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Multiplication describes prime factors and their repeated powers. Addition forces those three factorizations to fit one exact relation. The $abc$ conjecture says that this compatibility limits how large the sum can be compared with the distinct primes that appear.","\\BoundaryBox{\\textbf{Present boundary.} This work does not prove the general $abc$ conjecture. The strongest new determinant statements are formal, real analytic and $p$-adic local theorems. A nonzero global arithmetic realization with a uniform radical-scale height bound has not been constructed.}","The research has also made a substantive correction. If $Q=abc/\\rad(abc)$, then the defect relevant to the conjecture is $Q/(ab)=c/\\rad(abc)$. The target $Q=R^{o(1)}$ is false. Its missing factor $ab$ is restored explicitly in every final certificate.","\\section{Radical support and the exact conjecture}\\label{sec:foundations}\n\\subsection{The multiplicative quantity that ignores repetition}\nEvery positive integer has a unique prime factorization\n\\[\n n=\\prod_{p\\mid n}p^{e_p},\\qquad e_p\\ge1.\n\\]\nIts radical is\n\\[\n \\rad(n)=\\prod_{p\\mid n}p,\\qquad \\rad(1)=1.\n\\]\nThe radical retains which primes occur and discards how many times they occur. Thus $\\rad(72)=\\rad(2^3\\cdot3^2)=6$, whereas $72/6=12$ records the repeated-power contribution. If $n$ is squarefree, then $\\rad(n)=n$.","exact conjecture}\\label{sec:foundations}\n\\subsection{The multiplicative quantity that ignores repetition}\nEvery positive integer has a unique prime factorization\n\\[\n n=\\prod_{p\\mid n}p^{e_p},\\qquad e_p\\ge1.\n\\]\nIts radical is\n\\[\n \\rad(n)=\\prod_{p\\mid n}p,\\qquad \\rad(1)=1.\n\\]\nThe radical retains which primes occur and discards how many times they occur. Thus $\\rad(72)=\\rad(2^3\\cdot3^2)=6$, whereas $72/6=12$ records the repeated-power contribution. If $n$ is squarefree, then $\\rad(n)=n$.","he multiplicative quantity that ignores repetition}\nEvery positive integer has a unique prime factorization\n\\[\n n=\\prod_{p\\mid n}p^{e_p},\\qquad e_p\\ge1.\n\\]\nIts radical is\n\\[\n \\rad(n)=\\prod_{p\\mid n}p,\\qquad \\rad(1)=1.\n\\]\nThe radical retains which primes occur and discards how many times they occur. Thus $\\rad(72)=\\rad(2^3\\cdot3^2)=6$, whereas $72/6=12$ records the repeated-power contribution. If $n$ is squarefree, then $\\rad(n)=n$.","ty that ignores repetition}\nEvery positive integer has a unique prime factorization\n\\[\n n=\\prod_{p\\mid n}p^{e_p},\\qquad e_p\\ge1.\n\\]\nIts radical is\n\\[\n \\rad(n)=\\prod_{p\\mid n}p,\\qquad \\rad(1)=1.\n\\]\nThe radical retains which primes occur and discards how many times they occur. Thus $\\rad(72)=\\rad(2^3\\cdot3^2)=6$, whereas $72/6=12$ records the repeated-power contribution. If $n$ is squarefree, then $\\rad(n)=n$.","For a positive primitive additive triple\n\\[\n a+b=c,\\qquad \\gcd(a,b)=1,\n\\]\nall three pairs are coprime. For example, a divisor of $a$ and $c$ also divides $c-a=b$, hence is one. Therefore the prime supports are disjoint and\n\\[\n R=\\rad(abc)=R_aR_bR_c,\\qquad R_u=\\rad(u).\n\\]\nPut $M_u=u/R_u$. The exact decomposition is\n\\[\n abc=R M_aM_bM_c.\n\\]\nThe conjecture compares $c$ with $R$, rather than with $abc$ itself. Since $abc$ already contains $c$, an estimate against the full product would lose the intended content.","rimitive additive triple\n\\[\n a+b=c,\\qquad \\gcd(a,b)=1,\n\\]\nall three pairs are coprime. For example, a divisor of $a$ and $c$ also divides $c-a=b$, hence is one. Therefore the prime supports are disjoint and\n\\[\n R=\\rad(abc)=R_aR_bR_c,\\qquad R_u=\\rad(u).\n\\]\nPut $M_u=u/R_u$. The exact decomposition is\n\\[\n abc=R M_aM_bM_c.\n\\]\nThe conjecture compares $c$ with $R$, rather than with $abc$ itself. Since $abc$ already contains $c$, an estimate against the full product would lose the intended content.","e\n\\[\n a+b=c,\\qquad \\gcd(a,b)=1,\n\\]\nall three pairs are coprime. For example, a divisor of $a$ and $c$ also divides $c-a=b$, hence is one. Therefore the prime supports are disjoint and\n\\[\n R=\\rad(abc)=R_aR_bR_c,\\qquad R_u=\\rad(u).\n\\]\nPut $M_u=u/R_u$. The exact decomposition is\n\\[\n abc=R M_aM_bM_c.