\\begin{proof}\nFactor $n=\\prod_p p^{r_p}$. Multiplicativity expresses $\\tau(n)$ as a finite product of local blocks. An integral product is zero exactly when one factor is zero. Apply \\eqref{eq:tzero} to each factor.\n\\end{proof}","&\\text{and }\\tau(p)=0.\n\\end{aligned}\n\\]\nIn particular, Lehmer's conjecture is equivalent to nonvanishing at every prime.\n\\end{theorem}\n\\begin{proof}\nFactor $n=\\prod_p p^{r_p}$. Multiplicativity expresses $\\tau(n)$ as a finite product of local blocks. An integral product is zero exactly when one factor is zero. Apply \\eqref{eq:tzero} to each factor.\n\\end{proof}","One immediate corollary is that $\\tau(s^2)\\ne0$ for every $s$. This square theorem does not rely on the truth of Lehmer at primes. A possible prime zero would propagate only through odd exponents. Later the primitive residual will have the same zero criterion but a different proof of its square nonvanishing.","\\figplaceholder{prime_bound}{Actual prime coefficients through $512$, normalized by $2p^{11/2}$. Deligne's bound confines the values to $[-1,1]$. The bound is universal, whereas these plotted coefficients form a finite set. The permitted interval includes zero.}","\\section{A local parameter is not the global form}\nThe recurrence makes sense for every integer $a$. Setting $a=0$ gives a consistent sequence at one prime. Thus the recurrence alone cannot disallow a zero parameter. The actual modular form chooses its parameters through global geometry and modularity. A proof of Lehmer must use some part of that actual structure to forbid the parameter $a=0$ at every prime.","This distinction will recur in the local filtered-module construction and in the global arithmetic countermodel. They test whether the properties retained by a proposed argument already distinguish the actual $\\Delta$. When they admit zero parameters, they expose a missing premise rather than produce a counterexample to the conjecture.","\\chapter{The differential residual that survives extraction}\\label{ch:differential}\n\\section{A divisor convolution with the original cusp term}\nDefine the fourth divisor moment\n\\[\nS_4(n)=\\sum_{k=1}^{n-1}k^4\\sigma(k)\\sigma(n-k).\n\\]\nIt is a positive integer when $n>1$. The weights $k^4$ and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient:\n\\begin{equation}\\label{eq:s4general}\n840S_4(n)=n^4\\bigl[50\\sigma_3(n)+(35-84n)\\sigma(n)\\bigr]-\\tau(n).\n\\end{equation}\nThus the difference between two explicit divisor expressions is exactly $\\tau(n)$. Positivity of $S_4(n)$ does not imply that this difference is nonzero.","l that survives extraction}\\label{ch:differential}\n\\section{A divisor convolution with the original cusp term}\nDefine the fourth divisor moment\n\\[\nS_4(n)=\\sum_{k=1}^{n-1}k^4\\sigma(k)\\sigma(n-k).\n\\]\nIt is a positive integer when $n>1$. The weights $k^4$ and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient:\n\\begin{equation}\\label{eq:s4general}\n840S_4(n)=n^4\\bigl[50\\sigma_3(n)+(35-84n)\\sigma(n)\\bigr]-\\tau(n).\n\\end{equation}\nThus the difference between two explicit divisor expressions is exactly $\\tau(n)$. Positivity of $S_4(n)$ does not imply that this difference is nonzero.","igma(k)\\sigma(n-k).\n\\]\nIt is a positive integer when $n>1$. The weights $k^4$ and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient:\n\\begin{equation}\\label{eq:s4general}\n840S_4(n)=n^4\\bigl[50\\sigma_3(n)+(35-84n)\\sigma(n)\\bigr]-\\tau(n).\n\\end{equation}\nThus the difference between two explicit divisor expressions is exactly $\\tau(n)$. Positivity of $S_4(n)$ does not imply that this difference is nonzero.","is a positive integer when $n>1$. The weights $k^4$ and the two divisor sums suggest a route through elementary arithmetic. The exact identity, however, retains the cusp coefficient:\n\\begin{equation}\\label{eq:s4general}\n840S_4(n)=n^4\\bigl[50\\sigma_3(n)+(35-84n)\\sigma(n)\\bigr]-\\tau(n).\n\\end{equation}\nThus the difference between two explicit divisor expressions is exactly $\\tau(n)$. Positivity of $S_4(n)$ does not imply that this difference is nonzero.","At a prime $p$, $\\sigma(p)=1+p$ and $\\sigma_3(p)=1+p^3$. Put\n\\[\nC(p)=85-49p-84p^2+50p^3.\n\\]\nEquation \\eqref{eq:s4general} becomes\n\\begin{equation}\\label{eq:s4prime}\n\\tau(p)=p^4C(p)-840S_4(p).\n\\end{equation}\nFor $p>7$, the coefficient $840$ is a $p$-adic unit. Reducing modulo $p^4$ gives\n\\begin{equation}\\label{eq:p4equiv}\np^4\\mid S_4(p)\\quad\\Longleftrightarrow\\quad p^4\\mid\\tau(p).\n\\end{equation}\nThe proposed assertion that $p^4$ never divides $S_4(p)$ would therefore assert a uniform bound $v_p(\\tau(p))\\le3$. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.","1+p$ and $\\sigma_3(p)=1+p^3$. Put\n\\[\nC(p)=85-49p-84p^2+50p^3.\n\\]\nEquation \\eqref{eq:s4general} becomes\n\\begin{equation}\\label{eq:s4prime}\n\\tau(p)=p^4C(p)-840S_4(p).\n\\end{equation}\nFor $p>7$, the coefficient $840$ is a $p$-adic unit. Reducing modulo $p^4$ gives\n\\begin{equation}\\label{eq:p4equiv}\np^4\\mid S_4(p)\\quad\\Longleftrightarrow\\quad p^4\\mid\\tau(p).\n\\end{equation}\nThe proposed assertion that $p^4$ never divides $S_4(p)$ would therefore assert a uniform bound $v_p(\\tau(p))\\le3$. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.",")=1+p^3$. Put\n\\[\nC(p)=85-49p-84p^2+50p^3.\n\\]\nEquation \\eqref{eq:s4general} becomes\n\\begin{equation}\\label{eq:s4prime}\n\\tau(p)=p^4C(p)-840S_4(p).\n\\end{equation}\nFor $p>7$, the coefficient $840$ is a $p$-adic unit. Reducing modulo $p^4$ gives\n\\begin{equation}\\label{eq:p4equiv}\np^4\\mid S_4(p)\\quad\\Longleftrightarrow\\quad p^4\\mid\\tau(p).\n\\end{equation}\nThe proposed assertion that $p^4$ never divides $S_4(p)$ would therefore assert a uniform bound $v_p(\\tau(p))\\le3$. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.","ation \\eqref{eq:s4general} becomes\n\\begin{equation}\\label{eq:s4prime}\n\\tau(p)=p^4C(p)-840S_4(p).\n\\end{equation}\nFor $p>7$, the coefficient $840$ is a $p$-adic unit. Reducing modulo $p^4$ gives\n\\begin{equation}\\label{eq:p4equiv}\np^4\\mid S_4(p)\\quad\\Longleftrightarrow\\quad p^4\\mid\\tau(p).\n\\end{equation}\nThe proposed assertion that $p^4$ never divides $S_4(p)$ would therefore assert a uniform bound $v_p(\\tau(p))\\le3$. It is a sufficient strengthening of Lehmer, not a consequence of the displayed convolution.","\\section{Deriving the identity in the differential ring}\nSet\n\\[\nA(q)=\\sum_{n\\ge1}\\sigma(n)q^n,\\qquad\nB(q)=\\sum_{n\\ge1}\\sigma_3(n)q^n.\n\\]\nThen $E_2=1-24A$ and $E_4=1+240B$. The Ramanujan differential relations are\n\\begin{align}\n\\DQ E_2&=(E_2^2-E_4)/12,","\\section{Deriving the identity in the differential ring}\nSet\n\\[\nA(q)=\\sum_{n\\ge1}\\sigma(n)q^n,\\qquad\nB(q)=\\sum_{n\\ge1}\\sigma_3(n)q^n.\n\\]\nThen $E_2=1-24A$ and $E_4=1+240B$. The Ramanujan differential relations are\n\\begin{align}\n\\DQ E_2&=(E_2^2-E_4)/12,\\\\\n\\DQ E_4&=(E_2E_4-E_6)/3,\\\\\n\\DQ E_6&=(E_2E_6-E_4^2)/2.\n\\end{align}\nThese normalized relations are recorded in \\cite{Fonseca}, and the constant-one formula for $\\Delta$ is checked in \\cite{Stein}. They define a derivation on the polynomial ring $\\Q[E_2,E_4,E_6]$. Repeated differentiation and collection of terms yield\n\\begin{align}\nE_2\\DQ^2E_2&=4\\DQ^3E_2+\\frac3{10}\\DQ^2E_4,\\label{eq:d2}","\\DQ E_6&=(E_2E_6-E_4^2)/2.\n\\end{align}\nThese normalized relations are recorded in \\cite{Fonseca}, and the constant-one formula for $\\Delta$ is checked in \\cite{Stein}. They define a derivation on the polynomial ring $\\Q[E_2,E_4,E_6]$. Repeated differentiation and collection of terms yield\n\\begin{align}\nE_2\\DQ^2E_2&=4\\DQ^3E_2+\\frac3{10}\\DQ^2E_4,\\label{eq:d2}","\\DQ E_6&=(E_2E_6-E_4^2)/2.\n\\end{align}\nThese normalized relations are recorded in \\cite{Fonseca}, and the constant-one formula for $\\Delta$ is checked in \\cite{Stein}. They define a derivation on the polynomial ring $\\Q[E_2,E_4,E_6]$. Repeated differentiation and collection of terms yield\n\\begin{align}\nE_2\\DQ^2E_2&=4\\DQ^3E_2+\\frac3{10}\\DQ^2E_4,\\label{eq:d2}\\\\\nE_2\\DQ^4E_2&=\\frac{12}{5}\\DQ^5E_2+\\frac17\\DQ^4E_4-\\frac{24}{35}\\Delta.\\label{eq:d4}\n\\end{align}\nThese are polynomial identities in the differential ring. They can be verified by expanding each derivative using the three relations and checking a zero rational remainder. The cusp term in \\eqref{eq:d4} is essential.","Substituting $E_2=1-24A$, $E_4=1+240B$ into \\eqref{eq:d4} gives\n\\[\n-24\\DQ^4A+576A\\DQ^4A\n=-\\frac{288}{5}\\DQ^5A+\\frac{240}{7}\\DQ^4B-\\frac{24}{35}\\Delta.\n\\]\nMultiplication by $-35/24$ and rearrangement gives\n\\begin{equation}\\label{eq:projector}\n\\Delta=50\\DQ^4B+35\\DQ^4A-84\\DQ^5A-840A\\DQ^4A.\n\\end{equation}\nThe coefficient of $A\\DQ^4A$ at $q^n$ is precisely $S_4(n)$ after the summation variable is reversed if necessary. Coefficient extraction proves \\eqref{eq:s4general}.","For completeness, $\\Delta=(E_4^3-E_6^2)/1728$ has the expected product normalization. The differential relations imply $\\DQ\\Delta=E_2\\Delta$. The product in Chapter~\\ref{ch:engines} satisfies the same equation by logarithmic differentiation. Both begin with $q$. Coefficient recursion therefore identifies them as formal power series. This avoids treating a matching first coefficient as sufficient on its own.","\\section{The exact forbidden residue}\nCombine \\eqref{eq:p6} and \\eqref{eq:s4prime}. For $p>7$,\n\\begin{equation}\\label{eq:forbiddenconvolution}\n\\tau(p)=0\\quad\\Longleftrightarrow\\quad\n840S_4(p)\\equiv p^4C(p)\\pmod{p^6}.\n\\end{equation}\nIf $p^4\\mid S_4(p)$, division by $p^4$ is legitimate as an integer operation, and the same condition is\n\\[\n840\\frac{S_4(p)}{p^4}\\equiv85-49p\\pmod{p^2}.\n\\]\nThe terms $-84p^2$ and $50p^3$ have disappeared modulo $p^2$. This is a compact description of the remaining residue problem. It is not a contradiction. A successful proof through this formula must establish that the displayed residue is never attained by the actual convolution.","e exact forbidden residue}\nCombine \\eqref{eq:p6} and \\eqref{eq:s4prime}. For $p>7$,\n\\begin{equation}\\label{eq:forbiddenconvolution}\n\\tau(p)=0\\quad\\Longleftrightarrow\\quad\n840S_4(p)\\equiv p^4C(p)\\pmod{p^6}.\n\\end{equation}\nIf $p^4\\mid S_4(p)$, division by $p^4$ is legitimate as an integer operation, and the same condition is\n\\[\n840\\frac{S_4(p)}{p^4}\\equiv85-49p\\pmod{p^2}.\n\\]\nThe terms $-84p^2$ and $50p^3$ have disappeared modulo $p^2$. This is a compact description of the remaining residue problem. It is not a contradiction. A successful proof through this formula must establish that the displayed residue is never attained by the actual convolution.","ne \\eqref{eq:p6} and \\eqref{eq:s4prime}. For $p>7$,\n\\begin{equation}\\label{eq:forbiddenconvolution}\n\\tau(p)=0\\quad\\Longleftrightarrow\\quad\n840S_4(p)\\equiv p^4C(p)\\pmod{p^6}.\n\\end{equation}\nIf $p^4\\mid S_4(p)$, division by $p^4$ is legitimate as an integer operation, and the same condition is\n\\[\n840\\frac{S_4(p)}{p^4}\\equiv85-49p\\pmod{p^2}.\n\\]\nThe terms $-84p^2$ and $50p^3$ have disappeared modulo $p^2$. This is a compact description of the remaining residue problem. It is not a contradiction. A successful proof through this formula must establish that the displayed residue is never attained by the actual convolution.","At $p=2411$, retained exact values are\n\\[\n\\tau(p)=4542041100095889012,\\qquad\nS_4(p)=28168797627693083506845.\n\\]\nBoth have $p$-adic valuation $1$. The normalized residues are\n\\[\n\\frac{\\tau(p)}p\\equiv150\\pmod p,\\qquad\n\\frac{S_4(p)}p\\equiv947\\pmod p.\n\\]\nThey satisfy $-840\\cdot947\\equiv150\\pmod p$. This example shows that substantial cancellation can coexist with a clean exact identity. It neither refutes the stronger $p^4$ bound nor supplies it.","ed exact values are\n\\[\n\\tau(p)=4542041100095889012,\\qquad\nS_4(p)=28168797627693083506845.\n\\]\nBoth have $p$-adic valuation $1$. The normalized residues are\n\\[\n\\frac{\\tau(p)}p\\equiv150\\pmod p,\\qquad\n\\frac{S_4(p)}p\\equiv947\\pmod p.\n\\]\nThey satisfy $-840\\cdot947\\equiv150\\pmod p$. This example shows that substantial cancellation can coexist with a clean exact identity. It neither refutes the stronger $p^4$ bound nor supplies it.","\\section{Why a projector does not finish the proof}\nA differential projector can remove Eisenstein components and isolate a cusp component. That is a structural accomplishment: it explains where the coefficient sits in the algebra. But a projector selecting $\\Delta$ also selects its unknown prime coefficient. The coefficient $-1/840$ in \\eqref{eq:s4general} is a $p$-adic unit for $p>7$. There is no automatic extra factor of $p$ left behind.","not finish the proof}\nA differential projector can remove Eisenstein components and isolate a cusp component. That is a structural accomplishment: it explains where the coefficient sits in the algebra. But a projector selecting $\\Delta$ also selects its unknown prime coefficient. The coefficient $-1/840$ in \\eqref{eq:s4general} is a $p$-adic unit for $p>7$. There is no automatic extra factor of $p$ left behind.","\\chapter{Trace moments and a unit triangular inversion}\\label{ch:trace}\n\\section{The Hurwitz-moment description}\nAnother exact route uses Hurwitz class numbers. Let $H(D)$ denote the conventional weighted Hurwitz class number for a positive discriminant parameter $D$, with the standard exceptional automorphism weights. At a prime $p$ define\n\\begin{equation}\ns_{2j}(p)=\\frac12\\sum_{|t|<2\\sqrt p}t^{2j}H(4p-t^2).\n\\end{equation}\nHere the symbol $s_{2j}(p)$ is a trace moment and is unrelated to the coordinate power sums $s_j(v)$ introduced later. The argument of $H$ is positive in the displayed range. The normalization, including the factor $1/2$, must be kept fixed when a trace formula is imported.","inversion}\\label{ch:trace}\n\\section{The Hurwitz-moment description}\nAnother exact route uses Hurwitz class numbers. Let $H(D)$ denote the conventional weighted Hurwitz class number for a positive discriminant parameter $D$, with the standard exceptional automorphism weights. At a prime $p$ define\n\\begin{equation}\ns_{2j}(p)=\\frac12\\sum_{|t|<2\\sqrt p}t^{2j}H(4p-t^2).\n\\end{equation}\nHere the symbol $s_{2j}(p)$ is a trace moment and is unrelated to the coordinate power sums $s_j(v)$ introduced later. The argument of $H$ is positive in the displayed range. The normalization, including the factor $1/2$, must be kept fixed when a trace formula is imported.","exact route uses Hurwitz class numbers. Let $H(D)$ denote the conventional weighted Hurwitz class number for a positive discriminant parameter $D$, with the standard exceptional automorphism weights. At a prime $p$ define\n\\begin{equation}\ns_{2j}(p)=\\frac12\\sum_{|t|<2\\sqrt p}t^{2j}H(4p-t^2).\n\\end{equation}\nHere the symbol $s_{2j}(p)$ is a trace moment and is unrelated to the coordinate power sums $s_j(v)$ introduced later. The argument of $H$ is positive in the displayed range. The normalization, including the factor $1/2$, must be kept fixed when a trace formula is imported.","$H(D)$ denote the conventional weighted Hurwitz class number for a positive discriminant parameter $D$, with the standard exceptional automorphism weights. At a prime $p$ define\n\\begin{equation}\ns_{2j}(p)=\\frac12\\sum_{|t|<2\\sqrt p}t^{2j}H(4p-t^2).\n\\end{equation}\nHere the symbol $s_{2j}(p)$ is a trace moment and is unrelated to the coordinate power sums $s_j(v)$ introduced later. The argument of $H$ is positive in the displayed range. The normalization, including the factor $1/2$, must be kept fixed when a trace formula is imported.","criminant parameter $D$, with the standard exceptional automorphism weights. At a prime $p$ define\n\\begin{equation}\ns_{2j}(p)=\\frac12\\sum_{|t|<2\\sqrt p}t^{2j}H(4p-t^2).\n\\end{equation}\nHere the symbol $s_{2j}(p)$ is a trace moment and is unrelated to the coordinate power sums $s_j(v)$ introduced later. The argument of $H$ is positive in the displayed range. The normalization, including the factor $1/2$, must be kept fixed when a trace formula is imported.","The Eichler--Selberg trace identities give, for $r\\ge1$,\n\\begin{equation}\\label{eq:traceidentity}\n\\sum_{j=0}^r(-1)^{r-j}\\binom{r+j}{r-j}p^{r-j}s_{2j}(p)\n=-1-\\Tr T_{2r+2}(p),\n\\end{equation}\ntogether with $s_0(p)=p$. Here $T_k(p)$ is the Hecke operator at $p$ on $S_k(SL_2(\\Z))$. It is distinct from the scalar recurrence notation $T_r=\\tau(p^r)$ at a fixed prime. The trace normalization and Catalan inversion are checked against \\cite{LRTrace}, Sections 4--6. The cusp spaces in weights $4,6,8,10$ are zero. The weight-$12$ cusp space is spanned by $\\Delta$, so its Hecke trace is $\\tau(p)$.","identity}\n\\sum_{j=0}^r(-1)^{r-j}\\binom{r+j}{r-j}p^{r-j}s_{2j}(p)\n=-1-\\Tr T_{2r+2}(p),\n\\end{equation}\ntogether with $s_0(p)=p$. Here $T_k(p)$ is the Hecke operator at $p$ on $S_k(SL_2(\\Z))$. It is distinct from the scalar recurrence notation $T_r=\\tau(p^r)$ at a fixed prime. The trace normalization and Catalan inversion are checked against \\cite{LRTrace}, Sections 4--6. The cusp spaces in weights $4,6,8,10$ are zero. The weight-$12$ cusp space is spanned by $\\Delta$, so its Hecke trace is $\\tau(p)$.","\\binom{r+j}{r-j}p^{r-j}s_{2j}(p)\n=-1-\\Tr T_{2r+2}(p),\n\\end{equation}\ntogether with $s_0(p)=p$. Here $T_k(p)$ is the Hecke operator at $p$ on $S_k(SL_2(\\Z))$. It is distinct from the scalar recurrence notation $T_r=\\tau(p^r)$ at a fixed prime. The trace normalization and Catalan inversion are checked against \\cite{LRTrace}, Sections 4--6. The cusp spaces in weights $4,6,8,10$ are zero. The weight-$12$ cusp space is spanned by $\\Delta$, so its Hecke trace is $\\tau(p)$.","p)\n=-1-\\Tr T_{2r+2}(p),\n\\end{equation}\ntogether with $s_0(p)=p$. Here $T_k(p)$ is the Hecke operator at $p$ on $S_k(SL_2(\\Z))$. It is distinct from the scalar recurrence notation $T_r=\\tau(p^r)$ at a fixed prime. The trace normalization and Catalan inversion are checked against \\cite{LRTrace}, Sections 4--6. The cusp spaces in weights $4,6,8,10$ are zero. The weight-$12$ cusp space is spanned by $\\Delta$, so its Hecke trace is $\\tau(p)$.","ation}\ntogether with $s_0(p)=p$. Here $T_k(p)$ is the Hecke operator at $p$ on $S_k(SL_2(\\Z))$. It is distinct from the scalar recurrence notation $T_r=\\tau(p^r)$ at a fixed prime. The trace normalization and Catalan inversion are checked against \\cite{LRTrace}, Sections 4--6. The cusp spaces in weights $4,6,8,10$ are zero. The weight-$12$ cusp space is spanned by $\\Delta$, so its Hecke trace is $\\tau(p)$.","This leads to a lower triangular system for the moments. Its diagonal entries are all $1$. The inversion can recover the moments exactly, but the unit last diagonal also means that the last cusp trace survives with a unit coefficient.","\\section{The Catalan inverse}\nThe inverse matrix has coefficients\n\\begin{equation}\\label{eq:catalan}\n\\left[\\binom{2r}{r-j}-\\binom{2r}{r-j-1}\\right]p^{r-j},\n\\qquad0\\le j\\le r,\n\\end{equation}\nwhere a binomial coefficient outside its usual range is zero. The bracket is a ballot-number form of the Catalan triangle. It is an integer, and at $j=r$ it is $1$.","The resulting first moments are\n\\begin{align}\ns_2(p)&=p^2-1,\\\\\ns_4(p)&=2p^3-3p-1,\\\\\ns_6(p)&=5p^4-9p^2-5p-1,\\\\\ns_8(p)&=14p^5-28p^3-20p^2-7p-1,\\\\\ns_{10}(p)&=42p^6-90p^4-75p^3-35p^2-9p-1-\\tau(p).\n\\label{eq:s10}\n\\end{align}\nThe integer $42$ is the fifth Catalan number. It records the contribution from $s_0(p)=p$ after inversion. The lower-degree terms come from the constant $-1$ in the trace formula. The coefficient of $\\tau(p)$ is $-1$ because the last diagonal entry is $1$.","s_{10}(p)&=42p^6-90p^4-75p^3-35p^2-9p-1-\\tau(p).\n\\label{eq:s10}\n\\end{align}\nThe integer $42$ is the fifth Catalan number. It records the contribution from $s_0(p)=p$ after inversion. The lower-degree terms come from the constant $-1$ in the trace formula. The coefficient of $\\tau(p)$ is $-1$ because the last diagonal entry is $1$.","One can check the inversion without computing class numbers. Multiply the lower triangular matrices symbolically after factoring the appropriate powers of $p$. The binomial identities give the identity matrix. The retained exact test checks both products through size $32$, which is an implementation test for the universal binomial inversion rather than its proof.","\\section{A worked first row}\nAt $r=1$, \\eqref{eq:traceidentity} reads\n\\[\n-ps_0+s_2=-1.\n\\]\nSince $s_0=p$, this gives $s_2=p^2-1$. At $r=2$ it reads\n\\[\np^2s_0-3ps_2+s_4=-1.\n\\]\nSubstitution gives $s_4=2p^3-3p-1$. The same triangular procedure produces the next three rows. At the fifth row, the right side is $-1-\\tau(p)$ rather than $-1$. The inversion has not supplied a reason for that extra term to be nonzero.","This small calculation reveals the general mechanism. The early vanishing cusp spaces make the early moments polynomial in $p$. The first nonzero cusp space introduces the actual modular coefficient. That distinction is geometric, not a numerical defect of the inversion.","\\section{An exact reformulation can retain the hard part}\nEquation \\eqref{eq:s10} can be read as a formula for $\\tau(p)$ in terms of a weighted class-number moment. Setting its last term formally equal to zero is consistent with the triangular linear system. To exclude a zero one needs an additional arithmetic restriction on the actual Hurwitz sum, beyond the identities used to obtain the system.","For example, positivity of $H(D)$ makes $s_{10}(p)$ nonnegative. The polynomial on the right of \\eqref{eq:s10} with $\\tau(p)$ omitted is also compatible with nonnegativity for large $p$. A positivity statement of this general kind does not separate the two possible values. A useful inequality would have to be quantitatively strict in a way that excludes the exact polynomial value for every prime.","The historical check independently enumerated reduced positive definite binary quadratic forms for all primes through $499$ and for $p=2411$. There are $96$ prime rows in that scope. The resulting moments agree with the tau coefficients from the separate triangular engine. This gives exact evidence for the normalization and finite rows. It is not an exhaustive class-number theorem at all primes.","\\chapter{Five-core packets and signed integral geometry}\\label{ch:packets}\n\\section{From hooks to an integral shell}\nA partition is a finite weakly decreasing list of positive integers. Its Young diagram has one box for each unit in the sum of its parts. The hook of a box consists of that box, the boxes to its right and the boxes below it. A $5$-core is a partition with no hook length divisible by $5$. Equivalently, it has no hook of length $5$, the criterion relevant to the specialization below.","hapter{Five-core packets and signed integral geometry}\\label{ch:packets}\n\\section{From hooks to an integral shell}\nA partition is a finite weakly decreasing list of positive integers. Its Young diagram has one box for each unit in the sum of its parts. The hook of a box consists of that box, the boxes to its right and the boxes below it. A $5$-core is a partition with no hook length divisible by $5$. Equivalently, it has no hook of length $5$, the criterion relevant to the specialization below.","geometry}\\label{ch:packets}\n\\section{From hooks to an integral shell}\nA partition is a finite weakly decreasing list of positive integers. Its Young diagram has one box for each unit in the sum of its parts. The hook of a box consists of that box, the boxes to its right and the boxes below it. A $5$-core is a partition with no hook length divisible by $5$. Equivalently, it has no hook of length $5$, the criterion relevant to the specialization below.","The Nekrasov--Okounkov identity in Han's formulation is\n\\begin{equation}\\label{eq:hookidentity}\n\\prod_{m\\ge1}(1-q^m)^{z-1}\n=\\sum_\\lambda q^{|\\lambda|}\\prod_{h\\in H(\\lambda)}\\left(1-\\frac z{h^2}\\right).