\n\\]\nThe conjecture compares $c$ with $R$, rather than with $abc$ itself. Since $abc$ already contains $c$, an estimate against the full product would lose the intended content.","\\begin{conjecture}[Masser--Oesterl\\'e $abc$]\\label{conj:abc}\nFor every real $\\eps>0$ there is a constant $K_\\eps>0$, depending only on $\\eps$, such that every positive integer triple with $a+b=c$ and $\\gcd(a,b)=1$ satisfies\n\\[\n \\boxed{c\\le K_\\eps\\,\\rad(abc)^{1+\\eps}.}\n\\]\n\\end{conjecture}\nThe order of the quantifiers is the statement. First choose $\\eps$. Then choose a constant that works for every primitive triple. The constant may become large as $\\eps$ approaches zero. It may not depend on the particular triple.","nj:abc}\nFor every real $\\eps>0$ there is a constant $K_\\eps>0$, depending only on $\\eps$, such that every positive integer triple with $a+b=c$ and $\\gcd(a,b)=1$ satisfies\n\\[\n \\boxed{c\\le K_\\eps\\,\\rad(abc)^{1+\\eps}.}\n\\]\n\\end{conjecture}\nThe order of the quantifiers is the statement. First choose $\\eps$. Then choose a constant that works for every primitive triple. The constant may become large as $\\eps$ approaches zero. It may not depend on the particular triple.","This formulation is the one stated in Stewart and Yu's primary treatment of the conjecture~\\cite{StewartYu2001}. Replacing $\\le$ by $<$ does not change the all-$\\eps$ conjecture after enlarging the constant. Requiring $a\\le b$ also leaves it unchanged because the two inputs can be exchanged.","\\subsection{Equivalent output-sector formulations}\nThe form best adapted to this research separates the repeated depth of the output from the input prime support.\n\\begin{proposition}[Exact normalization]\\label{prop:defect}\nFor every primitive positive triple,\n\\[\n \\frac cR=\\frac{M_c}{R_aR_b},\\qquad\n Q:=\\frac{abc}{R},\\qquad \\frac Q{ab}=\\frac cR.\n\\]\nConsequently Conjecture~\\ref{conj:abc} is exactly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","output-sector formulations}\nThe form best adapted to this research separates the repeated depth of the output from the input prime support.\n\\begin{proposition}[Exact normalization]\\label{prop:defect}\nFor every primitive positive triple,\n\\[\n \\frac cR=\\frac{M_c}{R_aR_b},\\qquad\n Q:=\\frac{abc}{R},\\qquad \\frac Q{ab}=\\frac cR.\n\\]\nConsequently Conjecture~\\ref{conj:abc} is exactly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","ulations}\nThe form best adapted to this research separates the repeated depth of the output from the input prime support.\n\\begin{proposition}[Exact normalization]\\label{prop:defect}\nFor every primitive positive triple,\n\\[\n \\frac cR=\\frac{M_c}{R_aR_b},\\qquad\n Q:=\\frac{abc}{R},\\qquad \\frac Q{ab}=\\frac cR.\n\\]\nConsequently Conjecture~\\ref{conj:abc} is exactly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","input prime support.\n\\begin{proposition}[Exact normalization]\\label{prop:defect}\nFor every primitive positive triple,\n\\[\n \\frac cR=\\frac{M_c}{R_aR_b},\\qquad\n Q:=\\frac{abc}{R},\\qquad \\frac Q{ab}=\\frac cR.\n\\]\nConsequently Conjecture~\\ref{conj:abc} is exactly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","rt.\n\\begin{proposition}[Exact normalization]\\label{prop:defect}\nFor every primitive positive triple,\n\\[\n \\frac cR=\\frac{M_c}{R_aR_b},\\qquad\n Q:=\\frac{abc}{R},\\qquad \\frac Q{ab}=\\frac cR.\n\\]\nConsequently Conjecture~\\ref{conj:abc} is exactly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","actly equivalent to each all-$\\eps$ assertion\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","K_\\eps R_aR_bR^\\eps,\n \\qquad Q\\le K_\\eps abR^\\eps.\n\\]\n\\end{proposition}\n\\begin{proof}\nUse $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.\n\\end{proof}\nA local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.","\\subsection{Quality measures the exponent in one example}\nDefine the quality of a primitive triple by\n\\[\n q(a,b,c)=\\frac{\\log c}{\\log R}.\n\\]\nThe denominator is positive because $c\\ge2$ and $R\\ge2$. The logarithm base does not matter. The identity $c=R^q$ is exact, although decimal qualities are approximations.","A triple has $q>1$ exactly when $c>R$. It then has more output size than the global distinct-prime product. Such a triple is often called an $abc$ hit. A hit is an interesting example, not a counterexample to the conjecture. The constant $K_\\eps$ is part of the conjecture.","\\begin{center}\n\\begin{tabular}{@{}rrrrl@{}}\n\\toprule\n$a$&$b$&$c$&$R$&Factorization of the repeated terms","\\midrule\n$1$&$8$&$9$&$6$&$8=2^3$, $9=3^2$","$3$&$125$&$128$&$30$&$125=5^3$, $128=2^7$","$1$&$4374$&$4375$&$210$&$4374=2\\cdot3^7$, $4375=5^4\\cdot7$","$2$&$109\\cdot3^{10}$&$23^5$&$15042$&$23^5=6436343$","\\bottomrule\n\\end{tabular}\n\\end{center}\nThe last equality is $2+109\\cdot3^{10}=23^5$. Every factor in the table matters. Its radical is $2\\cdot3\\cdot109\\cdot23=15042$, giving quality about $1.629912$. The third triple has quality about $1.567887$. These are illustrative exact examples, without any claim that either is the largest known quality.","\\begin{proposition}[Finiteness formulation]\nConjecture~\\ref{conj:abc} is equivalent to the assertion that, for every $\\eps>0$, there are only finitely many primitive positive triples with $q(a,b,c)>1+\\eps$.\n\\end{proposition}\n\\begin{proof}\nAssume $abc$, using its bound with $\\eps/2$. A triple of quality above $1+\\eps$ satisfies\n\\[\n R^{1+\\eps}<c\\le K_{\\eps/2}R^{1+\\eps/2}.\n\\]\nHence $R^{\\eps/2}<K_{\\eps/2}$. The radical is bounded. The same conjectural inequality then bounds $c$, leaving finitely many positive triples.","ivalent to the assertion that, for every $\\eps>0$, there are only finitely many primitive positive triples with $q(a,b,c)>1+\\eps$.\n\\end{proposition}\n\\begin{proof}\nAssume $abc$, using its bound with $\\eps/2$. A triple of quality above $1+\\eps$ satisfies\n\\[\n R^{1+\\eps}<c\\le K_{\\eps/2}R^{1+\\eps/2}.\n\\]\nHence $R^{\\eps/2}<K_{\\eps/2}$. The radical is bounded. The same conjectural inequality then bounds $c$, leaving finitely many positive triples.","Conversely, fix $\\eps>0$ and assume the finiteness assertion. Outside its finite exceptional set, $c\\le R^{1+\\eps}$. Choose $K_\\eps$ to be at least one and at least every ratio $c/R^{1+\\eps}$ in the exceptional set. This single constant works for all triples.\n\\end{proof}\nThis equivalence explains why large searches can be valuable and still be logically limited. A computation can list hits in a finite range. It cannot certify the finiteness of the unsearched set.","\\subsection{Why the epsilon cannot be removed}\nThe uniform statement $c\\le KR$ for one absolute constant $K$ is false. There is an elementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induct","\\subsection{Why the epsilon cannot be removed}\nThe uniform statement $c\\le KR$ for one absolute constant $K$ is false. There is an elementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","t $c\\le KR$ for one absolute constant $K$ is false. There is an elementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","e absolute constant $K$ is false. There is an elementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","s false. There is an elementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","ementary family that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","y that proves this without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","his without using $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","g $abc$.\n\\begin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","gin{theorem}[Failure of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","re of the zero-epsilon estimate]\\label{thm:epsilon-zero}\nThe ratios $c/\\rad(abc)$ are unbounded among primitive positive additive triples.\n\\end{theorem}\n\\begin{proof}\nFor $n\\ge0$, take\n\\[\n a_n=1,\\quad b_n=4^{3^n}-1,\\quad c_n=4^{3^n}.\n\\]\nThese are primitive, $b_n$ is odd, and $\\rad(c_n)=2$. We claim\n\\[\n v_3(b_n)=n+1.\n\\]\nAt $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\\ge1$ and $3\\nmid u$, then\n\\[\n x^3-1=3^{k+1}u\\left(1+3^ku+3^{2k-1}u^2\\right).\n\\]\nThe bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.","Because $b_n$ contains $3^{n+1}$, its powerful part contains $3^n$. Thus\n\\[\n \\rad(b_n)\\le \\frac{b_n}{3^n},\\qquad\n R_n=2\\rad(b_n)\\le\\frac{2b_n}{3^n}.\n\\]\nTherefore\n\\[\n \\frac{c_n}{R_n}\\ge\\frac{3^n c_n}{2b_n}>\\frac{3^n}{2}\\longrightarrow\\infty.\n\\]\n\\end{proof}\nThe displayed lower bound does not contradict the positive-$\\eps$ conjecture. It does not exhibit infinitely many triples with a fixed positive quality gap. The output grows doubly exponentially with $n$, while the proved defect lower bound grows exponentially. A large defect lower bound and a fixed positive exponent gap are different assertions.","ins $3^n$. Thus\n\\[\n \\rad(b_n)\\le \\frac{b_n}{3^n},\\qquad\n R_n=2\\rad(b_n)\\le\\frac{2b_n}{3^n}.