\n\\end{equation}\nAt $z=25$, a partition having a hook of length $5$ contributes zero. The surviving terms are the $5$-cores. Multiplication by $q$ shows that the coefficient $\\tau(n)$ is the sum of the hook products over $5$-cores of size $n-1$.","identity in Han's formulation is\n\\begin{equation}\\label{eq:hookidentity}\n\\prod_{m\\ge1}(1-q^m)^{z-1}\n=\\sum_\\lambda q^{|\\lambda|}\\prod_{h\\in H(\\lambda)}\\left(1-\\frac z{h^2}\\right).\n\\end{equation}\nAt $z=25$, a partition having a hook of length $5$ contributes zero. The surviving terms are the $5$-cores. Multiplication by $q$ shows that the coefficient $\\tau(n)$ is the sum of the hook products over $5$-cores of size $n-1$.","formulation is\n\\begin{equation}\\label{eq:hookidentity}\n\\prod_{m\\ge1}(1-q^m)^{z-1}\n=\\sum_\\lambda q^{|\\lambda|}\\prod_{h\\in H(\\lambda)}\\left(1-\\frac z{h^2}\\right).\n\\end{equation}\nAt $z=25$, a partition having a hook of length $5$ contributes zero. The surviving terms are the $5$-cores. Multiplication by $q$ shows that the coefficient $\\tau(n)$ is the sum of the hook products over $5$-cores of size $n-1$.","Han's V-coding identifies these cores with\n\\begin{equation}\\label{eq:shell}\n\\Vn=\\left\\{v\\in\\Z^5:\nv_i\\equiv i\\pmod5,\\quad\\sum_{i=0}^4v_i=0,\\quad\n\\sum_{i=0}^4v_i^2=10n\\right\\}.\n\\end{equation}\nThe shift by $1$ in the partition size is important: the coding formula is $|\\lambda|=\\|v\\|^2/10-1$. Define\n\\begin{equation}\\label{eq:weight}\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=\\frac{D(v)}{288}.\n\\end{equation}\nHere $288=1!2!3!4!$. The hook product becomes $W(v)$, with positive prefactor for $t=5$. Therefore\n\\begin{equation}\\label{eq:packettau}\n\\boxed{\\tau(n)=\\sum_{v\\in\\Vn}W(v).}\n\\end{equation}\nThis is a classical identity with translated conventions, authenticated against Han's Theorems 1.1 and 1.2 \\cite{Han}. The finite shell is a new arena for the same coefficient, not a new coefficient definition.","ntifies these cores with\n\\begin{equation}\\label{eq:shell}\n\\Vn=\\left\\{v\\in\\Z^5:\nv_i\\equiv i\\pmod5,\\quad\\sum_{i=0}^4v_i=0,\\quad\n\\sum_{i=0}^4v_i^2=10n\\right\\}.\n\\end{equation}\nThe shift by $1$ in the partition size is important: the coding formula is $|\\lambda|=\\|v\\|^2/10-1$. Define\n\\begin{equation}\\label{eq:weight}\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=\\frac{D(v)}{288}.\n\\end{equation}\nHere $288=1!2!3!4!$. The hook product becomes $W(v)$, with positive prefactor for $t=5$. Therefore\n\\begin{equation}\\label{eq:packettau}\n\\boxed{\\tau(n)=\\sum_{v\\in\\Vn}W(v).}\n\\end{equation}\nThis is a classical identity with translated conventions, authenticated against Han's Theorems 1.1 and 1.2 \\cite{Han}. The finite shell is a new arena for the same coefficient, not a new coefficient definition.","\\label{eq:shell}\n\\Vn=\\left\\{v\\in\\Z^5:\nv_i\\equiv i\\pmod5,\\quad\\sum_{i=0}^4v_i=0,\\quad\n\\sum_{i=0}^4v_i^2=10n\\right\\}.\n\\end{equation}\nThe shift by $1$ in the partition size is important: the coding formula is $|\\lambda|=\\|v\\|^2/10-1$. Define\n\\begin{equation}\\label{eq:weight}\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=\\frac{D(v)}{288}.\n\\end{equation}\nHere $288=1!2!3!4!$. The hook product becomes $W(v)$, with positive prefactor for $t=5$. Therefore\n\\begin{equation}\\label{eq:packettau}\n\\boxed{\\tau(n)=\\sum_{v\\in\\Vn}W(v).}\n\\end{equation}\nThis is a classical identity with translated conventions, authenticated against Han's Theorems 1.1 and 1.2 \\cite{Han}. The finite shell is a new arena for the same coefficient, not a new coefficient definition.",":\nv_i\\equiv i\\pmod5,\\quad\\sum_{i=0}^4v_i=0,\\quad\n\\sum_{i=0}^4v_i^2=10n\\right\\}.\n\\end{equation}\nThe shift by $1$ in the partition size is important: the coding formula is $|\\lambda|=\\|v\\|^2/10-1$. Define\n\\begin{equation}\\label{eq:weight}\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=\\frac{D(v)}{288}.\n\\end{equation}\nHere $288=1!2!3!4!$. The hook product becomes $W(v)$, with positive prefactor for $t=5$. Therefore\n\\begin{equation}\\label{eq:packettau}\n\\boxed{\\tau(n)=\\sum_{v\\in\\Vn}W(v).}\n\\end{equation}\nThis is a classical identity with translated conventions, authenticated against Han's Theorems 1.1 and 1.2 \\cite{Han}. The finite shell is a new arena for the same coefficient, not a new coefficient definition.","=0}^4v_i=0,\\quad\n\\sum_{i=0}^4v_i^2=10n\\right\\}.\n\\end{equation}\nThe shift by $1$ in the partition size is important: the coding formula is $|\\lambda|=\\|v\\|^2/10-1$. Define\n\\begin{equation}\\label{eq:weight}\nD(v)=\\prod_{0\\le i<j\\le4}(v_i-v_j),\\qquad W(v)=\\frac{D(v)}{288}.\n\\end{equation}\nHere $288=1!2!3!4!$. The hook product becomes $W(v)$, with positive prefactor for $t=5$. Therefore\n\\begin{equation}\\label{eq:packettau}\n\\boxed{\\tau(n)=\\sum_{v\\in\\Vn}W(v).}\n\\end{equation}\nThis is a classical identity with translated conventions, authenticated against Han's Theorems 1.1 and 1.2 \\cite{Han}. The finite shell is a new arena for the same coefficient, not a new coefficient definition.","\\figplaceholder{fivecore_hooks}{The hook specialization retains $5$-cores and annihilates partitions with a hook of length $5$. The displayed size-$6$ examples illustrate the exact mechanism behind the coefficient shift $|\\lambda|=n-1$.}","\\section{Every packet is nonzero and integral}\nDistinct coordinate residues guarantee $v_i\\ne v_j$ for $i\\ne j$. Hence $D(v)$ and $W(v)$ are nonzero. Their signs can differ from packet to packet.","Integrality follows from an integer-valued basis. For every integer $x$, including negative integers, the generalized binomial coefficient $\\binom{x}{j}$ is an integer. Its leading coefficient as a polynomial in $x$ is $1/j!$. Thus\n\\begin{equation}\n\\det\\left[\\binom{v_i}{j}\\right]_{0\\le i,j\\le4}\n=\\frac{\\prod_{i<j}(v_j-v_i)}{\\prod_{j=0}^4j!}.\n\\end{equation}\nThere are ten pairs, so reversing every difference changes the sign by $(-1)^{10}=1$. The determinant is $W(v)$ and is integral. This proof is independent of the fact that the total in \\eqref{eq:packettau} is an integer.","teger-valued basis. For every integer $x$, including negative integers, the generalized binomial coefficient $\\binom{x}{j}$ is an integer. Its leading coefficient as a polynomial in $x$ is $1/j!$. Thus\n\\begin{equation}\n\\det\\left[\\binom{v_i}{j}\\right]_{0\\le i,j\\le4}\n=\\frac{\\prod_{i<j}(v_j-v_i)}{\\prod_{j=0}^4j!}.\n\\end{equation}\nThere are ten pairs, so reversing every difference changes the sign by $(-1)^{10}=1$. The determinant is $W(v)$ and is integral. This proof is independent of the fact that the total in \\eqref{eq:packettau} is an integer.","ger $x$, including negative integers, the generalized binomial coefficient $\\binom{x}{j}$ is an integer. Its leading coefficient as a polynomial in $x$ is $1/j!$. Thus\n\\begin{equation}\n\\det\\left[\\binom{v_i}{j}\\right]_{0\\le i,j\\le4}\n=\\frac{\\prod_{i<j}(v_j-v_i)}{\\prod_{j=0}^4j!}.\n\\end{equation}\nThere are ten pairs, so reversing every difference changes the sign by $(-1)^{10}=1$. The determinant is $W(v)$ and is integral. This proof is independent of the fact that the total in \\eqref{eq:packettau} is an integer.","Modulo $5$, the denominator $288$ is a unit and every difference has the fixed residue $i-j$. Hence\n\\begin{equation}\\label{eq:packetmod5}\nW(v)\\equiv\\frac{\\prod_{i<j}(i-j)}{288}=1\\pmod5.\n\\end{equation}\nEvery individual contribution is therefore a $5$-adic unit. A total divisible by $5$ arises from cancellation of those units, including their multiplicities.","\\section{Counting packets and restricting a possible zero}\nLet $\\chifive$ be the quadratic character modulo $5$, extended by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nT","\\section{Counting packets and restricting a possible zero}\nLet $\\chifive$ be the quadratic character modulo $5$, extended by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but stil","\\section{Counting packets and restricting a possible zero}\nLet $\\chifive$ be the quadratic character modulo $5$, extended by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","icting a possible zero}\nLet $\\chifive$ be the quadratic character modulo $5$, extended by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.",", extended by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","d by zero at multiples of $5$. The classical $5$-core count \\cite{GKS}, Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","Theorem 4 and equation (5.3), gives\n\\begin{equation}\\label{eq:packetcount}\n|\\Vn|=\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","\\sum_{d\\mid n}\\chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","chifive(d)\\frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","frac nd.\n\\end{equation}\nAt a prime,\n\\[\n|\\mathcal V_p|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","|=p+\\chifive(p).\n\\]\nTogether with \\eqref{eq:packetmod5}, this gives\n\\begin{equation}\\label{eq:primemod5}\n\\tau(p)\\equiv p+\\chifive(p)\\pmod5.\n\\end{equation}\nFor $p\\ne5$, $\\chifive(p)\\equiv p^2\\pmod5$, so this is also the familiar congruence $\\tau(p)\\equiv p(1+p)\\pmod5$ \\cite{BCO}. It excludes zeros at primes congruent to $1,2,3$ modulo $5$. Since $\\tau(5)=4830\\ne0$, the unexcluded prime family is\n\\[\np\\equiv4\\pmod5.\n\\]\nThe restriction is substantial but still leaves infinitely many primes.","The formula for the unweighted count must not be confused with the weighted sum. At $p=1249$ there are $1250$ packets, whose count has $5$-adic valuation $4$. The weighted coefficient has valuation $3$. At $p=499$, the count has valuation $3$ while the weighted coefficient has valuation $4$. Weights that are all $1$ modulo $5$ recover the first residue, but not all higher divisibility.","\\section{A complete shell enumerator}\nThe shell is finite because every coordinate has absolute value at most $\\sqrt{10n}$. Enumeration can reduce its dimension without discarding any solution. Choose $v_0,v_1,v_2$ in their residue classes within the bound. The remaining coordinates $x=v_3$, $y=v_4$ must satisfy\n\\[\nx+y=S=-v_0-v_1-v_2,\\qquad\nx^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.\n\\]\nThen\n\\[\n(x-y)^2=2T-S^2.\n\\]\nThe right side must be a nonnegative square. For each permitted sign of its square root, the formulas $x=(S+(x-y))/2$ and $y=(S-(x-y))/2$ give the candidates. Their integrality and residues modulo $5$ are checked directly. This is a complete quadratic solution, rather than a sampling procedure.","$\\sqrt{10n}$. Enumeration can reduce its dimension without discarding any solution. Choose $v_0,v_1,v_2$ in their residue classes within the bound. The remaining coordinates $x=v_3$, $y=v_4$ must satisfy\n\\[\nx+y=S=-v_0-v_1-v_2,\\qquad\nx^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.\n\\]\nThen\n\\[\n(x-y)^2=2T-S^2.\n\\]\nThe right side must be a nonnegative square. For each permitted sign of its square root, the formulas $x=(S+(x-y))/2$ and $y=(S-(x-y))/2$ give the candidates. Their integrality and residues modulo $5$ are checked directly. This is a complete quadratic solution, rather than a sampling procedure.","ration can reduce its dimension without discarding any solution. Choose $v_0,v_1,v_2$ in their residue classes within the bound. The remaining coordinates $x=v_3$, $y=v_4$ must satisfy\n\\[\nx+y=S=-v_0-v_1-v_2,\\qquad\nx^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.\n\\]\nThen\n\\[\n(x-y)^2=2T-S^2.\n\\]\nThe right side must be a nonnegative square. For each permitted sign of its square root, the formulas $x=(S+(x-y))/2$ and $y=(S-(x-y))/2$ give the candidates. Their integrality and residues modulo $5$ are checked directly. This is a complete quadratic solution, rather than a sampling procedure.","0,v_1,v_2$ in their residue classes within the bound. The remaining coordinates $x=v_3$, $y=v_4$ must satisfy\n\\[\nx+y=S=-v_0-v_1-v_2,\\qquad\nx^2+y^2=T=10n-v_0^2-v_1^2-v_2^2.\n\\]\nThen\n\\[\n(x-y)^2=2T-S^2.\n\\]\nThe right side must be a nonnegative square. For each permitted sign of its square root, the formulas $x=(S+(x-y))/2$ and $y=(S-(x-y))/2$ give the candidates. Their integrality and residues modulo $5$ are checked directly. This is a complete quadratic solution, rather than a sampling procedure.","A second enumerator can loop through four coordinates and solve the fifth from the sum. Agreement between these independent organizations is a useful bounded control. The historical packet-tree verifier compared them through $n=32$. Shell sums were then compared with an independent tau engine at a larger finite scope.","\\figplaceholder{shell_coordinates}{The shell $\\mathcal V_2$ is a concrete intersection of a zero-sum hyperplane, an integral residue condition and a sphere. Coordinates are labeled by their residues modulo $5$. A shell point is a packet, with its separate nonzero integral Vandermonde weight.}","\\section{The index-prime unit obstruction}\nIf $p>20$ is prime and $v\\in\\mathcal V_p$, then for each pair\n\\[\n0<(v_i-v_j)^2\\le2(v_i^2+v_j^2)\\le20p<p^2.\n\\]\nThus $p$ divides none of the differences and does not divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","If $p>20$ is prime and $v\\in\\mathcal V_p$, then for each pair\n\\[\n0<(v_i-v_j)^2\\le2(v_i^2+v_j^2)\\le20p<p^2.\n\\]\nThus $p$ divides none of the differences and does not divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","[\n0<(v_i-v_j)^2\\le2(v_i^2+v_j^2)\\le20p<p^2.\n\\]\nThus $p$ divides none of the differences and does not divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","_j^2)\\le20p<p^2.\n\\]\nThus $p$ divides none of the differences and does not divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","p$ divides none of the differences and does not divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","divide $288$. We obtain\n\\begin{proposition}\\label{prop:indexunit}\nFor every prime $p>20$ and every $v\\in\\mathcal V_p$, one has $v_p(W(v))=0$.\n\\end{proposition}\nAll packets are units at the index prime. There is no increasing termwise $p$-adic depth automatically built into this representation. At $p=2411$, for example, all $2412$ packet weights are $2411$-adic units while their sum has valuation $1$. The divisibility belongs to the cancellation, not to an individually deeper term.","\\chapter{A genuine auxiliary polynomial and its local roots}\\label{ch:auxiliary}\n\\section{A polynomial at each fixed shell}\nWrite $v_i=i+5c_i$ for $v\\in\\Vn$. Define\n\\begin{equation}\\label{eq:pn}\nP_n(T)=\\sum_{v\\in\\Vn}\\prod_{i<j}\\bigl[(i-j)+T(c_i-c_j)\\bigr]\n=\\sum_{j=0}^{10}B_j(n)T^j.\n\\end{equation}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","d its local roots}\\label{ch:auxiliary}\n\\section{A polynomial at each fixed shell}\nWrite $v_i=i+5c_i$ for $v\\in\\Vn$. Define\n\\begin{equation}\\label{eq:pn}\nP_n(T)=\\sum_{v\\in\\Vn}\\prod_{i<j}\\bigl[(i-j)+T(c_i-c_j)\\bigr]\n=\\sum_{j=0}^{10}B_j(n)T^j.\n\\end{equation}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","ial at each fixed shell}\nWrite $v_i=i+5c_i$ for $v\\in\\Vn$. Define\n\\begin{equation}\\label{eq:pn}\nP_n(T)=\\sum_{v\\in\\Vn}\\prod_{i<j}\\bigl[(i-j)+T(c_i-c_j)\\bigr]\n=\\sum_{j=0}^{10}B_j(n)T^j.\n\\end{equation}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","el{eq:pn}\nP_n(T)=\\sum_{v\\in\\Vn}\\prod_{i<j}\\bigl[(i-j)+T(c_i-c_j)\\bigr]\n=\\sum_{j=0}^{10}B_j(n)T^j.\n\\end{equation}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","c_i-c_j)\\bigr]\n=\\sum_{j=0}^{10}B_j(n)T^j.\n\\end{equation}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","n}\nFor a fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","fixed $n$, this is an integral polynomial of degree at most ten. It is obtained by summing products of ten linear factors. Evaluation at $T=5$ returns the Vandermonde numerators:\n\\begin{equation}\\label{eq:pn5}\nP_n(5)=288\\tau(n),\\qquad B_0(n)=288|\\Vn|.\n\\end{equation}\nThe variable $T$ is auxiliary. Nothing in the construction says that $\\tau(p)$ is one analytic function of the prime index $p$ in a $p$-adic or real variable.","Normalize the coefficient valuations by\n\\[\n\\mu_n=\\min_{B_j(n)\\ne0}\\bigl(j+v_5(B_j(n))\\bigr),\\qquad\nG_n(U)=5^{-\\mu_n}P_n(5U).\n\\]\nEvery coefficient of $G_n$ is integral, and at least one is a unit. Because $288$ is a $5$-adic unit,\n\\begin{equation}\\label{eq:gnvaluation}\nv_5(\\tau(n))=\\mu_n+v_5(G_n(1)).\n\\end{equation}\nThis formula remains meaningful if $\\tau(n)=0$, with valuation $\\infty$. The normalization does not assume nonvanishing: its constant coefficient comes from a positive packet count on a nonempty shell.","The bounded factorization experiment found all $300$ aggregate polynomials at primes $7\\le p\\le2000$ irreducible over $\\Q$. This observation is unused in the universal arguments. Excluding $U=1$ throughout the prime family still requires further arithmetic restrictions on the moments $B_j(p)$.","\\section{A simple root at \\texorpdfstring{$p=499$}{p=499}}\nThe exact coefficients $B_0,\\ldots,B_{10}$ at $499$ are\n\\begin{align*}\n(&144000,-360000,1063040,28418240,-148872808,","\\section{A simple root at \\texorpdfstring{$p=499$}{p=499}}\nThe exact coefficients $B_0,\\ldots,B_{10}$ at $499$ are\n\\begin{align*}\n(&144000,-360000,1063040,28418240,-148872808,\\\\\n&209474504,2422451240,16445000,8107402848,\\\\\n&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference","&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe roo","&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in","&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $4","&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the pri","&-11893316304,-1160632800).\n\\end{align*}\nTheir weighted valuations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","luations are\n\\[\n(3,5,3,4,4,5,7,11,8,9,12).\n\\]\nHence $\\mu_{499}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","}=3$ and reduction gives\n\\[\n\\overline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","erline G_{499}(U)=2+3U^2=3(U-1)(U+1)\\quad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","ad\\text{in }\\mathbf F_5[U].\n\\]\nAt the relevant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","evant value $U=1$,\n\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","\\[\nG_{499}(1)\\equiv10\\pmod{25},\\qquad G'_{499}(1)\\equiv1\\pmod5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","od5.\n\\]\nSubstitute $U=1+5d$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","$. The integral Taylor identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","r identity gives\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","ves\n\\[\nG_{499}(1+5d)\\equiv10+5d\\pmod{25}.\n\\]\nThe root in this residue class is therefore $\\rho\\equiv16\\pmod{25}$. Hensel's lemma lifts it uniquely to $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","o $\\Z_5$. The divided difference $(G(U)-G(\\rho))/(U-\\rho)$ is a unit on $U\\equiv1\\pmod5$, since its reduction is $G'(1)\\ne0$. Thus\n\\[\nv_5(G_{499}(U))=v_5(U-\\rho)\n\\]\non that residue class. At $U=1$ the difference from $\\rho$ has valuation $1$, giving\n\\[\nv_5(\\tau(499))=3+1=4.\n\\]\nThe root has been located in the auxiliary variable. This proves a finite coefficient valuation, without suggesting that $499$ is near a root in the prime-index variable.","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coeffici","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor ide","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is o","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check","\\section{A double residual root with no lift}\nAt $p=16249$, the shell has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","l has $16250$ packets. The weighted coefficient valuations of $P_p(5U)$ are\n\\[\n(4,5,4,4,6,5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","5,7,8,9,9,11),\n\\]\nso $\\mu_p=4$. The reduction is\n\\[\n\\overline G_p(U)=3+U^2+U^3=(U-1)^2(U-2)\\pmod5.\n\\]\nThe residual root at $1$ is double. Exact coefficient reduction gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","gives\n\\[\nG_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","_p(1)\\equiv15\\pmod{25},\\qquad G'_p(1)\\equiv0\\pmod5.\n\\]\nFor every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","For every $y\\in\\Z_5$, polynomial Taylor expansion has an integral remainder:\n\\[\nG_p(1+5y)=G_p(1)+5yG'_p(1)+25y^2H(y),\\qquad H\\in\\Z_5[y].\n\\]\nConsequently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","equently\n\\begin{equation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","quation}\\label{eq:nolift}\nG_p(1+5y)\\equiv15\\pmod{25}\\quad\\text{for every }y\\in\\Z_5.\n\\end{equation}\nNo point in this entire residue class is a root. Its valuation is exactly $1$, and\n\\[\nv_5(\\tau(16249))=4+1=5.\n\\]\nThe exact coefficient is $133116215274958169206250$. The universal statement in \\eqref{eq:nolift} follows from the Taylor identity and the two residues. A finite test at $625$ choices of $y$ is only a secondary code check.","\\figplaceholder{auxiliary_roots}{Two exact auxiliary root certificates. At $499$, the simple residual root near $1$ lifts and sits one $5$-adic digit away from the evaluation point. At $16249$, the double residual root has no lift in the class $1\\bmod5$, by an integral Taylor congruence. These are roots in $U$, not roots in the prime index.}","\\section{What a degree bound does not control}\nThe comparison cases are\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n$p$&Packets&$\\mu_p$&$v_5(G_p(1))$&$v_5(\\tau(p))$","\\bottomrule\n\\end{tabular}\n\\end{center}\nThe degree-$10$ bound controls the number of roots over an algebraic closure, counted with multiplicity. It does not give a uniform bound for a value's valuation. Even the degree-one polynomial $U-(1+5^k)$ has valuation $k$ at $U=1$. A root can lie arbitrarily close to the evaluation point as the polynomial changes.","Individual packet polynomials have rational finite roots in $[-4,4]$, since each root is $-(i-j)/(c_i-c_j)$ with a nonzero integer denominator. The aggregate polynomial need not retain that root region. Exact values at $p=31$ are\n\\[\nP_{31}(10)=-56767541184<0,\\qquad\nP_{31}(11)=4645170845280>0.\n\\]\nIt has a real root between $10$ and $11$. Summing signed real-rooted factors does not preserve their root bounds.","\\chapter{Symmetry and the harmonic theta mechanism}\\label{ch:theta}\n\\section{An involution that preserves the weight}\nThe coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","theta mechanism}\\label{ch:theta}\n\\section{An involution that preserves the weight}\nThe coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","heta}\n\\section{An involution that preserves the weight}\nThe coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","ion that preserves the weight}\nThe coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","weight}\nThe coordinate labels are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","s are residue labels, so negation alone does not preserve the standard shell. The correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","correct involution is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","tion is\n\\begin{equation}\\label{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","abel{eq:j}\nJ(v)_i=-v_{(-i\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","\\bmod5)}.\n\\end{equation}\nIt preserves the residues, sum and norm. Its coordinate permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","permutation is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","is $(1\\ 4)(2\\ 3)$, which is even. Negation contributes $(-1)^{10}=1$ to the degree-$10$ Vandermonde. Hence\n\\[\nW(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","(Jv)=W(v).\n\\]\nWrite $s_m(v)=\\sum_i v_i^m$. Then $s_m(Jv)=(-1)^ms_m(v)$. In particular, every odd homogeneous symmetric polynomial $Q$ satisfies $Q(Jv)=-Q(v)$. Pairing by $J$ gives\n\\begin{equation}\\label{eq:oddnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0.\n\\end{equation}\nAt a fixed point of $J$, such a $Q$ is zero, so the argument covers both orbit sizes. Odd corrections alone do not change the sum of an entire $J$-orbit. Even corrections will be the useful source of transfers between orbits.","\\section{Why the Vandermonde is harmonic}\nWork on the four-dimensional Euclidean hyperplane\n\\[\nH=\\{v\\in\\R^5:\\sum_i v_i=0\\}.\n\\]\nThe Laplacian on $H$ commutes with coordinate permutations. Since $D$ is alternating, its Laplacian is alternating and has degree $8$. Every alternating polynomial on $H$ vanishes when $v_i=v_j$. The ten distinct linear factors $v_i-v_j$ therefore divide it. A degree-$8$ polynomial cannot be divisible by their degree-$10$ product unless it is zero. Thus\n\\[\n\\Delta_H D=0.\n\\]\nThis degree argument is useful because it does not require expanding the ten-factor product.","on the four-dimensional Euclidean hyperplane\n\\[\nH=\\{v\\in\\R^5:\\sum_i v_i=0\\}.\n\\]\nThe Laplacian on $H$ commutes with coordinate permutations. Since $D$ is alternating, its Laplacian is alternating and has degree $8$. Every alternating polynomial on $H$ vanishes when $v_i=v_j$. The ten distinct linear factors $v_i-v_j$ therefore divide it. A degree-$8$ polynomial cannot be divisible by their degree-$10$ product unless it is zero. Thus\n\\[\n\\Delta_H D=0.\n\\]\nThis degree argument is useful because it does not require expanding the ten-factor product.","rplane\n\\[\nH=\\{v\\in\\R^5:\\sum_i v_i=0\\}.\n\\]\nThe Laplacian on $H$ commutes with coordinate permutations. Since $D$ is alternating, its Laplacian is alternating and has degree $8$. Every alternating polynomial on $H$ vanishes when $v_i=v_j$. The ten distinct linear factors $v_i-v_j$ therefore divide it. A degree-$8$ polynomial cannot be divisible by their degree-$10$ product unless it is zero. Thus\n\\[\n\\Delta_H D=0.\n\\]\nThis degree argument is useful because it does not require expanding the ten-factor product.","More generally, if $Q$ is symmetric and $DQ$ is harmonic homogeneous of degree $10+d$, its weighted shell theta series is\n\\begin{equation}\\label{eq:thetaq}\n\\Theta_Q(q)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q(v)\\right)q^n.\n\\end{equation}\nThe harmonic theta theorem initially supplies a vector-valued modular form. Obtaining a scalar full-level cusp form requires the following lattice specialization, rather than an unexplained assertion that every shell theta series has level one.","\\section{The regular discriminant cosets}\nLet $A_4=H\\cap\\Z^5$ and $L=\\sqrt5 A_4$. Its discriminant module is\n\\[\nL^*/L\\cong A_4^*/5A_4,\n\\qquad |L^*/L|=5^5=3125.\n\\]\nA representative of $L^*$ can be written\n\\[\nx=\\frac1{\\sqrt5}(a_i+t/5)_{i=0}^4,\n\\qquad0\\le t<5,\\qquad\\sum_i a_i=-t.\n\\]\nIf two $a_i$ have the same residue modulo $5$, the transposition of those coordinates fixes the discriminant coset. An alternating theta component on that coset must equal its own negative and is zero.","If all five residues are distinct, they are $0,1,2,3,4$ in some order. Their sum is zero modulo $5$, so $\\sum_i a_i=-t$ forces $t=0$. These regular cosets form one free $S_5$-orbit of size $120$. The sign representation occurs exactly once in the corresponding permutation module. It does not occur in an irregular orbit, since an odd stabilizer kills the alternating part. Thus the alternating theta vector lies on a one-dimensional sign line.","For a regular representative, $\\sum_i a_i^2$ is divisible by $5$, since the squared residues sum to $30$. It is even, since $a_i^2\\equiv a_i\\pmod2$ and $\\sum_i a_i=0$. The quadratic exponent $\\sum_i a_i^2/10$ is integral. The Weil operator $T$ therefore acts as $1$ on the sign line. The Weil action commutes with $S_5$, so this line is invariant. The rank is even, and the representation descends to $SL_2(\\Z)$. On the line the relation $(ST)^3=S^2$ reduces to $S^3=S^2$, giving $S=1$.","r a regular representative, $\\sum_i a_i^2$ is divisible by $5$, since the squared residues sum to $30$. It is even, since $a_i^2\\equiv a_i\\pmod2$ and $\\sum_i a_i=0$. The quadratic exponent $\\sum_i a_i^2/10$ is integral. The Weil operator $T$ therefore acts as $1$ on the sign line. The Weil action commutes with $S_5$, so this line is invariant. The rank is even, and the representation descends to $SL_2(\\Z)$. On the line the relation $(ST)^3=S^2$ reduces to $S^3=S^2$, giving $S=1$.","_i^2$ is divisible by $5$, since the squared residues sum to $30$. It is even, since $a_i^2\\equiv a_i\\pmod2$ and $\\sum_i a_i=0$. The quadratic exponent $\\sum_i a_i^2/10$ is integral. The Weil operator $T$ therefore acts as $1$ on the sign line. The Weil action commutes with $S_5$, so this line is invariant. The rank is even, and the representation descends to $SL_2(\\Z)$. On the line the relation $(ST)^3=S^2$ reduces to $S^3=S^2$, giving $S=1$.","$, since the squared residues sum to $30$. It is even, since $a_i^2\\equiv a_i\\pmod2$ and $\\sum_i a_i=0$. The quadratic exponent $\\sum_i a_i^2/10$ is integral. The Weil operator $T$ therefore acts as $1$ on the sign line. The Weil action commutes with $S_5$, so this line is invariant. The rank is even, and the representation descends to $SL_2(\\Z)$. On the line the relation $(ST)^3=S^2$ reduces to $S^3=S^2$, giving $S=1$.","ed residues sum to $30$. It is even, since $a_i^2\\equiv a_i\\pmod2$ and $\\sum_i a_i=0$. The quadratic exponent $\\sum_i a_i^2/10$ is integral. The Weil operator $T$ therefore acts as $1$ on the sign line. The Weil action commutes with $S_5$, so this line is invariant. The rank is even, and the representation descends to $SL_2(\\Z)$. On the line the relation $(ST)^3=S^2$ reduces to $S^3=S^2$, giving $S=1$.","The full-level scalar conclusion has now been justified by the discriminant calculation. The ordinary residue coset has $x=v/\\sqrt5$, and its exponent $\\|x\\|^2/2$ is exactly $\\|v\\|^2/10=n$. Scaling the homogeneous polynomial changes an overall scalar, not the modular weight or the zero conclusion.","\\section{The modularity contract and the cusp condition}\nThe authenticated harmonic theta theorem states that a homogeneous harmonic polynomial of degree $\\nu$ on a rank-$r$ positive definite even lattice gives a vector-valued modular form of weight $\\nu+r/2$ for the Weil representation \\cite{IW}. Here $r=4$ and $\\nu=10+d$. The project-specific sign-line argument above therefore gives\n\\begin{equation}\\label{eq:thetalevelone}\n\\Theta_Q\\in S_{12+d}(SL_2(\\Z))\n\\end{equation}\nfor the even-degree harmonic construction. The zero vector contributes no constant term. The full modular group has one cusp, so the form is cuspidal. This is the exact external-to-internal dependency: the general theorem supplies vector modularity, and the explicit coset argument supplies the scalar full-level specialization.","tion{The modularity contract and the cusp condition}\nThe authenticated harmonic theta theorem states that a homogeneous harmonic polynomial of degree $\\nu$ on a rank-$r$ positive definite even lattice gives a vector-valued modular form of weight $\\nu+r/2$ for the Weil representation \\cite{IW}. Here $r=4$ and $\\nu=10+d$. The project-specific sign-line argument above therefore gives\n\\begin{equation}\\label{eq:thetalevelone}\n\\Theta_Q\\in S_{12+d}(SL_2(\\Z))\n\\end{equation}\nfor the even-degree harmonic construction. The zero vector contributes no constant term. The full modular group has one cusp, so the form is cuspidal. This is the exact external-to-internal dependency: the general theorem supplies vector modularity, and the explicit coset argument supplies the scalar full-level specialization.","\\chapter{Universal null polynomials}\\label{ch:nulls}\n\\section{The correction that does not change any coefficient}\nA universally null polynomial is a symmetric polynomial $Q$ with\n\\begin{equation}\\label{eq:universalnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0\\quad\\text{for every }n\\ge1.\n\\end{equation}\nThis is much stronger than being null on one selected shell. If a finite family satisfies \\eqref{eq:universalnull}, any rational linear combination of it does too. Thus\n\\[\nF(v)=1+\\sum_j a_jQ_j(v)\n\\]\nhas the exact property\n\\begin{equation}\\label{eq:correctedtotal}\n\\sum_{v\\in\\Vn}W(v)F(v)=\\tau(n)\n\\end{equation}\nat every index, after the coefficients $a_j$ have been fixed. The choice of coefficients may depend on a target shell. The null identity does not thereby become merely a shell-specific identity.","n{The correction that does not change any coefficient}\nA universally null polynomial is a symmetric polynomial $Q$ with\n\\begin{equation}\\label{eq:universalnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0\\quad\\text{for every }n\\ge1.\n\\end{equation}\nThis is much stronger than being null on one selected shell. If a finite family satisfies \\eqref{eq:universalnull}, any rational linear combination of it does too. Thus\n\\[\nF(v)=1+\\sum_j a_jQ_j(v)\n\\]\nhas the exact property\n\\begin{equation}\\label{eq:correctedtotal}\n\\sum_{v\\in\\Vn}W(v)F(v)=\\tau(n)\n\\end{equation}\nat every index, after the coefficients $a_j$ have been fixed. The choice of coefficients may depend on a target shell. The null identity does not thereby become merely a shell-specific identity.","y null polynomial is a symmetric polynomial $Q$ with\n\\begin{equation}\\label{eq:universalnull}\n\\sum_{v\\in\\Vn}W(v)Q(v)=0\\quad\\text{for every }n\\ge1.\n\\end{equation}\nThis is much stronger than being null on one selected shell. If a finite family satisfies \\eqref{eq:universalnull}, any rational linear combination of it does too. Thus\n\\[\nF(v)=1+\\sum_j a_jQ_j(v)\n\\]\nhas the exact property\n\\begin{equation}\\label{eq:correctedtotal}\n\\sum_{v\\in\\Vn}W(v)F(v)=\\tau(n)\n\\end{equation}\nat every index, after the coefficients $a_j$ have been fixed. The choice of coefficients may depend on a target shell. The null identity does not thereby become merely a shell-specific identity.","Q(v)=0\\quad\\text{for every }n\\ge1.\n\\end{equation}\nThis is much stronger than being null on one selected shell. If a finite family satisfies \\eqref{eq:universalnull}, any rational linear combination of it does too. Thus\n\\[\nF(v)=1+\\sum_j a_jQ_j(v)\n\\]\nhas the exact property\n\\begin{equation}\\label{eq:correctedtotal}\n\\sum_{v\\in\\Vn}W(v)F(v)=\\tau(n)\n\\end{equation}\nat every index, after the coefficients $a_j$ have been fixed. The choice of coefficients may depend on a target shell. The null identity does not thereby become merely a shell-specific identity.","\\section{A low-degree limit}\nEvery alternating polynomial of total degree at most $13$ is $D$ times a symmetric quotient of degree at most $3$. On $s_1=0$, such a quotient is a linear combination of $1,s_2,s_3$. On a shell, $s_2=10n$, while the $s_3$ contribution vanishes under $J$. Its weighted sum is thus $(a+10bn)\\tau(n)$ for constants $a,b$.","This low-degree space cannot extract a new independent arithmetic residual. It is a scoped limitation on alternating polynomial refinements through total degree $13$. At quotient degree $4$, $s_4$ appears, and harmonic theta forms of higher weights introduce new moment identities. These will still require care: new cusp coefficients do not automatically give a nonzero lower bound for the original one.","\\section{An exact operator on symmetric power sums}\nDefine $\\mathcal L$ by\n\\[\n\\Delta_H(DQ)=D\\mathcal LQ.\n\\]\nThe Vandermonde is harmonic. Its logarithmic derivative and the product rule give, with $s_0=5$ and $s_1=0$,\n\\begin{equation}\\label{eq:lsm}\n\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2},\\qquad m\\ge2.\n\\end{equation}\nThe chain-rule metric is\n\\begin{equation}\\label{eq:gradient}\n\\nabla_Hs_m\\cdot\\nabla_Hs_n\n=mn\\left(s_{m+n-2}-\\frac{s_{m-1}s_{n-1}}5\\right).\n\\end{equation}\nFor a polynomial in $s_2,s_3,s_4,s_5$, these two formulas determine $\\mathcal LQ$ by the ordinary second-order product rule. Higher power sums are reduced through Newton's identities for five variables.","nde is harmonic. Its logarithmic derivative and the product rule give, with $s_0=5$ and $s_1=0$,\n\\begin{equation}\\label{eq:lsm}\n\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2},\\qquad m\\ge2.\n\\end{equation}\nThe chain-rule metric is\n\\begin{equation}\\label{eq:gradient}\n\\nabla_Hs_m\\cdot\\nabla_Hs_n\n=mn\\left(s_{m+n-2}-\\frac{s_{m-1}s_{n-1}}5\\right).\n\\end{equation}\nFor a polynomial in $s_2,s_3,s_4,s_5$, these two formulas determine $\\mathcal LQ$ by the ordinary second-order product rule. Higher power sums are reduced through Newton's identities for five variables.","the product rule give, with $s_0=5$ and $s_1=0$,\n\\begin{equation}\\label{eq:lsm}\n\\mathcal Ls_m\n=m\\sum_{a=0}^{m-2}s_as_{m-2-a}\n-\\frac{m(m-1)}5s_{m-2},\\qquad m\\ge2.\n\\end{equation}\nThe chain-rule metric is\n\\begin{equation}\\label{eq:gradient}\n\\nabla_Hs_m\\cdot\\nabla_Hs_n\n=mn\\left(s_{m+n-2}-\\frac{s_{m-1}s_{n-1}}5\\right).\n\\end{equation}\nFor a polynomial in $s_2,s_3,s_4,s_5$, these two formulas determine $\\mathcal LQ$ by the ordinary second-order product rule. Higher power sums are reduced through Newton's identities for five variables.","The apparent denominators from differentiating $D$ cancel pairwise. The quotient\n\\[\n\\frac{v_i^{m-1}-v_j^{m-1}}{v_i-v_j}\n\\]\nis a polynomial. Thus the operator really has rational polynomial coefficients on the symmetric ring. For example, \\eqref{eq:lsm} gives $\\mathcal Ls_2=48$. This agrees with the direct radial identity $\\Delta_H(s_2D)=48D$.","At each even invariant degree $d$, one lists monomials $s_2^as_3^bs_4^cs_5^e$ with $2a+3b+4c+5e=d$. Writing an unknown rational linear combination, the equation $\\mathcal LQ=0$ is a finite linear system. The initial theta coefficients supply further rational linear equations. No floating-point approximation is needed to define this kernel.","\\section{Why a finite initial segment suffices}\nThe level-one valence formula implies that a nonzero holomorphic modular form of weight $k$ has order at the cusp at most $k/12$ \\cite{Stein}. A cusp form whose coefficients of $q^1,\\ldots,q^{\\lfloor k/12\\rfloor}$ all vanish has order at least $\\lfloor k/12\\rfloor+1>k/12$. It must be zero.","For a harmonic $DQ$ of invariant degree $d$, the theta weight is $k=12+d$. Thus the constructor imposes\n\\[\n[q^m]\\Theta_Q=0,\\qquad1\\le m\\le\\left\\lfloor\\frac{12+d}{12}\\right\\rfloor.\n\\]\nHarmonicity supplies membership in the specified modular-form space, and valence supplies the implication from this finite certificate to all coefficients. Without harmonicity and the modularity theorem, testing the same initial segment would have no universal force.","The distinction between the three degree conventions is essential. The symmetric correction has invariant degree $d$. The alternating product $DQ$ has total degree $10+d$. Its theta form has modular weight $12+d$. A claim about corrections through degree $20$ means the first of these unless stated otherwise.","\\section{The first explicit corrections}\nThe construction gives\n\\begin{align}\\label{eq:q8}\nQ_8={}&6273s_2^4-33750s_2^2s_4-56800s_2s_3^2\\notag\\\\\n&+69600s_3s_5+45000s_4^2,\n\\end{align}\nand\n\\begin{align}\\label{eq:q10}\nQ_{10}={}&2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2\\notag","&+69600s_3s_5+45000s_4^2,\n\\end{align}\nand\n\\begin{align}\\label{eq:q10}\nQ_{10}={}&2091s_2^5-11250s_2^3s_4-145800s_2^2s_3^2\\notag\\\\\n&+409600s_2s_3s_5+15000s_2s_4^2\\notag\\\\\n&-30000s_3^2s_4-264000s_5^2.\n\\end{align}\nBoth $DQ_8$ and $DQ_{10}$ are harmonic. The first shell has the vector $(0,1,2,-2,-1)$, with\n\\[\ns_2=10,\\quad s_3=0,\\quad s_4=34,\\quad s_5=0.\n\\]\nBoth polynomials vanish there. Their theta weights are $20$ and $22$. The cusp spaces are respectively spanned by $\\Delta E_4^2$ and $\\Delta E_4E_6$, each with nonzero first coefficient. The zero first coefficient therefore kills the whole theta form. This is also the simplest valence argument for these two degrees.","&-30000s_3^2s_4-264000s_5^2.\n\\end{align}\nBoth $DQ_8$ and $DQ_{10}$ are harmonic. The first shell has the vector $(0,1,2,-2,-1)$, with\n\\[\ns_2=10,\\quad s_3=0,\\quad s_4=34,\\quad s_5=0.\n\\]\nBoth polynomials vanish there. Their theta weights are $20$ and $22$. The cusp spaces are respectively spanned by $\\Delta E_4^2$ and $\\Delta E_4E_6$, each with nonzero first coefficient. The zero first coefficient therefore kills the whole theta form. This is also the simplest valence argument for these two degrees.","The first retained family has fourteen corrections, with counts\n\\[\n8:1,\\quad10:1,\\quad12:1,\\quad14:2,\\quad16:2,\\quad18:3,\\quad20:4.\n\\]\nThe transport continuation adds $22$ more: four at degree $22$, five at $24$, six at $26$ and seven at $28$. The total is $36$. These are explicit permitted corrections, not a proof that every universally null polynomial lies in their span.","\\section{Moment identities that introduce other forms}\nThe harmonic polynomial $D(130s_4-47s_2^2)$ has weight-$16$ theta form. Its first coefficient is $-280$. Write $E_4\\Delta=\\sum a_{16}(n)q^n$. Then\n\\begin{equation}\\label{eq:moment16}\n13\\sum_{v\\in\\Vn}W(v)s_4(v)\n=470n^2\\tau(n)-28a_{16}(n).\n\\end{equation}\nLikewise $D(90s_2s_4-27s_2^3-320s_3^2)$ gives\n\\begin{align}\\label{eq:moment18}\n900n\\sum W(v)s_4(v)-27000n^3\\tau(n)\n-320\\sum W(v)s_3(v)^2\n=3600a_{18}(n),\n\\end{align}\nwhere $E_6\\Delta=\\sum a_{18}(n)q^n$ and the packet sums are on $\\Vn$. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer nonvanishing. A new modular moment is not automatically a sign or residue certificate.","D(130s_4-47s_2^2)$ has weight-$16$ theta form. Its first coefficient is $-280$. Write $E_4\\Delta=\\sum a_{16}(n)q^n$. Then\n\\begin{equation}\\label{eq:moment16}\n13\\sum_{v\\in\\Vn}W(v)s_4(v)\n=470n^2\\tau(n)-28a_{16}(n).\n\\end{equation}\nLikewise $D(90s_2s_4-27s_2^3-320s_3^2)$ gives\n\\begin{align}\\label{eq:moment18}\n900n\\sum W(v)s_4(v)-27000n^3\\tau(n)\n-320\\sum W(v)s_3(v)^2\n=3600a_{18}(n),\n\\end{align}\nwhere $E_6\\Delta=\\sum a_{18}(n)q^n$ and the packet sums are on $\\Vn$. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer nonvanishing. A new modular moment is not automatically a sign or residue certificate.","coefficient is $-280$. Write $E_4\\Delta=\\sum a_{16}(n)q^n$. Then\n\\begin{equation}\\label{eq:moment16}\n13\\sum_{v\\in\\Vn}W(v)s_4(v)\n=470n^2\\tau(n)-28a_{16}(n).\n\\end{equation}\nLikewise $D(90s_2s_4-27s_2^3-320s_3^2)$ gives\n\\begin{align}\\label{eq:moment18}\n900n\\sum W(v)s_4(v)-27000n^3\\tau(n)\n-320\\sum W(v)s_3(v)^2\n=3600a_{18}(n),\n\\end{align}\nwhere $E_6\\Delta=\\sum a_{18}(n)q^n$ and the packet sums are on $\\Vn$. These are exact identities. Their additional cusp coefficients must be controlled if the identities are used to infer nonvanishing. A new modular moment is not automatically a sign or residue certificate.","\\chapter{What an exact sign certificate proves}\\label{ch:certificates}\n\\section{The sufficient sign condition}\nTake a finite rational combination of universally null corrections and set $F=1+\\sum a_jQ_j$. If all real numbers $W(v)F(v)$ on a shell have one weak sign and at least one is nonzero, their sum is nonzero. By \\eqref{eq:correctedtotal}, that sum is $\\tau(n)$.","\\section{Six packets reduced to two at \\texorpdfstring{$n=7$}{n=7}}\nChoose\n\\[\na=\\frac{383}{1411344000000},\\qquad\nb=\\frac{139}{61158240000000},\\qquad\nF=1+aQ_8+bQ_{10}.\n\\]\nThere are three $J$-orbits, each of size two. Their representatives and weights are\n\\begin{center}\n\\begin{tabular}{lrr}\n\\toprule\nRepresentative $v$&$W(v)$&$F(v)$","\\midrule\n$(-5,-4,2,3,4)$&1176&0","$(-5,6,2,-2,-1)$&$-8624$&0","$(0,-4,7,-2,-1)$&$-924$&$299/33$","\\bottomrule\n\\end{tabular}\n\\end{center}\nFor the three rows, the $(Q_8,Q_{10})$ values are\n\\begin{align*}\n&(712800000,-525096000000),","\\bottomrule\n\\end{tabular}\n\\end{center}\nFor the three rows, the $(Q_8,Q_{10})$ values are\n\\begin{align*}\n&(712800000,-525096000000),\\\\\n&(-3790800000,12636000000),\\\\\n&(36288000000,-786240000000).\n\\end{align*}\nSubstitution verifies the displayed $F$ values exactly. Four packets disappear, and the remaining two are negative. Therefore\n\\begin{equation}\\label{eq:tau7cert}\n\\tau(7)=2(-924)\\frac{299}{33}=-16744.\n\\end{equation}\nThe solve can be performed by requiring only the first two rows to vanish. It need not use a preassigned value of $\\tau(7)$. The surviving value then supplies the coefficient, and the independent product engine checks it.","&(36288000000,-786240000000).\n\\end{align*}\nSubstitution verifies the displayed $F$ values exactly. Four packets disappear, and the remaining two are negative. Therefore\n\\begin{equation}\\label{eq:tau7cert}\n\\tau(7)=2(-924)\\frac{299}{33}=-16744.\n\\end{equation}\nThe solve can be performed by requiring only the first two rows to vanish. It need not use a preassigned value of $\\tau(7)$. The surviving value then supplies the coefficient, and the independent product engine checks it.","An integral form of the same certificate multiplies $F$ by $183474720000000$:\n\\[\n183474720000000+49790Q_8+417Q_{10}.\n\\]\nIts surviving value is $1662392160000000$. Clearing denominators is useful for a short exact checker, but does not change the sign argument.","\\figplaceholder{orbit7}{The three original $J$-orbit masses at $n=7$, each containing two packets. The exact certificate in the text removes the positive orbit and one negative orbit, then rescales the remaining negative one. The unchanged total is certified nonzero by the displayed rational coefficients.}","\\section{An exact obstruction in a specified smaller space}\nAt $n=31$, the seven retained corrections through degree $16$ do not admit a nontrivial common-sign certificate. The shell has sixteen $J$-orbits. Let $\\epsilon_i=\\sgn W(v_i)$ for their representatives. The retained exact dual has strictly positive rational entries $y_i$ satisfying\n\\[\n\\sum_i y_i=1,\\qquad\\sum_i y_i\\epsilon_i=0,\\qquad\n\\sum_i y_i\\epsilon_iQ_j(v_i)=0\n\\]\nfor all seven correction columns. Consequently, for every coefficient vector,\n\\[\n\\sum_i y_i\\epsilon_i\\left(1+\\sum_j a_jQ_j(v_i)\\right)=0.\n\\]\nIf every corrected packet had one weak sign and one were nonzero, every $\\epsilon_iF(v_i)$ would have that sign and one would be strict. A strictly positive weighted average could not be zero. This","\\section{An exact obstruction in a specified smaller space}\nAt $n=31$, the seven retained corrections through degree $16$ do not admit a nontrivial common-sign certificate. The shell has sixteen $J$-orbits. Let $\\epsilon_i=\\sgn W(v_i)$ for their representatives. The retained exact dual has strictly positive rational entries $y_i$ satisfying\n\\[\n\\sum_i y_i=1,\\qquad\\sum_i y_i\\epsilon_i=0,\\qquad\n\\sum_i y_i\\epsilon_iQ_j(v_i)=0\n\\]\nfor all seven correction columns. Consequently, for every coefficient vector,\n\\[\n\\sum_i y_i\\epsilon_i\\left(1+\\sum_j a_jQ_j(v_i)\\right)=0.\n\\]\nIf every corrected packet had one weak sign and one were nonzero, every $\\epsilon_iF(v_i)$ would have that sign and one would be strict. A strictly positive weighted average could not be zero. This contradiction proves infeasibility in the specified","\\section{An exact obstruction in a specified smaller space}\nAt $n=31$, the seven retained corrections through degree $16$ do not admit a nontrivial common-sign certificate. The shell has sixteen $J$-orbits. Let $\\epsilon_i=\\sgn W(v_i)$ for their representatives. The retained exact dual has strictly positive rational entries $y_i$ satisfying\n\\[\n\\sum_i y_i=1,\\qquad\\sum_i y_i\\epsilon_i=0,\\qquad\n\\sum_i y_i\\epsilon_iQ_j(v_i)=0\n\\]\nfor all seven correction columns. Consequently, for every coefficient vector,\n\\[\n\\sum_i y_i\\epsilon_i\\left(1+\\sum_j a_jQ_j(v_i)\\right)=0.\n\\]\nIf every corrected packet had one weak sign and one were nonzero, every $\\epsilon_iF(v_i)$ would have that sign and one would be strict. A strictly positive weighted average could not be zero. This contradiction proves infeasibility in the specified seven-column span, for either sign.","degree $16$ do not admit a nontrivial common-sign certificate. The shell has sixteen $J$-orbits. Let $\\epsilon_i=\\sgn W(v_i)$ for their representatives. The retained exact dual has strictly positive rational entries $y_i$ satisfying\n\\[\n\\sum_i y_i=1,\\qquad\\sum_i y_i\\epsilon_i=0,\\qquad\n\\sum_i y_i\\epsilon_iQ_j(v_i)=0\n\\]\nfor all seven correction columns. Consequently, for every coefficient vector,\n\\[\n\\sum_i y_i\\epsilon_i\\left(1+\\sum_j a_jQ_j(v_i)\\right)=0.\n\\]\nIf every corrected packet had one weak sign and one were nonzero, every $\\epsilon_iF(v_i)$ would have that sign and one would be strict. A strictly positive weighted average could not be zero. This contradiction proves infeasibility in the specified seven-column span, for either sign.","ir representatives. The retained exact dual has strictly positive rational entries $y_i$ satisfying\n\\[\n\\sum_i y_i=1,\\qquad\\sum_i y_i\\epsilon_i=0,\\qquad\n\\sum_i y_i\\epsilon_iQ_j(v_i)=0\n\\]\nfor all seven correction columns. Consequently, for every coefficient vector,\n\\[\n\\sum_i y_i\\epsilon_i\\left(1+\\sum_j a_jQ_j(v_i)\\right)=0.\n\\]\nIf every corrected packet had one weak sign and one were nonzero, every $\\epsilon_iF(v_i)$ would have that sign and one would be strict. A strictly positive weighted average could not be zero. This contradiction proves infeasibility in the specified seven-column span, for either sign.","The scope matters. The dual is not a theorem about every polynomial of degree at most $16$, unless completeness of the retained list in that larger class is separately proved. It is not a theorem about all universally null polynomials. It does not show that $\\tau(31)$ vanishes. It describes one exact finite correction space.","\\section{The obstruction can disappear when the space changes}\nAdding the retained corrections at degrees $18$ and $20$ gives an exact strictly negative certificate at $31$. The previous dual does not annihilate the larger family. This is an instructive example of a removable obstruction. It cannot support the assertion that every obstruction at every prime will be removable.","The retained strict rational certificates cover the following primes:\n\\begin{center}\n\\begin{tabular}{rrrrrrr}\n\\toprule\n$p$&2&3&5&7&11&13","Sign&$-$&$+$&$+$&$-$&$+$&$-$","\\midrule\n$p$&17&19&23&29&31&37","Sign&$-$&$+$&$+$&$+$&$-$&$-$","\\midrule\n$p$&47&59&&&&","Sign&$+$&$-$&&&&","\\bottomrule\n\\end{tabular}\n\\end{center}\nThe degrees used are not claimed optimal. In particular, no exact degree-$18$ impossibility at $31$ is asserted. The fresh release replay verifies all fourteen stored rational certificates and all $300$ packet margins in their combined finite scope.","\\figplaceholder{orbit31}{The sixteen original $J$-orbit masses at $31$. Eight are positive and eight negative. The coefficients sum to the nonzero total $-52843168$, but individual nonzero masses and a balance of sign counts do not force that total.}","\\section{Completeness retains the total}\nSuppose a shell has nonzero orbit masses $w_1,\\ldots,w_N$ and total $T$. Even allow every rational correction vector $q$ with $\\sum_iw_iq_i=0$. Then\n\\begin{equation}\\label{eq:fullsignequiv}\n\\exists q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","mpleteness retains the total}\nSuppose a shell has nonzero orbit masses $w_1,\\ldots,w_N$ and total $T$. Even allow every rational correction vector $q$ with $\\sum_iw_iq_i=0$. Then\n\\begin{equation}\\label{eq:fullsignequiv}\n\\exists q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","ins the total}\nSuppose a shell has nonzero orbit masses $w_1,\\ldots,w_N$ and total $T$. Even allow every rational correction vector $q$ with $\\sum_iw_iq_i=0$. Then\n\\begin{equation}\\label{eq:fullsignequiv}\n\\exists q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","sses $w_1,\\ldots,w_N$ and total $T$. Even allow every rational correction vector $q$ with $\\sum_iw_iq_i=0$. Then\n\\begin{equation}\\label{eq:fullsignequiv}\n\\exists q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","\\begin{equation}\\label{eq:fullsignequiv}\n\\exists q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","ts q:\\ w_i(1+q_i)\\text{ have one weak sign and are not all zero}\n\\quad\\Longleftrightarrow\\quad T\\ne0.\n\\end{equation}\nNecessity follows from conservation. For sufficiency, if $T\\ne0$, take\n\\[\nq_i=\\frac{T}{Nw_i}-1.\n\\]\nEvery corrected mass is $T/N$ and the correction has zero weighted total. This construction proves an equivalence. If used as a proof of $T\\ne0$ without independent information about $T$, it would be circular.","Thus a theorem that all zero-total correction vectors are available would explain the existence of a sign certificate once the total is known to survive. It would not independently establish that survival. A concrete sign certificate can still be decisive, as at $7$. A general completeness theorem must be combined with an independently established nonzero sign or arithmetic condition to settle every prime.","\\chapter{The exact dyadic mixed-sign boundary}\\label{ch:tree}\n\\section{Sectors and two different notions of separation}\nFor a packet put $c_i(v)=(v_i-i)/5$. At depth $k\\ge1$, write $M=2^k$. A sector is the residue of $c(v)$ modulo $M$, identifying the residues obtained from $v$ and $Jv$. One may represent it by the lexicographically smaller of $c(v)\\bmod M$ and $c(Jv)\\bmod M$. All sector sums retain the original packet multiplicities.","A sector is mixed if it contains both positive and negative weights. Sign-purity is weaker than resolving a sector into one $J$-orbit. Several different orbits can have the same sign. A theorem about the first mixed sector therefore answers a different question from a theorem about the first collision of distinct orbits.","The sign has an elementary coding description. For $i<j$, the difference\n\\[\nv_i-v_j=(i-j)+5(c_i-c_j)\n\\]\nis positive exactly when $c_i>c_j$. There are ten pairs, so\n\\begin{equation}\\label{eq:inversionsign}\n\\sgn W(v)=(-1)^{\\#\\{i<j:c_i>c_j\\}}.\n\\end{equation}\nEqual $c_i,c_j$ give a negative difference because $i-j<0$. Thus ties in the coding do not make a zero Vandermonde.","\\section{The sharp threshold}\n\\begin{theorem}\\label{thm:threshold}\nLet $k\\ge1$, $M=2^k$, and set\n\\begin{equation}\\label{eq:nk}\nN_k=\\left\\lceil\\frac{100M^2+10M+2}{35}\\right\\rceil\n=\\frac{20M^2+2M+\\gamma_k}{7},\n\\end{equation}\nwhere\n\\[\n\\gamma_k=\\begin{cases}\n7&k\\equiv1\\pmod3,","\\section{The sharp threshold}\n\\begin{theorem}\\label{thm:threshold}\nLet $k\\ge1$, $M=2^k$, and set\n\\begin{equation}\\label{eq:nk}\nN_k=\\left\\lceil\\frac{100M^2+10M+2}{35}\\right\\rceil\n=\\frac{20M^2+2M+\\gamma_k}{7},\n\\end{equation}\nwhere\n\\[\n\\gamma_k=\\begin{cases}\n7&k\\equiv1\\pmod3,\\\\\n1&k\\equiv2\\pmod3,\\\\\n6&k\\equiv0\\pmod3.\n\\end{cases}\n\\]\nEvery depth-$k$ sector in every shell $n<N_k$ is sign-pure. The shell $n=N_k$ has a mixed sector.\n\\end{theorem}\nThis threshold ranges over all positive integer shell indices. It does not assert that $N_k$ is prime. For example, $N_3=186$ is composite.","6&k\\equiv0\\pmod3.\n\\end{cases}\n\\]\nEvery depth-$k$ sector in every shell $n<N_k$ is sign-pure. The shell $n=N_k$ has a mixed sector.\n\\end{theorem}\nThis threshold ranges over all positive integer shell indices. It does not assert that $N_k$ is prime. For example, $N_3=186$ is composite.","\\section{The geometric lower bound}\nTake opposite-sign packets $v,w$ in one sector. Replace $w$ by $Jw$ if needed so that\n\\[\nw-v=5Md,\\qquad d\\in A_4\\setminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","opposite-sign packets $v,w$ in one sector. Replace $w$ by $Jw$ if needed so that\n\\[\nw-v=5Md,\\qquad d\\in A_4\\setminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","in one sector. Replace $w$ by $Jw$ if needed so that\n\\[\nw-v=5Md,\\qquad d\\in A_4\\setminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","ace $w$ by $Jw$ if needed so that\n\\[\nw-v=5Md,\\qquad d\\in A_4\\setminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","so that\n\\[\nw-v=5Md,\\qquad d\\in A_4\\setminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","tminus\\{0\\}.\n\\]\nSet $a=5M/2$ and $m=(v+w)/2$. Since $M$ is even, both are integral. Equal norms give\n\\begin{equation}\\label{eq:midpoint}\nm\\cdot d=0,\\qquad10n=\\|m\\|^2+a^2\\|d\\|^2.\n\\end{equation}\nThe squared norm of a nonzero integral zero-sum vector is a positive even integer. It cannot be $2$ here. Such a vector is $\\pm(e_i-e_j)$, and $m\\cdot d=0$ would force $v_i-v_j=\\pm5M$. This is divisible by $5$, contrary to the distinct coordinate residues.","If $\\|d\\|^2\\ge6$, \\eqref{eq:midpoint} gives\n\\[\nn\\ge\\frac{15}{4}M^2>\\frac{100M^2+10M+2}{35},\\qquad M\\ge2.\n\\]\nThe remaining case is $\\|d\\|^2=4$. Then $d$ has two coordinates $1$, two $-1$ and one $0$. Of the ten coordinate pairs, two have equal $d$ entries and eight have unequal entries. For a pair with $\\delta=d_i-d_j$,\n\\[\n(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\\delta^2.\n\\]\nThe two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of $D(v)$ and $D(w)$. None of these pair products is zero, since both packets have distinct coordinates. Hence at least one unequal pair has a positive product, giving","If $\\|d\\|^2\\ge6$, \\eqref{eq:midpoint} gives\n\\[\nn\\ge\\frac{15}{4}M^2>\\frac{100M^2+10M+2}{35},\\qquad M\\ge2.\n\\]\nThe remaining case is $\\|d\\|^2=4$. Then $d$ has two coordinates $1$, two $-1$ and one $0$. Of the ten coordinate pairs, two have equal $d$ entries and eight have unequal entries. For a pair with $\\delta=d_i-d_j$,\n\\[\n(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\\delta^2.\n\\]\nThe two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of $D(v)$ and $D(w)$. None of these pair products is zero, since both packets have distinct coordinates. Hence at least one unequal pair has a positive product, giving the integral inequality\n\\[\n|m_i-m_j|\\ge a|\\delta|+1.\n\\]","has two coordinates $1$, two $-1$ and one $0$. Of the ten coordinate pairs, two have equal $d$ entries and eight have unequal entries. For a pair with $\\delta=d_i-d_j$,\n\\[\n(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\\delta^2.\n\\]\nThe two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of $D(v)$ and $D(w)$. None of these pair products is zero, since both packets have distinct coordinates. Hence at least one unequal pair has a positive product, giving the integral inequality\n\\[\n|m_i-m_j|\\ge a|\\delta|+1.\n\\]","rdinates $1$, two $-1$ and one $0$. Of the ten coordinate pairs, two have equal $d$ entries and eight have unequal entries. For a pair with $\\delta=d_i-d_j$,\n\\[\n(v_i-v_j)(w_i-w_j)=(m_i-m_j)^2-a^2\\delta^2.\n\\]\nThe two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of $D(v)$ and $D(w)$. None of these pair products is zero, since both packets have distinct coordinates. Hence at least one unequal pair has a positive product, giving the integral inequality\n\\[\n|m_i-m_j|\\ge a|\\delta|+1.\n\\]","=(m_i-m_j)^2-a^2\\delta^2.\n\\]\nThe two equal-entry pairs have positive products. If all eight unequal-entry pairs had negative products, their product would also be positive, contradicting the opposite signs of $D(v)$ and $D(w)$. None of these pair products is zero, since both packets have distinct coordinates. Hence at least one unequal pair has a positive product, giving the integral inequality\n\\[\n|m_i-m_j|\\ge a|\\delta|+1.\n\\]","Project $e_i-e_j$ orthogonally to $d^\\perp$. Its squared norm is\n\\[\n2-\\frac{\\delta^2}{4}.\n\\]\nBecause $m\\perp d$, Cauchy's inequality gives\n\\[\n\\|m\\|^2\\ge\\frac{(a|\\delta|+1)^2}{2-\\delta^2/4}.\n\\]\nFor $|\\delta|=1$ this is $4(a+1)^2/7$. For $|\\delta|=2$ it is $(2a+1)^2$, a larger quantity. Therefore\n\\[\n10n\\ge4a^2+\\frac47(a+1)^2,\n\\]\nwhich simplifies to the lower bound inside the ceiling in \\eqref{eq:nk}. Taking the ceiling proves that no earlier shell can be mixed. Since powers of $2$ modulo $7$ cycle as $2,4,1$, direct division gives the displayed $\\gamma_k$ values.","a^2}{4}.\n\\]\nBecause $m\\perp d$, Cauchy's inequality gives\n\\[\n\\|m\\|^2\\ge\\frac{(a|\\delta|+1)^2}{2-\\delta^2/4}.\n\\]\nFor $|\\delta|=1$ this is $4(a+1)^2/7$. For $|\\delta|=2$ it is $(2a+1)^2$, a larger quantity. Therefore\n\\[\n10n\\ge4a^2+\\frac47(a+1)^2,\n\\]\nwhich simplifies to the lower bound inside the ceiling in \\eqref{eq:nk}. Taking the ceiling proves that no earlier shell can be mixed. Since powers of $2$ modulo $7$ cycle as $2,4,1$, direct division gives the displayed $\\gamma_k$ values.","Because $m\\perp d$, Cauchy's inequality gives\n\\[\n\\|m\\|^2\\ge\\frac{(a|\\delta|+1)^2}{2-\\delta^2/4}.\n\\]\nFor $|\\delta|=1$ this is $4(a+1)^2/7$. For $|\\delta|=2$ it is $(2a+1)^2$, a larger quantity. Therefore\n\\[\n10n\\ge4a^2+\\frac47(a+1)^2,\n\\]\nwhich simplifies to the lower bound inside the ceiling in \\eqref{eq:nk}. Taking the ceiling proves that no earlier shell can be mixed. Since powers of $2$ modulo $7$ cycle as $2,4,1$, direct division gives the displayed $\\gamma_k$ values.","erp d$, Cauchy's inequality gives\n\\[\n\\|m\\|^2\\ge\\frac{(a|\\delta|+1)^2}{2-\\delta^2/4}.\n\\]\nFor $|\\delta|=1$ this is $4(a+1)^2/7$. For $|\\delta|=2$ it is $(2a+1)^2$, a larger quantity. Therefore\n\\[\n10n\\ge4a^2+\\frac47(a+1)^2,\n\\]\nwhich simplifies to the lower bound inside the ceiling in \\eqref{eq:nk}. Taking the ceiling proves that no earlier shell can be mixed. Since powers of $2$ modulo $7$ cycle as $2,4,1$, direct division gives the displayed $\\gamma_k$ values.","ge\\frac{(a|\\delta|+1)^2}{2-\\delta^2/4}.\n\\]\nFor $|\\delta|=1$ this is $4(a+1)^2/7$. For $|\\delta|=2$ it is $(2a+1)^2$, a larger quantity. Therefore\n\\[\n10n\\ge4a^2+\\frac47(a+1)^2,\n\\]\nwhich simplifies to the lower bound inside the ceiling in \\eqref{eq:nk}. Taking the ceiling proves that no earlier shell can be mixed. Since powers of $2$ modulo $7$ cycle as $2,4,1$, direct division gives the displayed $\\gamma_k$ values.","\\section{Witnesses attaining the bound}\nAt $k=1$, take\n\\[\nv=(-5,-4,2,-2,9),\\qquad w=(5,6,-8,-2,-1).\n\\]\nThey belong to $\\mathcal V_{13}$, have opposite weights and differ coordinatewise by multiples of $10$. Thus they lie in one depth-one sector.","For $k\\ge2$, explicit affine families attain the lower bound. The following table lists $7v$ and $7w$, so that the displayed entries remain small integer affine expressions.\n\\begin{center}\\small\n\\begin{tabular}{c>{\\raggedright\\arraybackslash}p{0.39\\linewidth}>{\\raggedright\\arraybackslash}p{0.39\\linewidth}}\n\\toprule\n$k\\bmod3$&$7v$&$7w$","\\midrule\n1, $k\\ge4$&$(-20M+5,-10M-8,-10M-1,20M+16,20M-12)$&$(15M+5,-10M-8,25M-1,-15M+16,-15M-12)$","[3pt]\n2, $k\\ge2$&$(-25M-5,10M+2,-15M+4,15M-4,15M+3)$&$(10M-5,10M+2,20M+4,-20M-4,-20M+3)$\\\\[3pt]\n0, $k\\ge3$&$(-20M-15,10M-3,10M+4,20M+1,-20M+13)$&$(15M-15,-25M-3,10M+4,-15M+1,15M+13)$\\\\\n\\bottomrule\n\\end{tabular}\\end{center}\nThe appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are\n\\[\n\\sum_i(7v_i)^2=\\sum_i(7w_i)^2\n=1400M^2+140M+70\\gamma_k=490N_k.\n\\]\nIn the first two rows, $(w-v)/(5M)=(1,0,1,-1,-1)$. In the third it is $(1,-1,0,-1,1)$. Thus the packets share a sector.","\\bottomrule\n\\end{tabular}\\end{center}\nThe appropriate powers of $2$ make each entry divisible by $7$. After division, each vector has the standard residues, sum zero and norm squared $10N_k$. The norm identities before division are\n\\[\n\\sum_i(7v_i)^2=\\sum_i(7w_i)^2\n=1400M^2+140M+70\\gamma_k=490N_k.\n\\]\nIn the first two rows, $(w-v)/(5M)=(1,0,1,-1,-1)$. In the third it is $(1,-1,0,-1,1)$. Thus the packets share a sector.","Every coordinate difference is affine in $M$. Its sign is fixed on the indicated range. Checking the ten linear differences in each row proves that the two Vandermonde signs are opposite throughout that range. This is an all-depth argument using polynomial identities and linear inequalities, not an inference from a list of sample powers.","The first thresholds are\n\\[\n13, 47, 186, 737, 2935, 11722, 46849, 187319.\n\\]\nThe verifier checks witnesses through $k=64$ as an implementation control. The affine identities and lower-bound proof establish the theorem for every depth.","\\figplaceholder{mixed_threshold}{The first mixed shell $N_k$ at depths $1$ through $8$. These exact values belong to an all-depth geometric theorem. The index range is all positive integer shells, and the thresholds need not be prime.}","\\section{A geometric theorem without a nonzero total}\nRefining sectors eventually separates distinct $J$-orbits on a fixed finite shell. The sharp threshold describes how quickly opposite signs can share a sector. Neither assertion compares the total positive mass with the total negative mass. Geometric separation has not supplied an arithmetic inequality between them.","\\chapter{Why strict localization and automatic valuation fail}\\label{ch:failures}\n\\section{Conserved sectors can preserve opposite signs}\nPartition a shell into sectors $S_r$, with masses $t_r=\\sum_{v\\in S_r}W(v)$. Suppose every permitted correction is null inside every sector:\n\\[\n\\sum_{v\\in S_r}W(v)Q_j(v)=0.\n\\]\nThen $F=1+\\sum a_jQ_j$ leaves each corrected sector mass equal to $t_r$. If two sector masses have opposite nonzero signs, all corrected packets cannot have one weak sign. Summing within those two sectors proves the obstruction immediately.","At $p=7$, depth-one sector data are\n\\begin{center}\\small\n\\begin{tabular}{lrrr}\n\\toprule\nSector key&Packets&Mass&$\\sum WQ_8$","\\midrule\n$(0,0,0,0,0)$&2&$-1848$&$-67060224000000$","$(1,0,0,1,0)$&2&$-17248$&$65383718400000$","$(1,1,0,0,0)$&2&2352&1676505600000","\\bottomrule\n\\end{tabular}\\end{center}\nThe three correction sums total zero, but no individual one is zero. Thus the known universal null $Q_8$ is not sector-null when restricted by a parity indicator. The sign certificate at $7$ works by transferring mass between sectors. A completion that preserves every one of these sectors would forbid the needed transfer.","This is a failure of the proposed localization principle, not a failure of the universal null identity. Multiplying $Q$ by an arbitrary sector indicator changes its theta series and may destroy universal nullness. A correct localized operation must be constructed by solving for combinations whose parent effects cancel, as in Chapter~\\ref{ch:transport}.","\\section{Aggregate parity identities have a larger scope}\nThe number of odd coordinates of a packet is even. Since squares modulo $4$ are $0$ or $1$,\n\\[\n\\#\\{i:v_i\\text{ odd}\\}\\equiv10n\\equiv2n\\pmod4.\n\\]\nFor odd $n$ there are two odd coordinates. For even $n$ there are zero or four. Let $A,B,C$ be the weighted generating series for these three aggregate classes, respectively.","The all-even class scales by $2$. The bijection $u_j=v_{2j\\bmod5}/2$ has a coordinate permutation of sign $-1$, so\n\\[\nA(q)=-2^{10}\\Delta(q^4).\n\\]\nUsing the Hecke recurrence at $2$ gives\n\\begin{align}\nB(q)&=-24\\Delta(q^2)-2^{10}\\Delta(q^4),","The all-even class scales by $2$. The bijection $u_j=v_{2j\\bmod5}/2$ has a coordinate permutation of sign $-1$, so\n\\[\nA(q)=-2^{10}\\Delta(q^4).\n\\]\nUsing the Hecke recurrence at $2$ gives\n\\begin{align}\nB(q)&=-24\\Delta(q^2)-2^{10}\\Delta(q^4),\\\\\nC(q)&=\\Delta(q)+24\\Delta(q^2)+2^{11}\\Delta(q^4).\n\\end{align}\nFor a homogeneous universally null correction, scaling kills its aggregate $A$ series, and odd and even support then kill the corresponding aggregate $B$ and $C$ series. This does not imply nullity inside each of the six finer parity sectors making up class $C$. The explicit $p=7$ table is a counterexample to that stronger inference.","C(q)&=\\Delta(q)+24\\Delta(q^2)+2^{11}\\Delta(q^4).\n\\end{align}\nFor a homogeneous universally null correction, scaling kills its aggregate $A$ series, and odd and even support then kill the corresponding aggregate $B$ and $C$ series. This does not imply nullity inside each of the six finer parity sectors making up class $C$. The explicit $p=7$ table is a counterexample to that stronger inference.","At sufficient depth, a sector is one $J$-orbit. For an even symmetric $Q$, its weighted correction sum on the orbit is its nonzero mass times $Q(v)$. Sector nullity then forces $Q(v)=0$ there. Imposing sector-nullity at every depth eventually removes every effective even correction. Refinement adds constraints rather than creating new available transfers.","\\section{A unique least valuation is a sufficient criterion}\nIf one summand in an integer sum has strictly smaller $\\ell$-adic valuation than all others, the sum has that least valuation and is nonzero. This follows by dividing by the least power of $\\ell$ and observing one unit plus terms divisible by $\\ell$.","Terminal packet resolution does not guarantee this hypothesis. Every packet is $1$ modulo $5$, and each $J$-orbit has size $1$ or $2$. Every orbit mass is therefore a $5$-adic unit. At an index prime $p>20$, Proposition~\\ref{prop:indexunit} makes every orbit mass a $p$-adic unit as well. When several orbits occur, there is no unique least valuation at either of these primes.","At $p=31$, depth $2$ already resolves all sixteen $J$-orbits. Eight masses are positive and eight negative. Four have the minimum $2$-adic valuation $3$. At depth $1$, two sector masses tie at that minimum. The unique-minimum test therefore fails at every positive depth in this example, although\n\\[\n\\tau(31)=-52843168,\\qquad v_2(\\tau(31))=5.\n\\]\nFurther residue information about tied sums may be useful. The counterexample excludes the proposed automatic criterion, rather than every possible valuation argument.","\\section{A prime-index interpolation that cannot be integral}\nAn earlier proposed shortcut treated\n\\[\nF(u)=\\frac{\\tau(2^{14}u-1)}{2^{14}}\n\\]\nat prime arguments as the restriction of a single integral $2$-adic power series. Any convergent integral restricted power series is $1$-Lipschitz: subtracting its values at $x,y$ factors by $x-y$ with an integral convergent quotient.","Take the prime indices\n\\[\n(u_1,p_1)=(15,245759),\\qquad(u_2,p_2)=(47,770047).\n\\]\nThe exact values are\n\\begin{align*}\n\\tau(245759)&=77900595298037436891536424960,","Take the prime indices\n\\[\n(u_1,p_1)=(15,245759),\\qquad(u_2,p_2)=(47,770047).\n\\]\nThe exact values are\n\\begin{align*}\n\\tau(245759)&=77900595298037436891536424960,\\\\\n\\tau(770047)&=-379872617393156753656047295381504.\n\\end{align*}\nTheir residues modulo $2^{19}$ are $393216$ and $147456$. Dividing by $2^{14}$ gives\n\\[\nF(15)\\equiv24\\pmod{32},\\qquad F(47)\\equiv9\\pmod{32}.\n\\]\nThus $v_2(F(47)-F(15))=0$, while $v_2(47-15)=5$. This violates the necessary Lipschitz condition and refutes the stated integral interpolation model.","\\tau(770047)&=-379872617393156753656047295381504.\n\\end{align*}\nTheir residues modulo $2^{19}$ are $393216$ and $147456$. Dividing by $2^{14}$ gives\n\\[\nF(15)\\equiv24\\pmod{32},\\qquad F(47)\\equiv9\\pmod{32}.\n\\]\nThus $v_2(F(47)-F(15))=0$, while $v_2(47-15)=5$. This violates the necessary Lipschitz condition and refutes the stated integral interpolation model.","Different valuations alone would not disprove analyticity. The obstruction is this divided-difference failure. The example does not exclude models with denominators, finer residue classes or additional variables. It also does not concern the legitimate auxiliary polynomial $G_n(U)$ at a fixed shell. The two constructions use different variables and different hypotheses.","\\section{Exact lifting equalities also need proof}\nAnother proposed shortcut was\n\\[\nv_\\ell(\\tau(p))=v_\\ell(p+1)\n\\]\nfor candidate primes $p\\ne\\ell$ with $p\\equiv-1\\pmod\\ell$, at $\\ell=2,3,5$. All three equalities fail. Exhaustive search through $150000$ gives the first failures within these respective candidate domains:\n\\begin{center}\n\\begin{tabular}{rrrr}\n\\toprule\n$\\ell$&$p$&$v_\\ell(\\tau(p))$&$v_\\ell(p+1)$","\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $p=1249$, the $5$-adic discrepancy goes the other way: the valuations are $3$ and $4$. The decisive integers were independently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search t","\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $p=1249$, the $5$-adic discrepancy goes the other way: the valuations are $3$ and $4$. The decisive integers were independently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search through $1","\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $p=1249$, the $5$-adic discrepancy goes the other way: the valuations are $3$ and $4$. The decisive integers were independently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search through $150000$. All decisive values were independently checked. The","\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $p=1249$, the $5$-adic discrepancy goes the other way: the valuations are $3$ and $4$. The decisive integers were independently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search through $150000$. All decisive values were independently checked. The equaliti","\\bottomrule\n\\end{tabular}\n\\end{center}\nAt $p=1249$, the $5$-adic discrepancy goes the other way: the valuations are $3$ and $4$. The decisive integers were independently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search through $150000$. All decisive values were independently checked. The equalities fail, while Lehmer is not refuted.}","ndependently checked through the Eisenstein convolution. These are counterexamples to the auxiliary valuation equalities, never to Lehmer. They show that higher cancellation cannot be read off from $p+1$ by the stated formula.\n\\figplaceholder{lifting_failures}{The first failures of the three proposed valuation equalities in their stated candidate domains, found by exhaustive prime search through $150000$. All decisive values were independently checked. The equalities fail, while Lehmer is not refuted.}","\\chapter{Primitive content and conserved arithmetic sectors}\\label{ch:content}\n\\section{The coordinate gcd is a real invariant}\nFor $v\\in\\Vn$, let $g(v)$ be the positive gcd of its five coordinates. Since the coordinates occupy all residues modulo $5$, one has $5\\nmid g(v)$. Call $g(v)$ the content of the packet. A primitive packet has content $1$.","The content is compatible with the shell in a precise way. If $g=g(v)$, define\n\\begin{equation}\\label{eq:contentscale}\nu_j=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","ent is compatible with the shell in a precise way. If $g=g(v)$, define\n\\begin{equation}\\label{eq:contentscale}\nu_j=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","cise way. If $g=g(v)$, define\n\\begin{equation}\\label{eq:contentscale}\nu_j=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","$g=g(v)$, define\n\\begin{equation}\\label{eq:contentscale}\nu_j=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","scale}\nu_j=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","=\\frac{v_{gj\\bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","bmod5}}g,\\qquad j=0,\\ldots,4.\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.",".\n\\end{equation}\nThe coordinate permutation restores the standard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","tandard residues after division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","r division. The vector $u$ has sum zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","zero and gcd $1$. Its norm squared is $10n/g^2$. For every integral standard-residue vector of sum zero, the norm is divisible by $5$, because its squared residues sum to $30$, and by $2$, because squares and coordinates agree modulo $2$. Thus $10n/g^2$ is $10$ times an integer, and\n\\[\ng^2\\mid n.\n\\]\nThe map is a bijection from exact-content-$g$ packets on shell $n$ to primitive packets on shell $n/g^2$.","The permutation $j\\mapsto gj\\bmod5$ has sign $\\chifive(g)$. For the four units modulo $5$, multiplication by $1$ is the identity, multiplication by $2$ or $3$ is a four-cycle on the nonzero residues, and multiplication by $4$ is two transpositions. Scaling the degree-$10$ Vandermonde therefore gives\n\\begin{equation}\\label{eq:wscale}\nW(v)=\\chifive(g)g^{10}W(u).\n\\end{equation}\nThe character is not an arbitrary normalization. It records the sign of the residue-restoring permutation.","\\section{The primitive residual and its inverse}\nDefine\n\\begin{equation}\\label{eq:residualdef}\nR(n)=\\sum_{\\substack{v\\in\\Vn\\\\g(v)=1}}W(v).\n\\end{equation}\nPartitioning the packets by content and using \\eqref{eq:wscale} yields\n\\begin{equation}\\label{eq:contentdecomposition}\n\\tau(n)=\\sum_{g^2\\mid n}\\chifive(g)g^{10}R(n/g^2).\n\\end{equation}\nTerms with $5\\mid g$ have character zero and correspond to no actual content sector. They may be included in the formula without adding contributions.","g(v)=1}}W(v).\n\\end{equation}\nPartitioning the packets by content and using \\eqref{eq:wscale} yields\n\\begin{equation}\\label{eq:contentdecomposition}\n\\tau(n)=\\sum_{g^2\\mid n}\\chifive(g)g^{10}R(n/g^2).\n\\end{equation}\nTerms with $5\\mid g$ have character zero and correspond to no actual content sector. They may be included in the formula without adding contributions.","The square-supported arithmetic kernel in \\eqref{eq:contentdecomposition} has inverse\n\\[\nd^2\\longmapsto\\mu(d)\\chifive(d)d^{10}.\n\\]\nIndeed, at a square index $m^2$, the convolution of the two kernels is\n\\[\n\\chifive(m)m^{10}\\sum_{d\\mid m}\\mu(d),\n\\]\nwhich is $1$ at $m=1$ and $0$ otherwise. At nonsquare indices both kernels have no support. Consequently\n\\begin{equation}\\label{eq:mobius}\n\\boxed{R(n)=\\sum_{d^2\\mid n}\\mu(d)\\chifive(d)d^{10}\\tau(n/d^2).}\n\\end{equation}\nBoth formulas are exact for every positive integer. They are not fitted from the first few shells.","At $n=1$, $R(1)=1$. At a prime, only $d=1$ occurs in \\eqref{eq:mobius}, giving\n\\begin{equation}\\label{eq:rprime}\nR(p)=\\tau(p).\n\\end{equation}\nThe primitive extraction has no radial contribution to remove at a prime. This is the simplest reason it cannot, by itself, settle the prime nonvanishing obligation.","\\section{Homogeneous universal nulls preserve each content}\n\\begin{theorem}\\label{thm:homcontent}\nLet $Q$ be a symmetric homogeneous polynomial of degree $d$ that is universally null. Then its weighted sum is zero separately in every exact-content sector on every shell.\n\\end{theorem}\n\\begin{proof}\nLet\n\\[\nR_Q(n)=\\sum_{\\substack{v\\in\\Vn","\\section{Homogeneous universal nulls preserve each content}\n\\begin{theorem}\\label{thm:homcontent}\nLet $Q$ be a symmetric homogeneous polynomial of degree $d$ that is universally null. Then its weighted sum is zero separately in every exact-content sector on every shell.\n\\end{theorem}\n\\begin{proof}\nLet\n\\[\nR_Q(n)=\\sum_{\\substack{v\\in\\Vn\\\\g(v)=1}}W(v)Q(v).\n\\]\nSymmetry removes the residue permutation from $Q$, and homogeneity gives $Q(v)=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","g(v)=1}}W(v)Q(v).\n\\]\nSymmetry removes the residue permutation from $Q$, and homogeneity gives $Q(v)=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","}}W(v)Q(v).\n\\]\nSymmetry removes the residue permutation from $Q$, and homogeneity gives $Q(v)=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","he residue permutation from $Q$, and homogeneity gives $Q(v)=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","n from $Q$, and homogeneity gives $Q(v)=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","=g^dQ(u)$. The scaling bijection yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","yields\n\\begin{equation}\\label{eq:thetacontent}\n\\sum_{v\\in\\Vn}W(v)Q(v)\n=\\sum_{g^2\\mid n}\\chifive(g)g^{10+d}R_Q(n/g^2).\n\\end{equation}\nThe left side is zero. At $n=1$, only $g=1$ contributes, so $R_Q(1)=0$. Inductively, the $g=1$ term is $R_Q(n)$ and every nonzero term with $g>1$ has smaller argument. Thus $R_Q(n)=0$ for every $n$. The contribution of exact content $g$ is its scaling factor times $R_Q(n/g^2)$, so it is zero as well.\n\\end{proof}","\\section{A composite obstruction with opposite conserved masses}\nAt $n=121$, the contents are $1$ and $11$. The exact masses are\n\\[\nR(121)=-25439104668,\\qquad11^{10}R(1)=25937424601.\n\\]\nTheir sum is\n\\[\n\\tau(121)=498319933.\n\\]\nOne can verify these values independently from $\\tau(11)=534612$:\n\\[\n\\tau(121)=\\tau(11)^2-11^{11},\\qquad\nR(121)=\\tau(121)-11^{10}.\n\\]\nThe two content masses have opposite nonzero signs. Every retained homogeneous universal-null correction preserves them separately. Hence no correction in that entire permitted span can make all real corrected packet weights have one common sign at $121$. Increasing the degree cannot remove this obstruction.",")=498319933.\n\\]\nOne can verify these values independently from $\\tau(11)=534612$:\n\\[\n\\tau(121)=\\tau(11)^2-11^{11},\\qquad\nR(121)=\\tau(121)-11^{10}.\n\\]\nThe two content masses have opposite nonzero signs. Every retained homogeneous universal-null correction preserves them separately. Hence no correction in that entire permitted span can make all real corrected packet weights have one common sign at $121$. Increasing the degree cannot remove this obstruction.","The obstruction is composite-specific. A prime shell has content $1$ only. There the conservation law adds no invariant beyond the total. It must not be promoted into an impossibility theorem for prime-specific sign constructions.","\\figplaceholder{content121}{The two exact-content masses at $121$ have opposite signs. Their sum survives, but every permitted finite symmetric polynomial universal null conserves the two masses separately. A common-sign correction at this composite shell is therefore impossible in that class.}","\\section{A rank bound from content}\nSuppose a shell has $N$ $J$-orbits occupying $c$ nonempty content classes. A transfer vector produced by a homogeneous universal null has zero sum in each class. That subspace has dimension $N-c$, so every such transfer matrix has rank at most $N-c$.","The constructed family through degree $28$ attains this bound at selected composite shells:\n\\begin{center}\n\\begin{tabular}{rrrr}\n\\toprule\n$n$&$J$-orbits&Content classes&Rank","\\midrule\n9&4&$1,3$&2","49&22&$1,7$&20","81&31&$1,3,9$&28","\\chapter{Refinement and exact transfer spaces}\\label{ch:transport}\n\\section{Mass vectors and aggregation}\nFix a shell, a finite rational family $Q_1,\\ldots,Q_h$ of universally null corrections, and the occupied sectors at depth $k$. Let their number be $s_k$. Define\n\\[\nt_{k,r}=\\sum_{v\\in S_{k,r}}W(v),\\qquad\n(M_k)_{rj}=\\sum_{v\\in S_{k,r}}W(v)Q_j(v).\n\\]\nThe vector $t_k$ contains sector masses. The columns of $M_k$ are the effects of the permitted corrections. Coefficients $a\\in\\Q^h$ change the mass vector to $t_k+M_ka$.","finement and exact transfer spaces}\\label{ch:transport}\n\\section{Mass vectors and aggregation}\nFix a shell, a finite rational family $Q_1,\\ldots,Q_h$ of universally null corrections, and the occupied sectors at depth $k$. Let their number be $s_k$. Define\n\\[\nt_{k,r}=\\sum_{v\\in S_{k,r}}W(v),\\qquad\n(M_k)_{rj}=\\sum_{v\\in S_{k,r}}W(v)Q_j(v).\n\\]\nThe vector $t_k$ contains sector masses. The columns of $M_k$ are the effects of the permitted corrections. Coefficients $a\\in\\Q^h$ change the mass vector to $t_k+M_ka$.","Mass vectors and aggregation}\nFix a shell, a finite rational family $Q_1,\\ldots,Q_h$ of universally null corrections, and the occupied sectors at depth $k$. Let their number be $s_k$. Define\n\\[\nt_{k,r}=\\sum_{v\\in S_{k,r}}W(v),\\qquad\n(M_k)_{rj}=\\sum_{v\\in S_{k,r}}W(v)Q_j(v).\n\\]\nThe vector $t_k$ contains sector masses. The columns of $M_k$ are the effects of the permitted corrections. Coefficients $a\\in\\Q^h$ change the mass vector to $t_k+M_ka$.","hell, a finite rational family $Q_1,\\ldots,Q_h$ of universally null corrections, and the occupied sectors at depth $k$. Let their number be $s_k$. Define\n\\[\nt_{k,r}=\\sum_{v\\in S_{k,r}}W(v),\\qquad\n(M_k)_{rj}=\\sum_{v\\in S_{k,r}}W(v)Q_j(v).\n\\]\nThe vector $t_k$ contains sector masses. The columns of $M_k$ are the effects of the permitted corrections. Coefficients $a\\in\\Q^h$ change the mass vector to $t_k+M_ka$.","Every depth-$k+1$ sector belongs to exactly one parent at depth $k$. Let $A$ be the corresponding $0$--$1$ aggregation matrix. Summing child contributions gives the exact identities\n\\begin{equation}\\label{eq:aggregate}\nt_k=At_{k+1},\\qquad M_k=AM_{k+1}.\n\\end{equation}\nEvery occupied parent has a child. Each correction column has global total zero.","Put\n\\[\nH_k=\\ker(\\mathbf1^T:\\Q^{s_k}\\to\\Q),\\qquad\nT_k=\\im M_k,\\qquad E_k=H_k/T_k.\n\\]\nThe quotient $E_k$ is the vector space of total-preserving changes that are unavailable from the chosen family. It is not a count of failed sign certificates. A sign certificate may exist even when $E_k$ is nonzero.","\\section{The exact refinement sequence}\n\\begin{theorem}\\label{thm:refinement}\nThere is a short exact sequence\n\\begin{equation}\\label{eq:refinement}\n0\\longrightarrow\n\\frac{\\ker A}{M_{k+1}(\\ker M_k)}\n\\longrightarrow E_{k+1}\\longrightarrow E_k\\longrightarrow0.\n\\end{equation}\nConsequently, if $r_k=\\rank M_k$,\n\\begin{equation}\\label{eq:defectgrowth}\n\\dim E_{k+1}-\\dim E_k\n=(s_{k+1}-s_k)-(r_{k+1}-r_k).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe aggregation map sends $H_{k+1}$ onto $H_k$. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.","inement}\nThere is a short exact sequence\n\\begin{equation}\\label{eq:refinement}\n0\\longrightarrow\n\\frac{\\ker A}{M_{k+1}(\\ker M_k)}\n\\longrightarrow E_{k+1}\\longrightarrow E_k\\longrightarrow0.\n\\end{equation}\nConsequently, if $r_k=\\rank M_k$,\n\\begin{equation}\\label{eq:defectgrowth}\n\\dim E_{k+1}-\\dim E_k\n=(s_{k+1}-s_k)-(r_{k+1}-r_k).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe aggregation map sends $H_{k+1}$ onto $H_k$. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.","is a short exact sequence\n\\begin{equation}\\label{eq:refinement}\n0\\longrightarrow\n\\frac{\\ker A}{M_{k+1}(\\ker M_k)}\n\\longrightarrow E_{k+1}\\longrightarrow E_k\\longrightarrow0.\n\\end{equation}\nConsequently, if $r_k=\\rank M_k$,\n\\begin{equation}\\label{eq:defectgrowth}\n\\dim E_{k+1}-\\dim E_k\n=(s_{k+1}-s_k)-(r_{k+1}-r_k).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe aggregation map sends $H_{k+1}$ onto $H_k$. Given a zero-total parent vector, place each parent entry in one selected child and zero in its other children. The resulting child vector has zero total and maps to the required vector.","Equation \\eqref{eq:aggregate} gives $A(T_{k+1})=T_k$, so aggregation induces a surjective map between the quotients. Its kernel is the quotient of $\\ker A$ by $T_{k+1}\\cap\\ker A$. A transfer $M_{k+1}a$ lies in $\\ker A$ exactly when\n\\[\nAM_{k+1}a=M_ka=0.\n\\]\nThus $T_{k+1}\\cap\\ker A=M_{k+1}(\\ker M_k)$, which proves \\eqref{eq:refinement}. Finally $\\dim E_k=s_k-1-r_k$, giving \\eqref{eq:defectgrowth}.\n\\end{proof}","This is ordinary exact linear algebra applied to the packet data. Refinement introduces $s_{k+1}-s_k$ new detail directions. Effective new column rank resolves $r_{k+1}-r_k$ of them. The formula says exactly how much of the new geometric detail remains unavailable.","\\section{Localization by cancellation, rather than indicators}\nLet the columns of a rational matrix $N$ form a basis of $\\ker M_k$. Then\n\\begin{equation}\\label{eq:detailmatrix}\nB_k=M_{k+1}N,\\qquad AB_k=0.\n\\end{equation}\nThe columns of $B_k$ transfer mass only within the parents: their effects on every parent cancel. They are still combinations of universal null polynomials. This is the valid repair of the failed sector-indicator proposal.","\\section{The exact calculation at 31}\nThe shell at $31$ has six coarse occupied sectors and sixteen terminal $J$-orbit sectors. The constructed ranks are\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\nDegree cutoff&Coarse rank&Orbit rank&Orbit defect&Detail rank","\\bottomrule\n\\end{tabular}\n\\end{center}\nThe zero-total coarse space has dimension $5$, and the zero-total orbit space has dimension $15$. At degree $22$, the available columns fill both. The detail space has dimension $15-5=10$.","The stored two-stage certificate first eliminates five coarse masses, using only five elimination equations. It then uses coefficient vectors in $\\ker M_1$ to eliminate all but one fine mass while preserving every coarse total. The survivor is\n\\[\n-52843168.\n\\]\nThe product engine independently identifies this with $\\tau(31)$. The solve does not require a preassigned tau value in its elimination equations. The final survivor remains an exact finite calculation, not a formula that independently forces a nonzero survivor at every prime.","The degree-$20$ family has an orbit-transfer defect of dimension $1$, but it already supplied a strictly negative sign certificate at $31$. Filling every zero-total direction is stronger than needed for that finite sign task. Conversely, filling them at degree $22$ does not by itself prove a nonzero total. These two facts prevent rank from being misinterpreted as a nonvanishing test.","\\figplaceholder{ranks31}{The exact ranks at $31$. The coarse rank is already full at degree $14$, while the terminal orbit rank reaches $15$ at degree $22$. A degree-$20$ sign certificate exists despite its one-dimensional transfer defect. Rank and nonvanishing are different statements.}","\\section{Other finite saturation profiles}\nFull zero-total orbit rank is obtained in the retained family at\n\\begin{center}\n\\begin{tabular}{rrr}\n\\toprule\nPrime&$J$-orbits&Degree cutoff","\\bottomrule\n\\end{tabular}\n\\end{center}\nThese cutoffs are not asserted optimal among all polynomials. The first five corrections through degree $14$ also had full coarse zero-total rank at all $94$ odd primes below $500$. Among the $250$ odd indices below $500$, the only coarse failures were $9$ and $81$. The content split explains genuine conserved directions in those composite examples.","The fresh transport replay verifies $144$ refinement ranks and the complete rational certificate at $31$. Its results authenticate these finite matrices. A uniform theorem about every prime, every degree requirement or every content-preserving transfer is not obtained by these rank profiles.","\\section{What further completeness would mean}\nA plausible geometric question is whether all-degree universal nulls span every content-preserving transfer at each shell. Chapter~\\ref{ch:interpolation} explains why ordinary polynomial interpolation does not settle it. Even if such completeness were proved, the conserved total at a prime would remain $\\tau(p)$.","\\chapter{Every primitive square survives}\\label{ch:even}\n\\section{The local numerator introduced by content}\nThe transform in \\eqref{eq:mobius} is a Dirichlet convolution of multiplicative functions, so $R$ is multiplicative. At a prime $p$, write\n\\[\na=\\tau(p),\\qquad b=p^{11},\\qquad c=\\chifive(p)p^{10}.\n\\]\nThe local generating factor is\n\\begin{equation}\\label{eq:rlocal}\n\\boxed{\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-cX^2}{1-aX+bX^2}.}\n\\end{equation}\nIt follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,\n\\[\nR(1)=1,\\qquad R(p)=a,\\qquad\nR(p^r)=T_r-cT_{r-2}\\quad(r\\ge2).\n\\]\nThe subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.","write\n\\[\na=\\tau(p),\\qquad b=p^{11},\\qquad c=\\chifive(p)p^{10}.\n\\]\nThe local generating factor is\n\\begin{equation}\\label{eq:rlocal}\n\\boxed{\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-cX^2}{1-aX+bX^2}.}\n\\end{equation}\nIt follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,\n\\[\nR(1)=1,\\qquad R(p)=a,\\qquad\nR(p^r)=T_r-cT_{r-2}\\quad(r\\ge2).\n\\]\nThe subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.","r is\n\\begin{equation}\\label{eq:rlocal}\n\\boxed{\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-cX^2}{1-aX+bX^2}.}\n\\end{equation}\nIt follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,\n\\[\nR(1)=1,\\qquad R(p)=a,\\qquad\nR(p^r)=T_r-cT_{r-2}\\quad(r\\ge2).\n\\]\nThe subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.","}\\label{eq:rlocal}\n\\boxed{\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-cX^2}{1-aX+bX^2}.}\n\\end{equation}\nIt follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,\n\\[\nR(1)=1,\\qquad R(p)=a,\\qquad\nR(p^r)=T_r-cT_{r-2}\\quad(r\\ge2).\n\\]\nThe subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.","local}\n\\boxed{\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-cX^2}{1-aX+bX^2}.}\n\\end{equation}\nIt follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,\n\\[\nR(1)=1,\\qquad R(p)=a,\\qquad\nR(p^r)=T_r-cT_{r-2}\\quad(r\\ge2).\n\\]\nThe subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.","The even primitive blocks are not the same as the even tau blocks. For example,\n\\[\nR(p^2)=a^2-p^{10}(p+\\chifive(p)).\n\\]\nAn argument that $\\tau(p^{2m})\\ne0$ does not alone imply the corresponding difference is nonzero. We now prove the primitive statement directly.","\\begin{theorem}\\label{thm:primitiveeven}\nFor every prime $p$ and every integer $m\\ge1$,\n\\[\nR(p^{2m})\\ne0.\n\\]\nConsequently $R(s^2)\\ne0$ for every positive integer $s$.\n\\end{theorem}","The proof is stronger than the actual-tau specialization in a useful way. For $p\\ne5$, it works for every integer recurrence parameter $a$ with $a\\equiv p+\\chifive(p)\\pmod5$. At $p=5$, it works for every integer $a$. It therefore includes a hypothetical parameter $a=0$ whenever that parameter is compatible with the congruence. The square theorem has not secretly assumed prime nonvanishing.","\\section{A residue eliminates three prime classes}\nSuppose $p\\ne5$ and put $\\chi=\\chifive(p)$. Modulo $5$, $\\chi\\equiv p^2$, so the two recurrence roots may be taken as $p$ and $\\chi$:\n\\[\np+\\chi\\equiv a,\\qquad p\\chi\\equiv p^{11}\\pmod5.\n\\]\nThe complete homogeneous recurrence coefficient is\n\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","lasses}\nSuppose $p\\ne5$ and put $\\chi=\\chifive(p)$. Modulo $5$, $\\chi\\equiv p^2$, so the two recurrence roots may be taken as $p$ and $\\chi$:\n\\[\np+\\chi\\equiv a,\\qquad p\\chi\\equiv p^{11}\\pmod5.\n\\]\nThe complete homogeneous recurrence coefficient is\n\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","ay be taken as $p$ and $\\chi$:\n\\[\np+\\chi\\equiv a,\\qquad p\\chi\\equiv p^{11}\\pmod5.\n\\]\nThe complete homogeneous recurrence coefficient is\n\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","chi$:\n\\[\np+\\chi\\equiv a,\\qquad p\\chi\\equiv p^{11}\\pmod5.\n\\]\nThe complete homogeneous recurrence coefficient is\n\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","icient is\n\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","\\[\nT_r\\equiv C_r=\\sum_{i=0}^r p^{r-i}\\chi^i\\pmod5.\n\\]\nAlso $\\chi p^{10}\\equiv1\\pmod5$. Since $\\chi^2=1$, for $r\\ge2$ the sum telescopes:\n\\begin{equation}\\label{eq:rmod5}\nR(p^r)\\equiv C_r-C_{r-2}\n=p^{r-1}(p+\\chi)\\pmod5.\n\\end{equation}\nThe same expression holds at $r=1$. It is nonzero for $p\\equiv1,2,3\\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\\equiv4\\pmod5$ remain.","\\section{The recurrence polynomial and untied valuations}\nFor $p\\equiv4\\pmod5$, $\\chi=1$ and $a\\equiv0\\pmod5$. The exact recurrence expansion is\n\\begin{equation}\\label{eq:trexpansion}\nT_r=\\sum_{j=0}^{\\lfloor r/2\\rfloor}\n(-1)^j\\binom{r-j}{j}b^ja^{r-2j}.\n\\end{equation}\nIt follows by induction: multiplying the previous row by $a$ and subtracting $b$ times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are $1$, so a strictly least valuation at either end cannot be raised by a coefficient divisible by $p$.",". The exact recurrence expansion is\n\\begin{equation}\\label{eq:trexpansion}\nT_r=\\sum_{j=0}^{\\lfloor r/2\\rfloor}\n(-1)^j\\binom{r-j}{j}b^ja^{r-2j}.\n\\end{equation}\nIt follows by induction: multiplying the previous row by $a$ and subtracting $b$ times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are $1$, so a strictly least valuation at either end cannot be raised by a coefficient divisible by $p$.","or r/2\\rfloor}\n(-1)^j\\binom{r-j}{j}b^ja^{r-2j}.\n\\end{equation}\nIt follows by induction: multiplying the previous row by $a$ and subtracting $b$ times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are $1$, so a strictly least valuation at either end cannot be raised by a coefficient divisible by $p$.","First suppose $a\\ne0$, and write $t=v_p(a)$. In $T_{2m}$, the term indexed by $j$ has valuation at least\n\\[\n2mt+j(11-2t).\n\\]\nIf $t\\le4$, the $j=0$ term $a^{2m}$ has strictly least valuation $2mt$. Likewise $T_{2m-2}$ has valuation $(2m-2)t$. The correction $p^{10}T_{2m-2}$ has valuation\n\\[\n2mt+(10-2t)>2mt.\n\\]\nThus\n\\begin{equation}\\label{eq:eventlow}\nv_p(R(p^{2m}))=2mt,\\qquad t\\le4.\n\\end{equation}","If $t\\ge6$, the constant-in-$a$ term $(-b)^m$ is uniquely least-valued in $T_{2m}$, with valuation $11m$. The analogous term gives $v_p(T_{2m-2})=11(m-1)$. The correction has valuation $11m-1$, now smaller than the first block. Hence\n\\begin{equation}\\label{eq:eventhigh}\nv_p(R(p^{2m}))=11m-1,\\qquad t\\ge6.\n\\end{equation}\nActual nonzero tau parameters at $p\\ge5$ cannot have $t\\ge6$ by Deligne's bound. The wider argument is nonetheless a useful exact control for the recurrence class.","If $a=0$, do not assign it a finite $t$. Direct substitution gives\n\\begin{equation}\\label{eq:evenzeroa}\nR(p^{2m})=(-1)^mp^{11m-1}(p+1),\n\\end{equation}\nwhich is nonzero. This branch is indispensable: a theorem purporting to survive a hypothetical prime zero must actually evaluate that case.","\\section{The tied valuation \\texorpdfstring{$t=5$}{t=5}}\nThe preceding endpoint arguments leave $t=5$. Write $a=p^5u$ with $p\\nmid u$, and set\n\\[\nH_r(u)=p^{-5r}T_r.\n\\]\nEquation \\eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2","\\section{The tied valuation \\texorpdfstring{$t=5$}{t=5}}\nThe preceding endpoint arguments leave $t=5$. Write $a=p^5u$ with $p\\nmid u$, and set\n\\[\nH_r(u)=p^{-5r}T_r.\n\\]\nEquation \\eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","t=5$. Write $a=p^5u$ with $p\\nmid u$, and set\n\\[\nH_r(u)=p^{-5r}T_r.\n\\]\nEquation \\eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","$a=p^5u$ with $p\\nmid u$, and set\n\\[\nH_r(u)=p^{-5r}T_r.\n\\]\nEquation \\eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","5r}T_r.\n\\]\nEquation \\eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","ref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is\n\\[\nH_0=1,\\qquad H_1=u,\\qquad H_r=uH_{r-1}-pH_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","H_{r-2}.\n\\]\nThe primitive block becomes\n\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","\\begin{equation}\\label{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","{eq:tied}\nR(p^{2m})=p^{10m}\\bigl(H_{2m}(u)-H_{2m-2}(u)\\bigr).\n\\end{equation}\nThe expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are\n\\begin{equation}\\label{eq:fm}\nF_m(0)=(-1)^mp^{m-1}(p+1),\\qquad\nF_m(x)\\equiv x^{m-1}(x-1)\\pmod p.\n\\end{equation}\nThe leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.","Suppose $R(p^{2m})=0$. Then $x=u^2$ is a positive integer root of the monic integer polynomial $F_m$. The integer-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","p^{2m})=0$. Then $x=u^2$ is a positive integer root of the monic integer polynomial $F_m$. The integer-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","u^2$ is a positive integer root of the monic integer polynomial $F_m$. The integer-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","ger root of the monic integer polynomial $F_m$. The integer-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","The integer-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","-root theorem gives $x\\mid p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","p^{m-1}(p+1)$. Because $p\\nmid u$, one obtains $x\\mid p+1$. Reduction in \\eqref{eq:fm} gives $x\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","\\equiv1\\pmod p$. Since $1\\le x\\le p+1$, the only possibilities are\n\\[\nu^2=1\\quad\\text{or}\\quad u^2=p+1.\n\\]\nThe first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives\n\\[\np=(|u|-1)(|u|+1).\n\\]\nFor a prime this forces $|u|=2$ and $p=3$, outside the family $p\\equiv4\\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.","For $m=1$, the polynomial is simply $F_1(x)=x-(p+1)$. For larger $m$, its degree grows, but the same constant-term and residue argument controls any integral square root. That is why no bound on $m$ is needed.","\\section{The prime 5 and multiplication over squares}\nAt $p=5$, $\\chifive(5)=0$, so $R(5^r)=T_r$. If $a=0$, its even blocks are $(-5^{11})^m$. If $a\\ne0$, then $2v_5(a)$ is even and cannot equal $11$. In the expansion \\eqref{eq:trexpansion}, either the first term or the constant-in-$a$ term has a unique least valuation at an even exponent. Thus no even block vanishes.","All prime cases have now been covered. If $s=\\prod p^{m_p}$, multiplicativity gives\n\\[\nR(s^2)=\\prod_pR(p^{2m_p}).\n\\]\nEvery factor is nonzero, and $R(1)=1$. This proves Theorem~\\ref{thm:primitiveeven}.","The theorem gives survival, not a common sign. For example,\n\\[\nR(4)=-448,\\qquad R(9)=-54594,\\qquad R(25)=-25499225,\n\\]\nwhile other square values can have different signs. Nor does it furnish a fixed auxiliary-prime unit: a nonzero integer can be divisible by arbitrarily high preassigned moduli at suitable varying primes. Chapter~\\ref{ch:congruences} makes that last observation into an actual-prime theorem.","\\figplaceholder{square_residuals}{Exact primitive residuals at square indices $s^2$ for $1\\le s\\le22$, displayed on a signed logarithmic scale. Their survival is a consequence of the universal theorem, whose proof also allows a compatible zero prime parameter. The graph illustrates the theorem rather than proves it.}","\\chapter{Odd primitive blocks retain exactly the prime obstruction}\\label{ch:odd}\n\\section{Polynomial division before evaluation}\nThe odd Hecke polynomial $T_{2m+1}$ is divisible by $a$. For $m\\ge1$, so is\n\\[\nT_{2m+1}-cT_{2m-1}.\n\\]\nDefine its quotient as an integer polynomial in $a^2$ before substituting the actual value $a=\\tau(p)$. This matters at a hypothetical zero: the quotient exists algebraically even when numerical division by $a$ would be undefined.","ctly the prime obstruction}\\label{ch:odd}\n\\section{Polynomial division before evaluation}\nThe odd Hecke polynomial $T_{2m+1}$ is divisible by $a$. For $m\\ge1$, so is\n\\[\nT_{2m+1}-cT_{2m-1}.\n\\]\nDefine its quotient as an integer polynomial in $a^2$ before substituting the actual value $a=\\tau(p)$. This matters at a hypothetical zero: the quotient exists algebraically even when numerical division by $a$ would be undefined.","\\begin{theorem}\\label{thm:oddmultiplier}\nFor every odd prime $p\\ne5$ and every $m\\ge0$, an odd integer $U_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconsta","\\begin{theorem}\\label{thm:oddmultiplier}\nFor every odd prime $p\\ne5$ and every $m\\ge0$, an odd integer $U_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent t","\\begin{theorem}\\label{thm:oddmultiplier}\nFor every odd prime $p\\ne5$ and every $m\\ge0$, an odd integer $U_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2","\\begin{theorem}\\label{thm:oddmultiplier}\nFor every odd prime $p\\ne5$ and every $m\\ge0$, an odd integer $U_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","very $m\\ge0$, an odd integer $U_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","_{p,m}$ satisfies\n\\[\nR(p^{2m+1})=\\tau(p)U_{p,m}.\n\\]\nThe multiplier is defined by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","by the recurrence polynomial, including when $\\tau(p)=0$. If $\\tau(p)\\ne0$, then\n\\[\nv_2(R(p^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","^{2m+1}))=v_2(\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","\\tau(p)).\n\\]\n\\end{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","nd{theorem}\n\\begin{proof}\nFor $m=0$, take $U_{p,0}=1$. For $m\\ge1$, use the quotient\n\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","\\[\nU(a^2)=\\frac{T_{2m+1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","1}-cT_{2m-1}}a\n\\]\nas polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \\eqref{eq:trexpansion}, it is\n\\begin{equation}\\label{eq:oddconstant}\n(-1)^mb^{m-1}\\bigl((m+1)b+mc\\bigr).\n\\end{equation}\nFor an odd prime $p\\ne5$, both $b=p^{11}$ and $c=\\chifive(p)p^{10}$ are odd. The bracket in \\eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.","The actual $a=\\tau(p)$ is even by \\eqref{eq:parity}. Every nonconstant term in the polynomial in $a^2$ is even after evaluation. Hence the evaluated quotient is odd. It cannot be zero. Multiplication by $a$ gives the theorem, and multiplying a nonzero integer by an odd integer preserves its $2$-adic valuation.\n\\end{proof}","This proof uses parity rather than a delicate size estimate for the quotient. It also exhibits the zero parameter honestly: if $a=0$, every odd primitive block is zero and every even primitive block remains nonzero by Chapter~\\ref{ch:even}.","\\section{The primes 2 and 5}\nAt $p=2$, the residue formula \\eqref{eq:rmod5} gives\n\\[\nR(2^r)\\equiv2^{r-1}(2-1)\\not\\equiv0\\pmod5\n\\]\nfor every $r\\ge1$. Thus no local block at $2$ vanishes.","At $p=5$, the character is zero and $R(5^r)=\\tau(5^r)$. The actual parameter is $\\tau(5)=4830$, with $v_5(a)=1$. In \\eqref{eq:trexpansion}, the $j=0$ term has valuation $r$ and every other term has valuation at least $r+9j$. Hence\n\\[\nv_5(R(5^r))=r\n\\]\nfor every positive exponent, including odd ones.","\\section{The exact zero-equivalence theorem}\n\\begin{theorem}\\label{thm:zeroequivalence}\nFor every positive integer $n$,\n\\[\nR(n)=0\\quad\\Longleftrightarrow\\quad\\tau(n)=0.\n\\]\nMore precisely, either side is zero exactly when some prime occurring to an odd exponent in $n$ has $\\tau(p)=0$.\n\\end{theorem}\n\\begin{proof}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for","\\section{The exact zero-equivalence theorem}\n\\begin{theorem}\\label{thm:zeroequivalence}\nFor every positive integer $n$,\n\\[\nR(n)=0\\quad\\Longleftrightarrow\\quad\\tau(n)=0.\n\\]\nMore precisely, either side is zero exactly when some prime occurring to an odd exponent in $n$ has $\\tau(p)=0$.\n\\end{theorem}\n\\begin{proof}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","roequivalence}\nFor every positive integer $n$,\n\\[\nR(n)=0\\quad\\Longleftrightarrow\\quad\\tau(n)=0.\n\\]\nMore precisely, either side is zero exactly when some prime occurring to an odd exponent in $n$ has $\\tau(p)=0$.\n\\end{theorem}\n\\begin{proof}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","o exactly when some prime occurring to an odd exponent in $n$ has $\\tau(p)=0$.\n\\end{theorem}\n\\begin{proof}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","y when some prime occurring to an odd exponent in $n$ has $\\tau(p)=0$.\n\\end{theorem}\n\\begin{proof}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","f}\nThe primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","imitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\\ref{thm:primereduction} gives the same criterion for $\\tau(n)$.\n\\end{proof}","\\section{The theorem's gain and its limit}\nThe primitive square theorem and the odd multiplier theorem classify the new local factors. They show that the extraction does not add an accidental vanishing caused by the numerator $1-\\chifive(p)p^{10}X^2$. They also give a precise dyadic valuation law at every odd nonzero block.","The remaining obstruction is concentrated in the first prime parameter. For $p\\equiv4\\pmod5$, the hypotheses used in the even theorem permit $a=0$. The resulting blocks are\n\\[\n\\begin{aligned}\nR(p^{2m})&=(-1)^mp^{11m-1}(p+1),&&m\\ge1,","The remaining obstruction is concentrated in the first prime parameter. For $p\\equiv4\\pmod5$, the hypotheses used in the even theorem permit $a=0$. The resulting blocks are\n\\[\n\\begin{aligned}\nR(p^{2m})&=(-1)^mp^{11m-1}(p+1),&&m\\ge1,\\\\\nR(p^{2m+1})&=0,&&m\\ge0.\n\\end{aligned}\n\\]\nThe constant block is $R(1)=1$. This is consistent with the congruence, parity and recurrence inputs. Therefore those inputs and the new square theorem cannot jointly establish $a\\ne0$. They have isolated its role.","R(p^{2m+1})&=0,&&m\\ge0.\n\\end{aligned}\n\\]\nThe constant block is $R(1)=1$. This is consistent with the congruence, parity and recurrence inputs. Therefore those inputs and the new square theorem cannot jointly establish $a\\ne0$. They have isolated its role.","At a prime shell there is no imprimitive part, so the universal target is still\n\\[\n\\sum_{\\substack{v\\in\\mathcal V_p\\\\g(v)=1}}W(v)\n=R(p)=\\tau(p)\\ne0.\n\\]\nThe content geometry has clarified the radial arithmetic. The prime total still needs a global or uniform arithmetic mechanism excluding cancellation to zero.","\\chapter{All finite symmetric polynomial nulls conserve content}\\label{ch:pcall}\n\\section{Removing the homogeneity restriction}\nThe earlier content proof used $Q(gv)=g^dQ(v)$. A finite nonhomogeneous polynomial does not have one scaling degree. In principle, cancellation between its homogeneous pieces might make it universally null while allowing individual pieces to have nonzero sums. The following theorem excludes that escape.","\\begin{theorem}\\label{thm:pcall}\nLet $Q$ be any finite symmetric polynomial over $\\C$, restricted to $\\sum_i v_i=0$. Suppose\n\\[\n\\sum_{v\\in\\Vn}W(v)Q(v)=0\n\\]\nfor every positive integer $n$. Write $Q=\\sum_dQ_d$ in homogeneous degrees. Then every $Q_d$ is separately universally null. Moreover, the weighted $Q$-sum is zero in each exact-content sector of every shell.\n\\end{theorem}","\\section{Harmonic decomposition of an alternating polynomial}\nFor a homogeneous $Q_d$, put $P=DQ_d$ and $M=10+d$. The standard decomposition into harmonic components on the rank-four Euclidean hyperplane is\n\\begin{equation}\\label{eq:harmdecomp}\nP=\\sum_{j\\ge0}s_2^jh_{M-2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus onl","\\section{Harmonic decomposition of an alternating polynomial}\nFor a homogeneous $Q_d$, put $P=DQ_d$ and $M=10+d$. The standard decomposition into harmonic components on the rank-four Euclidean hyperplane is\n\\begin{equation}\\label{eq:harmdecomp}\nP=\\sum_{j\\ge0}s_2^jh_{M-2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ c","\\section{Harmonic decomposition of an alternating polynomial}\nFor a homogeneous $Q_d$, put $P=DQ_d$ and $M=10+d$. The standard decomposition into harmonic components on the rank-four Euclidean hyperplane is\n\\begin{equation}\\label{eq:harmdecomp}\nP=\\sum_{j\\ge0}s_2^jh_{M-2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","c decomposition of an alternating polynomial}\nFor a homogeneous $Q_d$, put $P=DQ_d$ and $M=10+d$. The standard decomposition into harmonic components on the rank-four Euclidean hyperplane is\n\\begin{equation}\\label{eq:harmdecomp}\nP=\\sum_{j\\ge0}s_2^jh_{M-2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","ernating polynomial}\nFor a homogeneous $Q_d$, put $P=DQ_d$ and $M=10+d$. The standard decomposition into harmonic components on the rank-four Euclidean hyperplane is\n\\begin{equation}\\label{eq:harmdecomp}\nP=\\sum_{j\\ge0}s_2^jh_{M-2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","2j},\n\\end{equation}\nwhere $h_\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","\\ell$ is harmonic homogeneous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","ous of degree $\\ell$. It can be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","an be constructed recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","ted recursively from the Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","e Laplacian. For a harmonic $h_\\ell$,\n\\begin{equation}\\label{eq:radiallap}\n\\Delta_H(s_2^jh_\\ell)=4j(\\ell+j+1)s_2^{j-1}h_\\ell.\n\\end{equation}\nThe nonzero coefficients in this formula give uniqueness of the decomposition. The operator commutes with coordinate permutations, so every $h_\\ell$ is alternating when $P$ is alternating. Hence $D$ divides every nonzero $h_\\ell$, and $\\ell\\ge10$. Thus only $j\\le d/2$ can occur.","When $d$ is odd, the shell sum vanishes directly under $J$ as in \\eqref{eq:oddnull}, separately inside each content class. We may therefore concentrate on even $d$, when all harmonic degrees in \\eqref{eq:harmdecomp} are even and the scalar harmonic theta argument applies.","\\section{Radial factors become derivatives}\nFor an alternating harmonic $h_\\ell$, write\n\\[\n\\theta_h(\\tau)=\\sum_{\\substack{v_i\\equiv i\\ (5)","\\section{Radial factors become derivatives}\nFor an alternating harmonic $h_\\ell$, write\n\\[\n\\theta_h(\\tau)=\\sum_{\\substack{v_i\\equiv i\\ (5)\\\\\\sum v_i=0}}\nh(v)\\exp\\left(2\\pi i\\tau\\frac{\\|v\\|^2}{10}\\right).\n\\]\nChapter~\\ref{ch:theta} establishes that it is a scalar cusp form of weight $\\ell+2$. The normalization includes only a fixed scalar if one divides by $288$.","sum v_i=0}}\nh(v)\\exp\\left(2\\pi i\\tau\\frac{\\|v\\|^2}{10}\\right).\n\\]\nChapter~\\ref{ch:theta} establishes that it is a scalar cusp form of weight $\\ell+2$. The normalization includes only a fixed scalar if one divides by $288$.","Each application of $\\DQ=q\\,d/dq$ multiplies the shell coefficient by $\\|v\\|^2/10$. Therefore\n\\begin{equation}\\label{eq:radialderivative}\n\\theta_{s_2^jh}=10^j\\DQ^j\\theta_h.\n\\end{equation}\nDifferentiating a modular transformation law shows that $\\DQ^jf$, for modular weight $w$, is quasimodular of weight $w+2j$ and depth at most $j$. Combining \\eqref{eq:harmdecomp} and \\eqref{eq:radialderivative}, the homogeneous shell series\n\\[\nF_d(\\tau)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q_d(v)\\right)q^n\n\\]\nis quasimodular of weight\n\\[\n(M-2j+2)+2j=12+d,\n\\]\nindependent of the radial index $j$. Its depth is at most $d/2$. Polynomial growth and Gaussian decay justify the termwise differentiations.","dq$ multiplies the shell coefficient by $\\|v\\|^2/10$. Therefore\n\\begin{equation}\\label{eq:radialderivative}\n\\theta_{s_2^jh}=10^j\\DQ^j\\theta_h.\n\\end{equation}\nDifferentiating a modular transformation law shows that $\\DQ^jf$, for modular weight $w$, is quasimodular of weight $w+2j$ and depth at most $j$. Combining \\eqref{eq:harmdecomp} and \\eqref{eq:radialderivative}, the homogeneous shell series\n\\[\nF_d(\\tau)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q_d(v)\\right)q^n\n\\]\nis quasimodular of weight\n\\[\n(M-2j+2)+2j=12+d,\n\\]\nindependent of the radial index $j$. Its depth is at most $d/2$. Polynomial growth and Gaussian decay justify the termwise differentiations.","lderivative}\n\\theta_{s_2^jh}=10^j\\DQ^j\\theta_h.\n\\end{equation}\nDifferentiating a modular transformation law shows that $\\DQ^jf$, for modular weight $w$, is quasimodular of weight $w+2j$ and depth at most $j$. Combining \\eqref{eq:harmdecomp} and \\eqref{eq:radialderivative}, the homogeneous shell series\n\\[\nF_d(\\tau)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q_d(v)\\right)q^n\n\\]\nis quasimodular of weight\n\\[\n(M-2j+2)+2j=12+d,\n\\]\nindependent of the radial index $j$. Its depth is at most $d/2$. Polynomial growth and Gaussian decay justify the termwise differentiations.","uation}\nDifferentiating a modular transformation law shows that $\\DQ^jf$, for modular weight $w$, is quasimodular of weight $w+2j$ and depth at most $j$. Combining \\eqref{eq:harmdecomp} and \\eqref{eq:radialderivative}, the homogeneous shell series\n\\[\nF_d(\\tau)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q_d(v)\\right)q^n\n\\]\nis quasimodular of weight\n\\[\n(M-2j+2)+2j=12+d,\n\\]\nindependent of the radial index $j$. Its depth is at most $d/2$. Polynomial growth and Gaussian decay justify the termwise differentiations.","modular transformation law shows that $\\DQ^jf$, for modular weight $w$, is quasimodular of weight $w+2j$ and depth at most $j$. Combining \\eqref{eq:harmdecomp} and \\eqref{eq:radialderivative}, the homogeneous shell series\n\\[\nF_d(\\tau)=\\sum_{n\\ge1}\\left(\\sum_{v\\in\\Vn}W(v)Q_d(v)\\right)q^n\n\\]\nis quasimodular of weight\n\\[\n(M-2j+2)+2j=12+d,\n\\]\nindependent of the radial index $j$. Its depth is at most $d/2$. Polynomial growth and Gaussian decay justify the termwise differentiations.","\\section{Separating the quasimodular weights}\nWe need only an elementary weight-separation statement. Suppose finitely many quasimodular functions $F_k$ of distinct nonnegative integer weights satisfy\n\\[\nF_k(\\gamma z)=\\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r},\n\\qquad f_{k,0}=F_k,\\qquad p_k\\le k,\n\\]\nfor $\\gamma=\\left[\\begin{smallmatrix}a&b","\\section{Separating the quasimodular weights}\nWe need only an elementary weight-separation statement. Suppose finitely many quasimodular functions $F_k$ of distinct nonnegative integer weights satisfy\n\\[\nF_k(\\gamma z)=\\sum_{r=0}^{p_k}f_{k,r}(z)c^r(cz+d)^{k-r},\n\\qquad f_{k,0}=F_k,\\qquad p_k\\le k,\n\\]\nfor $\\gamma=\\left[\\begin{smallmatrix}a&b\\\\c&d\\end{smallmatrix}\\right]\\in SL_2(\\Z)$. Constants from the usual quasimodular convention can be absorbed into the coefficient functions. Assume $\\sum_kF_k=0$ identically.","c&d\\end{smallmatrix}\\right]\\in SL_2(\\Z)$. Constants from the usual quasimodular convention can be absorbed into the coefficient functions. Assume $\\sum_kF_k=0$ identically.","Fix $z$ in the upper half-plane. The polynomial\n\\begin{equation}\\label{eq:weightpolynomial}\nP_z(c,d)=\\sum_k\\sum_r f_{k,r}(z)c^r(cz+d)^{k-r}\n\\end{equation}\nvanishes at every relatively prime integer pair $(c,d)$, because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime $c$. Infinitely many integers $d$ are coprime to it, so the polynomial in $d$ vanishes identically. Each of its coefficient polynomials in $c$ then vanishes at infinitely many primes, so is identically zero. Hence $P_z$ is the zero polynomial.","ix $z$ in the upper half-plane. The polynomial\n\\begin{equation}\\label{eq:weightpolynomial}\nP_z(c,d)=\\sum_k\\sum_r f_{k,r}(z)c^r(cz+d)^{k-r}\n\\end{equation}\nvanishes at every relatively prime integer pair $(c,d)$, because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime $c$. Infinitely many integers $d$ are coprime to it, so the polynomial in $d$ vanishes identically. Each of its coefficient polynomials in $c$ then vanishes at infinitely many primes, so is identically zero. Hence $P_z$ is the zero polynomial.","al\n\\begin{equation}\\label{eq:weightpolynomial}\nP_z(c,d)=\\sum_k\\sum_r f_{k,r}(z)c^r(cz+d)^{k-r}\n\\end{equation}\nvanishes at every relatively prime integer pair $(c,d)$, because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime $c$. Infinitely many integers $d$ are coprime to it, so the polynomial in $d$ vanishes identically. Each of its coefficient polynomials in $c$ then vanishes at infinitely many primes, so is identically zero. Hence $P_z$ is the zero polynomial.","m_r f_{k,r}(z)c^r(cz+d)^{k-r}\n\\end{equation}\nvanishes at every relatively prime integer pair $(c,d)$, because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime $c$. Infinitely many integers $d$ are coprime to it, so the polynomial in $d$ vanishes identically. Each of its coefficient polynomials in $c$ then vanishes at infinitely many primes, so is identically zero. Hence $P_z$ is the zero polynomial.","rime integer pair $(c,d)$, because every such pair is the bottom row of a modular matrix. These pairs are Zariski dense. To see this without a geometric theorem, fix any positive prime $c$. Infinitely many integers $d$ are coprime to it, so the polynomial in $d$ vanishes identically. Each of its coefficient polynomials in $c$ then vanishes at infinitely many primes, so is identically zero. Hence $P_z$ is the zero polynomial.","Every summand belonging to weight $k$ is homogeneous of total degree $k$ in $(c,d)$. Different homogeneous components of a zero polynomial vanish separately. Evaluate the degree-$k$ component at $(c,d)=(0,1)$. Only $r=0$ survives, giving $F_k(z)=0$. Since $z$ was arbitrary, every $F_k$ is zero.","Apply this to the finite even homogeneous degrees with weights $12+d$. Universal nullness of $Q$ says their sum is zero, after the odd pieces have already vanished. Thus every $F_d$ is zero. Equality of Fourier coefficients proves that every homogeneous $Q_d$ is universally null.","The argument has two independent mechanisms. Modular transformation laws prevent different homogeneous degrees from canceling globally. The arithmetic scaling identity prevents a homogeneous null from moving mass between exact contents. Both are needed. Merely applying the homogeneous scaling formula to a nonhomogeneous $Q$ would skip the first mechanism.","\\section{The strengthened composite boundary}\nAt $n=121$, the opposed content masses remain conserved under every finite symmetric polynomial universal-null correction. This is now true without a degree cutoff, a prescribed finite family or an initial homogeneity assumption. Any real corrected weights of one common sign would force both content totals to have that sign, contradicting the exact values.","The theorem closes one possible escape from the content invariant. It does not constrain all nonpolynomial operations, and it does not produce a new prime invariant. At a prime there is still one content class, and its total is still $\\tau(p)$.","\\chapter{Interpolation is not universal nullness}\\label{ch:interpolation}\n\\section{Coordinates that identify exactly the orbits}\nFix a nonempty shell and put $C=10n$. Consider the normalized even symmetric invariants\n\\begin{equation}\\label{eq:invariantmap}\nI(v)=\\left(\\frac{s_4}{C^2},\\frac{s_3^2}{C^3},\n\\frac{s_3s_5}{C^4},\\frac{s_5^2}{C^5}\\right).\n\\end{equation}\nThey are unchanged by $J$.","\\begin{proposition}\\label{prop:invariantseparation}\nOn a fixed shell, $I(v)=I(w)$ if and only if $w=v$ or $w=Jv$.\n\\end{proposition}\n\\begin{proof}\nEquality gives equal $s_4$ and equal rank-one outer products of the real vectors $(s_3(v),s_5(v))$ and $(s_3(w),s_5(w))$. Those vectors agree up to a common sign, including the zero case. Thus the first five power sums of $v$ and either $w$ or $-w$ agree: $s_1=0$, $s_2=C$ and the remaining three by the preceding comparison.","gin{proposition}\\label{prop:invariantseparation}\nOn a fixed shell, $I(v)=I(w)$ if and only if $w=v$ or $w=Jv$.\n\\end{proposition}\n\\begin{proof}\nEquality gives equal $s_4$ and equal rank-one outer products of the real vectors $(s_3(v),s_5(v))$ and $(s_3(w),s_5(w))$. Those vectors agree up to a common sign, including the zero case. Thus the first five power sums of $v$ and either $w$ or $-w$ agree: $s_1=0$, $s_2=C$ and the remaining three by the preceding comparison.","sition}\\label{prop:invariantseparation}\nOn a fixed shell, $I(v)=I(w)$ if and only if $w=v$ or $w=Jv$.\n\\end{proposition}\n\\begin{proof}\nEquality gives equal $s_4$ and equal rank-one outer products of the real vectors $(s_3(v),s_5(v))$ and $(s_3(w),s_5(w))$. Those vectors agree up to a common sign, including the zero case. Thus the first five power sums of $v$ and either $w$ or $-w$ agree: $s_1=0$, $s_2=C$ and the remaining three by the preceding comparison.","Newton's identities recover the elementary symmetric polynomials from those power sums. The coordinate multisets therefore agree. Each multiset has exactly one coordinate in each residue class modulo $5$, so equal multisets imply $v=w$. Negated multisets imply $w_i=-v_{-i\\bmod5}=J(v)_i$. The converse follows from the even invariance of all four coordinates in \\eqref{eq:invariantmap}.\n\\end{proof}","This removes a potential geometric explanation for an evaluation defect. If an all-degree universal-null family has an additional annihilator on a shell, it cannot arise from two distinct $J$-orbits secretly having the same even symmetric coordinates. It must reflect a further arithmetic or modular restriction.","\\section{An explicit homogeneous interpolation basis}\nLet $v_1,\\ldots,v_N$ represent the distinct $J$-orbits. For an integer $t\\ge0$, define the degree-$10$ symmetric polynomial\n\\begin{equation}\\label{eq:et}\nE_t=s_4s_2^3+t\\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.\n\\end{equation}\nFor two distinct orbits, $E_t(v_i)-E_t(v_j)$ is a nonzero polynomial in $t$ of degree at most $3$, by Proposition~\\ref{prop:invariantseparation}. It has at most three roots. Among all pairs there are at most $3N(N-1)/2$ forbidden values. Hence some integer\n\\[\n0\\le t\\le\\frac{3N(N-1)}2\n\\]\nseparates all $e_i=E_t(v_i)$.","nt the distinct $J$-orbits. For an integer $t\\ge0$, define the degree-$10$ symmetric polynomial\n\\begin{equation}\\label{eq:et}\nE_t=s_4s_2^3+t\\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.\n\\end{equation}\nFor two distinct orbits, $E_t(v_i)-E_t(v_j)$ is a nonzero polynomial in $t$ of degree at most $3$, by Proposition~\\ref{prop:invariantseparation}. It has at most three roots. Among all pairs there are at most $3N(N-1)/2$ forbidden values. Hence some integer\n\\[\n0\\le t\\le\\frac{3N(N-1)}2\n\\]\nseparates all $e_i=E_t(v_i)$.",", define the degree-$10$ symmetric polynomial\n\\begin{equation}\\label{eq:et}\nE_t=s_4s_2^3+t\\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.\n\\end{equation}\nFor two distinct orbits, $E_t(v_i)-E_t(v_j)$ is a nonzero polynomial in $t$ of degree at most $3$, by Proposition~\\ref{prop:invariantseparation}. It has at most three roots. Among all pairs there are at most $3N(N-1)/2$ forbidden values. Hence some integer\n\\[\n0\\le t\\le\\frac{3N(N-1)}2\n\\]\nseparates all $e_i=E_t(v_i)$.","l\n\\begin{equation}\\label{eq:et}\nE_t=s_4s_2^3+t\\,s_3^2s_2^2+t^2s_3s_5s_2+t^3s_5^2.\n\\end{equation}\nFor two distinct orbits, $E_t(v_i)-E_t(v_j)$ is a nonzero polynomial in $t$ of degree at most $3$, by Proposition~\\ref{prop:invariantseparation}. It has at most three roots. Among all pairs there are at most $3N(N-1)/2$ forbidden values. Hence some integer\n\\[\n0\\le t\\le\\frac{3N(N-1)}2\n\\]\nseparates all $e_i=E_t(v_i)$.","For that $t$, set\n\\begin{equation}\\label{eq:li}\nL_i(v)=\\prod_{j\\ne i}\n\\frac{E_t(v)-e_j(s_2(v)/C)^5}{e_i-e_j}.\n\\end{equation}\nEach factor is homogeneous of degree $10$, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree $10(N-1)$. On the shell, $s_2/C=1$, so\n\\[\nL_i(v_j)=\\delta_{ij}.\n\\]\nConsequently, any prescribed rational orbit values $a_1,\\ldots,a_N$ are interpolated by $Q=\\sum_ia_iL_i$. If $N=1$, an ordinary constant suffices.","label{eq:li}\nL_i(v)=\\prod_{j\\ne i}\n\\frac{E_t(v)-e_j(s_2(v)/C)^5}{e_i-e_j}.\n\\end{equation}\nEach factor is homogeneous of degree $10$, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree $10(N-1)$. On the shell, $s_2/C=1$, so\n\\[\nL_i(v_j)=\\delta_{ij}.\n\\]\nConsequently, any prescribed rational orbit values $a_1,\\ldots,a_N$ are interpolated by $Q=\\sum_ia_iL_i$. If $N=1$, an ordinary constant suffices.","prod_{j\\ne i}\n\\frac{E_t(v)-e_j(s_2(v)/C)^5}{e_i-e_j}.\n\\end{equation}\nEach factor is homogeneous of degree $10$, and the denominators are nonzero rational numbers. The resulting polynomial is symmetric homogeneous of degree $10(N-1)$. On the shell, $s_2/C=1$, so\n\\[\nL_i(v_j)=\\delta_{ij}.\n\\]\nConsequently, any prescribed rational orbit values $a_1,\\ldots,a_N$ are interpolated by $Q=\\sum_ia_iL_i$. If $N=1$, an ordinary constant suffices.","\\section{A shell-null interpolant can fail on the first shell}\nLet $m_i$ be the nonzero mass of the $i$th orbit. Interpolation makes the map $Q\\mapsto(m_iQ(v_i))$ onto $\\Q^N$. Thus any zero-total transfer on this one shell can be produced by a homogeneous polynomial.","The condition that its weighted sum vanish on this shell is weaker than requiring it to vanish on every shell. The natural interpolation that kills all but the first orbit at $n=7$ has degree $20$ and zero correction total at that target. Its corrected survivor is $-16744$. Yet its weighted correction sum on shell $1$ is\n\\[\n-\\frac{449}{42371287350}\\ne0.\n\\]\nThe analogous degree-$150$ construction at $31$ also fails already at shell $1$. These exact failures occur at small shells where different, universal nulls are known to supply sign certificates.","condition that its weighted sum vanish on this shell is weaker than requiring it to vanish on every shell. The natural interpolation that kills all but the first orbit at $n=7$ has degree $20$ and zero correction total at that target. Its corrected survivor is $-16744$. Yet its weighted correction sum on shell $1$ is\n\\[\n-\\frac{449}{42371287350}\\ne0.\n\\]\nThe analogous degree-$150$ construction at $31$ also fails already at shell $1$. These exact failures occur at small shells where different, universal nulls are known to supply sign certificates.","vanish on this shell is weaker than requiring it to vanish on every shell. The natural interpolation that kills all but the first orbit at $n=7$ has degree $20$ and zero correction total at that target. Its corrected survivor is $-16744$. Yet its weighted correction sum on shell $1$ is\n\\[\n-\\frac{449}{42371287350}\\ne0.\n\\]\nThe analogous degree-$150$ construction at $31$ also fails already at shell $1$. These exact failures occur at small shells where different, universal nulls are known to supply sign certificates.","\\section{Harmonic projection does not preserve values}\nOne might try to interpolate first and then take a harmonic projection. There is no general principle preserving the target-shell values under this operation. Since $D$ is harmonic of degree $10$ on a rank-four space,\n\\[\n\\Delta_H(s_2D)=48D.\n\\]\nThe degree-$12$ top harmonic component of $s_2D$ is zero: the entire polynomial is a radial factor times a lower-degree harmonic polynomial. Yet $s_2D$ is nonzero at every packet on a nonempty shell.","e values}\nOne might try to interpolate first and then take a harmonic projection. There is no general principle preserving the target-shell values under this operation. Since $D$ is harmonic of degree $10$ on a rank-four space,\n\\[\n\\Delta_H(s_2D)=48D.\n\\]\nThe degree-$12$ top harmonic component of $s_2D$ is zero: the entire polynomial is a radial factor times a lower-degree harmonic polynomial. Yet $s_2D$ is nonzero at every packet on a nonempty shell.","More generally, the top harmonic projection of $s_2^kD$ at degree $10+2k$ is zero for every $k>0$. Radial information can disappear. Replacing an interpolant by its top harmonic piece may therefore destroy exactly the values it was designed to have.","\\section{The strongest conceivable content fullness}\nLet the orbits be grouped by exact content and let $T_g$ be the mass of each group. Suppose, as an unproved hypothesis, that all-degree universal null transfers span every vector with zero sum in each content class. Then a strictly positive corrected mass on every orbit exists exactly when every nonempty $T_g$ is positive. Necessity is conservation. For sufficiency, distribute $T_g$ equally among its orbits, then use fullness to realize the difference from the original mass vector.","content fullness}\nLet the orbits be grouped by exact content and let $T_g$ be the mass of each group. Suppose, as an unproved hypothesis, that all-degree universal null transfers span every vector with zero sum in each content class. Then a strictly positive corrected mass on every orbit exists exactly when every nonempty $T_g$ is positive. Necessity is conservation. For sufficiency, distribute $T_g$ equally among its orbits, then use fullness to realize the difference from the original mass vector.","The negative version is identical. For a weakly nonnegative certificate with at least one strict mass, the criterion is $T_g\\ge0$ for every class and $\\sum_gT_g>0$.","At a prime there is one class. Under even this maximal fullness hypothesis, the existence of a common-sign certificate is therefore equivalent to $\\tau(p)\\ne0$. A concrete independently verified certificate can prove nonvanishing. The mere availability of every conserved redistribution cannot decide whether the conserved total vanishes.","\\chapter{Every fixed congruence depth has actual prime survivors}\\label{ch:congruences}\n\\section{A theorem about actual primes}\nCongruences can disprove a zero at a particular prime, but no fixed finite congruence depth excludes every prime. The following statement concerns the actual coefficients of $\\Delta$, not a synthetic recurrence.","\\begin{theorem}\\label{thm:actualsurvivors}\nFor every integer $M\\ge1$, a set of rational primes of positive natural density satisfies\n\\begin{equation}\\label{eq:actualsurvivors}\np\\equiv-1\\pmod{5M},\\qquad\\tau(p)\\equiv0\\pmod M.\n\\end{equation}\nAt every such prime, simultaneously for every $r\\ge1$,\n\\begin{equation}\\label{eq:allrcongruence}\nR(p^r)\\equiv0\\pmod M.\n\\end{equation}\n\\end{theorem}","\\section{Integral representations and complex conjugation}\nFor each prime $\\ell$, the representation attached to $\\Delta$ has the arithmetic-Frobenius normalization\n\\[\n\\rho_\\ell:G_{\\Q}\\longrightarrow GL_2(\\Q_\\ell),\n\\]\nis continuous and unramified outside $\\ell$, and satisfies\n\\begin{equation}\\label{eq:frobpoly}\n\\det(X-\\rho_\\ell(\\Frob_p))=X^2-\\tau(p)X+p^{11},\n\\qquad p\\ne\\ell.\n\\end{equation}\nIts determinant is the eleventh power of the $\\ell$-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the primary sources \\cite{DeligneModular,SerreBourbaki}. Both conventions agree at complex conjugation for the argument below.","omplex conjugation}\nFor each prime $\\ell$, the representation attached to $\\Delta$ has the arithmetic-Frobenius normalization\n\\[\n\\rho_\\ell:G_{\\Q}\\longrightarrow GL_2(\\Q_\\ell),\n\\]\nis continuous and unramified outside $\\ell$, and satisfies\n\\begin{equation}\\label{eq:frobpoly}\n\\det(X-\\rho_\\ell(\\Frob_p))=X^2-\\tau(p)X+p^{11},\n\\qquad p\\ne\\ell.\n\\end{equation}\nIts determinant is the eleventh power of the $\\ell$-adic cyclotomic character. The original cohomological formulation uses a dual convention for Frobenius. The trace and determinant contract in this normalization is explicitly checked in the primary sources \\cite{DeligneModular,SerreBourbaki}. Both conventions agree at complex conjugation for the argument below.","A continuous representation of the compact group $G_{\\Q}$ has a stable $\\Z_\\ell$ lattice. To see this directly, take an initial lattice $L_0$. Compactness bounds all its translates inside $\\ell^{-a}L_0$ for some $a$. The module\n\\[\nL=\\sum_{g\\in G_{\\Q}}\\rho_\\ell(g)L_0\n\\]\nis contained in $\\ell^{-a}L_0$ and contains $L_0$. It is finitely generated, since a submodule of a finite free $\\Z_\\ell$ module is finitely generated, and it is invariant by construction. Reductions modulo $\\ell^e$ are therefore defined. No image-surjectivity theorem is required.","l$ lattice. To see this directly, take an initial lattice $L_0$. Compactness bounds all its translates inside $\\ell^{-a}L_0$ for some $a$. The module\n\\[\nL=\\sum_{g\\in G_{\\Q}}\\rho_\\ell(g)L_0\n\\]\nis contained in $\\ell^{-a}L_0$ and contains $L_0$. It is finitely generated, since a submodule of a finite free $\\Z_\\ell$ module is finitely generated, and it is invariant by construction. Reductions modulo $\\ell^e$ are therefore defined. No image-surjectivity theorem is required.","Let $c$ be complex conjugation. Then $c^2=1$ and the cyclotomic character sends $c$ to $-1$. Consequently $\\det\\rho_\\ell(c)=-1$. In characteristic zero an involution is diagonalizable, with eigenvalues among $1,-1$. The determinant forces one of each, so\n\\begin{equation}\\label{eq:complexconjtrace}\n\\Tr\\rho_\\ell(c)=0\n\\end{equation}\nas an exact $\\Z_\\ell$ identity. This reasoning also covers $\\ell=2$: the trace is computed before reduction. Diagonalizing an involution in characteristic $2$ would not justify the same step.","-1$. Consequently $\\det\\rho_\\ell(c)=-1$. In characteristic zero an involution is diagonalizable, with eigenvalues among $1,-1$. The determinant forces one of each, so\n\\begin{equation}\\label{eq:complexconjtrace}\n\\Tr\\rho_\\ell(c)=0\n\\end{equation}\nas an exact $\\Z_\\ell$ identity. This reasoning also covers $\\ell=2$: the trace is computed before reduction. Diagonalizing an involution in characteristic $2$ would not justify the same step.","\\section{The actual joint image}\nFactor $M=\\prod_{\\ell^e\\Vert M}\\ell^e$ and form the single joint finite map\n\\begin{equation}\\label{eq:jointimage}\n\\Psi:G_{\\Q}\\longrightarrow\n\\left(\\prod_{\\ell^e\\Vert M}GL_2(\\Z/\\ell^e\\Z)\\right)\n\\times(\\Z/5M\\Z)^\\times.\n\\end{equation}\nIts components are the integral representation reductions and the cyclotomic character modulo $5M$. Let $G$ be its actual image. The kernel defines a finite Galois extension of $\\Q$. The image need not be the whole displayed product. Relations between different representations and determinants are allowed.","\\prod_{\\ell^e\\Vert M}\\ell^e$ and form the single joint finite map\n\\begin{equation}\\label{eq:jointimage}\n\\Psi:G_{\\Q}\\longrightarrow\n\\left(\\prod_{\\ell^e\\Vert M}GL_2(\\Z/\\ell^e\\Z)\\right)\n\\times(\\Z/5M\\Z)^\\times.\n\\end{equation}\nIts components are the integral representation reductions and the cyclotomic character modulo $5M$. Let $G$ be its actual image. The kernel defines a finite Galois extension of $\\Q$. The image need not be the whole displayed product. Relations between different representations and determinants are allowed.","Let $C$ be the conjugacy class of $\\Psi(c)$ in $G$. Chebotarev gives unramified primes whose Frobenius lies in $C$, with density $|C|/|G|>0$ \\cite{Chebotarev}. The trace in each representation component is zero by \\eqref{eq:complexconjtrace}, and the cyclotomic component is $-1$. Thus every $\\ell^e$ dividing $M$ divides $\\tau(p)$, while $p\\equiv-1\\pmod{5M}$. This proves \\eqref{eq:actualsurvivors}.","For $M=1$, the representation product is empty. The cyclotomic quotient modulo $5$ suffices. Finitely many ramified primes are omitted in every case, which does not affect positive density.","\\figplaceholder{actual_survivors}{The simultaneous congruence proof uses the actual image of a joint representation. Complex conjugation supplies one existing trace-zero element with cyclotomic value $-1$, and Chebotarev supplies matching prime classes. No independence of the representation factors is assumed.}","\\section{All primitive exponents vanish modulo the fixed modulus}\nAt a prime in \\eqref{eq:actualsurvivors}, one has $\\chifive(p)=1$, $p^{10}\\equiv1\\pmod M$, $p^{11}\\equiv-1\\pmod M$ and $\\tau(p)\\equiv0\\pmod M$. In the formal power series ring $(\\Z/M\\Z)[[X]]$, the primitive factor becomes\n\\[\n\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-p^{10}X^2}{1-\\tau(p)X+p^{11}X^2}\n\\equiv\\frac{1-X^2}{1-X^2}=1.\n\\]\nThe denominator has constant coefficient $1$, so it is invertible even when $M$ is composite or even. No field hypothesis is needed.","nts vanish modulo the fixed modulus}\nAt a prime in \\eqref{eq:actualsurvivors}, one has $\\chifive(p)=1$, $p^{10}\\equiv1\\pmod M$, $p^{11}\\equiv-1\\pmod M$ and $\\tau(p)\\equiv0\\pmod M$. In the formal power series ring $(\\Z/M\\Z)[[X]]$, the primitive factor becomes\n\\[\n\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-p^{10}X^2}{1-\\tau(p)X+p^{11}X^2}\n\\equiv\\frac{1-X^2}{1-X^2}=1.\n\\]\nThe denominator has constant coefficient $1$, so it is invertible even when $M$ is composite or even. No field hypothesis is needed.","}\nAt a prime in \\eqref{eq:actualsurvivors}, one has $\\chifive(p)=1$, $p^{10}\\equiv1\\pmod M$, $p^{11}\\equiv-1\\pmod M$ and $\\tau(p)\\equiv0\\pmod M$. In the formal power series ring $(\\Z/M\\Z)[[X]]$, the primitive factor becomes\n\\[\n\\sum_{r\\ge0}R(p^r)X^r\n=\\frac{1-p^{10}X^2}{1-\\tau(p)X+p^{11}X^2}\n\\equiv\\frac{1-X^2}{1-X^2}=1.\n\\]\nThe denominator has constant coefficient $1$, so it is invertible even when $M$ is composite or even. No field hypothesis is needed.","Coefficient extraction gives \\eqref{eq:allrcongruence}. The constant coefficient remains $R(1)=1$. Equivalently, the reduced Hecke recurrence has $T_0=1$, $T_1=0$ and $T_r=T_{r-2}$. Even $T_r$ are $1$, odd $T_r$ are $0$, and the primitive subtraction kills every positive exponent.","This includes nonzero even primitive blocks. Theorem~\\ref{thm:primitiveeven} says they are nonzero integers. The present theorem says that at suitable varying primes they are divisible by any preassigned $M$. The two statements are compatible.","\\section{The quantifiers cannot be reversed}\nGiven a fixed finite list of moduli $m_1,\\ldots,m_s$, apply Theorem~\\ref{thm:actualsurvivors} to their least common multiple. Infinitely many actual primes defeat every test in that list whose decisive step is finding a nonzero residue of $\\tau(p)$ or any $R(p^r)$. This remains true even if the test is allowed all positive local exponents.","The assertion is\n\\[\n\\forall M\\ \\exists\\text{ a positive-density set of suitable }p.\n\\]\nIt is not\n\\[\n\\exists p\\ \\forall M:\\ M\\mid\\tau(p).\n\\]\nThe surviving primes may change with $M$. No passage to an exact integer zero is available by reversing the quantifiers.","Nor does the theorem exclude a modulus chosen adaptively from $p$, such as the $p^6$ criterion. It excludes a fixed finite collection of nonzero-residue tests of the stated type. Congruences used for other purposes and global arguments remain possible.","The fresh arithmetic replay searched twenty-four selected moduli through $10000$. Twenty-two had a simultaneous witness, producing $704$ checked primitive congruences through exponent $32$. The moduli $11$ and $19$ had no witness in that bounded range. Their absence is consistent with a positive-density theorem and does not refute it. Synthetic recurrence tests add $24576$ exact checks, but the universal prime-existence proof is the Galois argument above, not those counts.","\\begin{theorem}\\label{thm:localphi}\nFor every prime $p$, there is a weakly admissible two-dimensional filtered $\\varphi$-module over $\\Q_p$, with monodromy $N=0$, filtration jumps $0,11$ and Frobenius characteristic polynomial $X^2+p^{11}$.\n\\end{theorem}","This is a local existence theorem. It neither identifies the object with the restriction of $\\Delta$'s global representation nor asserts the existence of a level-one modular form with a prescribed zero coefficient.","\\section{The matrix and filtration}\nTake $D=\\Q_pe_1+\\Q_pe_2$ and define\n\\[\n\\varphi(e_1)=e_2,\\qquad\\varphi(e_2)=-p^{11}e_1,\\qquad N=0.\n\\]\nIn the displayed basis,\n\\[\n\\varphi=\\begin{pmatrix}0&-p^{11}","\\section{The matrix and filtration}\nTake $D=\\Q_pe_1+\\Q_pe_2$ and define\n\\[\n\\varphi(e_1)=e_2,\\qquad\\varphi(e_2)=-p^{11}e_1,\\qquad N=0.\n\\]\nIn the displayed basis,\n\\[\n\\varphi=\\begin{pmatrix}0&-p^{11}\\\\1&0\\end{pmatrix},\\qquad\n\\Tr\\varphi=0,\\qquad\\det\\varphi=p^{11},\\qquad\n\\varphi^2=-p^{11}I.\n\\]\nChoose any $\\Q_p$-line $L\\subset D$ and put\n\\[\n\\Fil^iD=\n\\begin{cases}\nD&i\\le0,","1&0\\end{pmatrix},\\qquad\n\\Tr\\varphi=0,\\qquad\\det\\varphi=p^{11},\\qquad\n\\varphi^2=-p^{11}I.\n\\]\nChoose any $\\Q_p$-line $L\\subset D$ and put\n\\[\n\\Fil^iD=\n\\begin{cases}\nD&i\\le0,","1&0\\end{pmatrix},\\qquad\n\\Tr\\varphi=0,\\qquad\\det\\varphi=p^{11},\\qquad\n\\varphi^2=-p^{11}I.\n\\]\nChoose any $\\Q_p$-line $L\\subset D$ and put\n\\[\n\\Fil^iD=\n\\begin{cases}\nD&i\\le0,\\\\\nL&1\\le i\\le11,\\\\\n0&i\\ge12.\n\\end{cases}\n\\]\nThe two nonzero associated graded pieces are $\\operatorname{gr}^0D=D/L$ and $\\operatorname{gr}^{11}D=L$, each of dimension $1$.","0&i\\ge12.\n\\end{cases}\n\\]\nThe two nonzero associated graded pieces are $\\operatorname{gr}^0D=D/L$ and $\\operatorname{gr}^{11}D=L$, each of dimension $1$.","\\section{Checking weak admissibility}\nFor the chosen cohomological convention, define\n\\[\nt_H(D)=\\sum_i i\\dim\\operatorname{gr}^iD,\\qquad\nt_N(D)=v_p(\\det\\varphi).\n\\]\nWeak admissibility requires $t_H(D)=t_N(D)$ and $t_H(D')\\le t_N(D')$ for every $\\varphi$-stable subobject $D'$, with the induced filtration. The total equality is\n\\[\nt_H(D)=11=t_N(D).\n\\]","There is no proper nonzero $\\varphi$-stable $\\Q_p$-subspace. In dimension two such a subspace would be a line, and $\\varphi$ on it would have an eigenvalue $\\lambda\\in\\Q_p$ satisfying\n\\[\n\\lambda^2=-p^{11}.\n\\]\nThis would require $2v_p(\\lambda)=11$, impossible for an integer-valued valuation on $\\Q_p^\\times$. Thus the proper-subobject inequalities are vacuous. The zero object imposes equality $0=0$, and the whole object satisfies the total equality. The filtered module is weakly admissible.","The proof works at every $p$, including $2$. It does not rely on a sign argument over $\\Q_p$ or on quadratic-residue classifications. The odd determinant valuation is sufficient.","\\section{The realization theorem and conventions}\nColmez--Fontaine's theorem makes a weakly admissible filtered module admissible \\cite{CF}. In the present setting $K=K_0=\\Q_p$ and the base Frobenius is the identity. The theorem supplies a two-dimensional semistable $\\Q_p$-representation realizing the filtered module. Since $N=0$, the comparison is crystalline, using $B_{\\mathrm{cris}}=B_{\\mathrm{st}}^{N=0}$.","The filtration jumps $0,11$ and determinant $p^{11}$ are stated in the cohomological convention. Dualizing changes the Galois and Hodge sign conventions. The arithmetic-Frobenius representation in Chapter~\\ref{ch:congruences} and this local filtered object should not be identified merely because a polynomial has been written in a familiar form. The construction's role is compatibility, not that missing identification.","\\section{Purity and slopes}\nThe Newton slopes are $11/2,11/2$. The complex roots of $X^2+p^{11}$ are\n\\[\n\\pm i\\,p^{11/2}.\n\\]\nThey are algebraic integers, since they solve a monic integral polynomial, and each has the expected weight-$11$ absolute value. Thus zero trace is compatible simultaneously with integrality, purity, Deligne's size bound and weak admissibility.","The Newton polygon lies above the Hodge polygon with jumps $0,11$, and the total endpoints agree. There is no contradiction in those polygons. A local condition that is satisfied by this object cannot, by itself, forbid zero trace.","For the actual nonzero $\\tau(p)$ at $p\\ge5$, Deligne's estimate forces $v_p(\\tau(p))\\le5$. A value with valuation at least $6$ would be zero by the size bound. This does not exclude the zero case. It is precisely the case the conjecture asks us to eliminate.","\\figplaceholder{trace_phi}{The constructed rank-two filtered Frobenius module has trace zero, determinant $p^{11}$ and filtration jumps $0,11$. An odd valuation excludes a rational eigenline, making the proper-subobject inequalities vacuous. This local compatibility is distinct from an actual global identification with $\\Delta$.}","Finite matrix checks can authenticate the formula $\\varphi^2=-p^{11}I$. The release replay checks it at all $1229$ primes through $10000$. The absence of a rational eigenline at every prime follows from the valuation proof, not from that finite matrix range. Conversely, neither the finite checks nor the realization theorem provides the additional global exclusion needed for Lehmer.","\\begin{theorem}\\label{thm:countermodel}\nFor every real $B\\ge1$ and integer $M\\ge1$, there exist a prime $q>B$ and a multiplicative integer sequence $f$ with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","every real $B\\ge1$ and integer $M\\ge1$, there exist a prime $q>B$ and a multiplicative integer sequence $f$ with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","$M\\ge1$, there exist a prime $q>B$ and a multiplicative integer sequence $f$ with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","a multiplicative integer sequence $f$ with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","ative integer sequence $f$ with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","with the following properties:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","ies:\n\\begin{enumerate}\n\\item $f(1)=1$ and $f(n)=\\tau(n)$ for every $n\\le B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","B$.\n\\item $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every positive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","sitive integer $n$.\n\\item Every weight-$12$ prime-power recurrence holds:\n\\[\nf(p^r)=f(p)f(p^{r-1})-p^{11}f(p^{r-2}),\\qquad r\\ge2.\n\\]\n\\item $|f(n)|\\le d(n)n^{11/2}$, where $d(n)$ counts the positive divisors of $n$.\n\\item $f(n)$ is odd exactly when $n$ is an odd square, and $f(n)\\equiv n\\sigma(n)\\pmod5$.\n\\item The arithmetic primitive transform $R_f$ is nonzero at every square.\n\\item Nevertheless $f(q)=R_f(q)=0$.\n\\end{enumerate}\n\\end{theorem}","This is a countermodel to an implication from the listed premises alone. It is not a counterexample to Lehmer for the fixed actual form. The sequence is not unconditionally described as nonmodular, because if the chosen actual $\\tau(q)$ were already zero, the construction could coincide with $\\tau$.","\\section{Choosing an actual prime}\nApply Theorem~\\ref{thm:actualsurvivors} with $L=10M$. Its positive-density set contains a prime $q>B$ with\n\\[\nq\\equiv-1\\pmod{50M},\\qquad\\tau(q)\\equiv0\\pmod{10M}.\n\\]\nEnlarging $q$ avoids the finitely many prime divisors of $10M$. In particular $q$ is odd, $q\\ne5$ and $\\chifive(q)=1$.","Set prime parameters\n\\[\na_p=\\tau(p)\\quad(p\\ne q),\\qquad a_q=0.\n\\]\nFor every prime define integral local blocks\n\\[\nF_{p,0}=1,\\qquad F_{p,1}=a_p,\\qquad\nF_{p,r}=a_pF_{p,r-1}-p^{11}F_{p,r-2}.\n\\]\nFor $n=\\prod p^{r_p}$, define $f(n)=\\prod_pF_{p,r_p}$. Unique factorization makes this a well-defined multiplicative integer sequence with the stated recurrence.","\\section{Agreement at every fixed modulus and finite prefix}\nOnly the parameter at $q$ has changed. Every $n\\le B$ is coprime to $q$, so $f(n)=\\tau(n)$ there. At $q$, the old and new first parameters agree modulo $10M$. Induction in the integral recurrence gives\n\\[\nF_{q,r}\\equiv\\tau(q^r)\\pmod{10M}\n\\]\nfor every exponent. All other local blocks agree exactly. Multiplying local congruences gives $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every $n$, not only through a finite range.","ulus and finite prefix}\nOnly the parameter at $q$ has changed. Every $n\\le B$ is coprime to $q$, so $f(n)=\\tau(n)$ there. At $q$, the old and new first parameters agree modulo $10M$. Induction in the integral recurrence gives\n\\[\nF_{q,r}\\equiv\\tau(q^r)\\pmod{10M}\n\\]\nfor every exponent. All other local blocks agree exactly. Multiplying local congruences gives $f(n)\\equiv\\tau(n)\\pmod{10M}$ for every $n$, not only through a finite range.","The parity criterion follows from agreement modulo $2$, and the modulo-$5$ congruence follows from agreement modulo $5$. Thus even globally quantified versions of these retained properties do not remove the zero parameter.","One can incorporate any finite list of fixed congruence moduli by taking their common multiple in $M$. One can incorporate any finite coefficient check by increasing $B$. The resulting sequence and prime may change with $B$ and $M$. No statement that one altered sequence agrees with $\\tau$ modulo every modulus simultaneously is made.","\\section{The full size bound survives}\nAt primes other than $q$, the Hecke roots have absolute value $p^{11/2}$ by Deligne. Formula \\eqref{eq:rootrec} therefore gives\n\\[\n|\\tau(p^r)|\\le(r+1)p^{11r/2}.\n\\]\nAt the modified prime,\n\\[\nF_{q,2m}=(-q^{11})^m,\\qquad F_{q,2m+1}=0.\n\\]\nThese obey the same local bound at every exponent, including zero. Multiplying the local estimates gives\n\\[\n|f(n)|\\le\\prod_{p^{r_p}\\Vert n}(r_p+1)p^{11r_p/2}\n=d(n)n^{11/2}.\n\\]\nA zero coefficient is fully compatible with an upper bound. The global estimate has not detected the modified parameter.","\\section{All primitive squares still survive}\nDefine the arithmetic transform\n\\[\nR_f(n)=\\sum_{d^2\\mid n}\\mu(d)\\chifive(d)d^{10}f(n/d^2).\n\\]\nIt is multiplicative. Its prime blocks are $R_f(p)=a_p$, and for $r\\ge2$,\n\\[\nR_f(p^r)=F_{p,r}-\\chifive(p)p^{10}F_{p,r-2}.\n\\]\nFor $p\\ne q$, these are the actual $R$ blocks and their even terms are nonzero. At $q$, direct substitution and $\\chifive(q)=1$ give\n\\[\nR_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\\ne0,\\qquad m\\ge1.\n\\]\nThus every positive-exponent factor of $R_f(s^2)$ is nonzero, and $R_f(1)=1$, so the transform is nonzero at every square. Meanwhile $R_f(q)=a_q=0$.","^2\\mid n}\\mu(d)\\chifive(d)d^{10}f(n/d^2).\n\\]\nIt is multiplicative. Its prime blocks are $R_f(p)=a_p$, and for $r\\ge2$,\n\\[\nR_f(p^r)=F_{p,r}-\\chifive(p)p^{10}F_{p,r-2}.\n\\]\nFor $p\\ne q$, these are the actual $R$ blocks and their even terms are nonzero. At $q$, direct substitution and $\\chifive(q)=1$ give\n\\[\nR_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\\ne0,\\qquad m\\ge1.\n\\]\nThus every positive-exponent factor of $R_f(s^2)$ is nonzero, and $R_f(1)=1$, so the transform is nonzero at every square. Meanwhile $R_f(q)=a_q=0$.","^{10}f(n/d^2).\n\\]\nIt is multiplicative. Its prime blocks are $R_f(p)=a_p$, and for $r\\ge2$,\n\\[\nR_f(p^r)=F_{p,r}-\\chifive(p)p^{10}F_{p,r-2}.\n\\]\nFor $p\\ne q$, these are the actual $R$ blocks and their even terms are nonzero. At $q$, direct substitution and $\\chifive(q)=1$ give\n\\[\nR_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\\ne0,\\qquad m\\ge1.\n\\]\nThus every positive-exponent factor of $R_f(s^2)$ is nonzero, and $R_f(1)=1$, so the transform is nonzero at every square. Meanwhile $R_f(q)=a_q=0$.",")=a_p$, and for $r\\ge2$,\n\\[\nR_f(p^r)=F_{p,r}-\\chifive(p)p^{10}F_{p,r-2}.\n\\]\nFor $p\\ne q$, these are the actual $R$ blocks and their even terms are nonzero. At $q$, direct substitution and $\\chifive(q)=1$ give\n\\[\nR_f(q^{2m})=(-1)^mq^{11m-1}(q+1)\\ne0,\\qquad m\\ge1.\n\\]\nThus every positive-exponent factor of $R_f(s^2)$ is nonzero, and $R_f(1)=1$, so the transform is nonzero at every square. Meanwhile $R_f(q)=a_q=0$.","\\section{The logical conclusion and the remaining information}\nThe theorem refutes\n\\[\n\\text{the listed arithmetic properties}\\quad\\Longrightarrow\\quad\n\\text{every coefficient is nonzero}.\n\\]\nThe construction does not impose a modular-form realization, the packet-shell identity or the compatible global Galois representations of $\\Delta$. Those are potential sources of additional information, rather than properties already contained in the weaker premises.","There is no need to know whether $\\tau(q)$ is nonzero. If it is nonzero, $f$ differs from $\\tau$. If it is zero, $f=\\tau$ and Lehmer is already false. In either case, an arithmetic sequence with the listed properties and a zero exists. Calling $f$ nonmodular without the extra nonzero assumption would be logically stronger than what has been proved.","\\chapter*{A note to the reader}\n\\addcontentsline{toc}{chapter}{A note to the reader}\nRamanujan's tau function is easy to define and remarkably hard to keep away from zero. Its coefficients arise from one fixed infinite product,\n\\[\n\\Delta(q)=q\\prod_{m\\ge1}(1-q^m)^{24}=\\sum_{n\\ge1}\\tau(n)q^n.\n\\]\nThe question in this volume is whether every one of those integer coefficients survives. Lehmer's conjecture says that it does. A great deal of structure is known, including strong upper bounds, exact recurrences, congruences and large verified ranges. None of these facts, by itself, provides a lower bound that excludes an exact zero at every prime.","\\begin{claimboundary}\nThis edition does not prove Lehmer's conjecture and does not present an integer counterexample. It proves and explains auxiliary statements, records exact finite certificates and identifies the remaining obligation for the actual modular form $\\Delta$. Written proofs, source-dependent theorems and bounded computations retain their distinct meanings throughout.\n\\end{claimboundary}","The primitive residual $R(n)$ plays a central role. It removes packets having a common coordinate factor. We will prove that $R(s^2)$ never vanishes and that $R(n)=0$ exactly when $\\tau(n)=0$. These are useful structural theorems. They do not dispose of a prime zero, because $R(p)=\\tau(p)$. At a prime the radial extraction leaves the original coefficient untouched.","Some prerequisites are used as contracts rather than reproved from the foundations. These include the modularity and Hecke properties of $\\Delta$, harmonic theta modularity, the level-one valence formula, Deligne's representations and bound, Chebotarev's theorem, and the filtered-module classification. Their precise applications and normalizations are stated where used. The bibliography and source appendix identify the relevant primary or author-hosted passages. A reader may take those contracts as established and verify the project-specific algebra in the main text.","The text uses $v_\\ell(x)$ for the exponent of the prime $\\ell$ in a nonzero integer $x$, with $v_\\ell(0)=\\infty$. The quadratic character $\\chifive$ is zero on multiples of $5$, is $1$ on residues $1,4$, and is $-1$ on residues $2,3$. A shell is a set of integral vectors with a fixed norm and prescribed coordinate residues. A packet is one vector in that shell. Its weight can be positive or negative, but is never zero."]}