\n\\]\nTherefore\n\\[\n \\frac{c_n}{R_n}\\ge\\frac{3^n c_n}{2b_n}>\\frac{3^n}{2}\\longrightarrow\\infty.\n\\]\n\\end{proof}\nThe displayed lower bound does not contradict the positive-$\\eps$ conjecture. It does not exhibit infinitely many triples with a fixed positive quality gap. The output grows doubly exponentially with $n$, while the proved defect lower bound grows exponentially. A large defect lower bound and a fixed positive exponent gap are different assertions.","\\subsection{A known unconditional bound}\nStewart and Yu proved an effective bound of a different scale~\\cite[Theorem 1]{StewartYu2001}: there is an effectively computable absolute $C>0$ such that every primitive positive triple satisfies\n\\[\n c<\\exp\\!\\left(CR^{1/3}(\\log R)^3\\right).\n\\]\nIt controls the height in terms of the radical unconditionally. Its exponential dependence is much larger than the power $R^{1+\\eps}$ required by $abc$. It therefore provides a benchmark, without closing the gap this research addresses.","ion{A known unconditional bound}\nStewart and Yu proved an effective bound of a different scale~\\cite[Theorem 1]{StewartYu2001}: there is an effectively computable absolute $C>0$ such that every primitive positive triple satisfies\n\\[\n c<\\exp\\!\\left(CR^{1/3}(\\log R)^3\\right).\n\\]\nIt controls the height in terms of the radical unconditionally. Its exponential dependence is much larger than the power $R^{1+\\eps}$ required by $abc$. It therefore provides a benchmark, without closing the gap this research addresses.","bound}\nStewart and Yu proved an effective bound of a different scale~\\cite[Theorem 1]{StewartYu2001}: there is an effectively computable absolute $C>0$ such that every primitive positive triple satisfies\n\\[\n c<\\exp\\!\\left(CR^{1/3}(\\log R)^3\\right).\n\\]\nIt controls the height in terms of the radical unconditionally. Its exponential dependence is much larger than the power $R^{1+\\eps}$ required by $abc$. It therefore provides a benchmark, without closing the gap this research addresses.","\\subsection{A published claim and a continuing verification dispute}\nMochizuki's four IUT papers were published in 2021. The fourth paper states an application to $abc$~\\cite{MochizukiIUTIV}. Publication and establishment of the claimed proof are distinct questions. Scholze and Stix documented an objection to the comparison underlying IUT III, Corollary 3.12, and Mochizuki published a response disputing their simplifications~\\cite{ScholzeStix2018,MochizukiComments2018}.","A 2026 Project LANA verification report withholds a verdict on whether IUT proves $abc$. It identifies a needed compatibility between the relevant analytic and anabelian comparison maps~\\cite{LANA2026}. That report also distinguishes its objection from accepting every identification in the earlier criticism. This edition follows that precise status: a published proof claim and a documented, unresolved verification dispute. None of the project proofs below depends on treating IUT as an established proof of $abc$, and none claims to repair its disputed comparison.","compatibility between the relevant analytic and anabelian comparison maps~\\cite{LANA2026}. That report also distinguishes its objection from accepting every identification in the earlier criticism. This edition follows that precise status: a published proof claim and a documented, unresolved verification dispute. None of the project proofs below depends on treating IUT as an established proof of $abc$, and none claims to repair its disputed comparison.","\\clearpage\n\\section{The polynomial theorem and the missing integer analogue}\\label{sec:mason}\n\\subsection{Roots play the role of distinct primes}\nFor a nonzero polynomial $F$ over an algebraically closed field, let $\\operatorname{rad}(F)$ be the product of its distinct monic linear factors. Its degree counts distinct roots. Write $n_0(F)=\\deg\\operatorname{rad}(F)$.","The analogy is direct. Prime powers $p^e$ become repeated roots $(X-\\alpha)^e$. A derivative lowers the multiplicity by one. The divisor left over after that operation is $F/\\operatorname{rad}(F)$, just as $n/\\rad(n)$ is the repeated-power part of an integer.","\\begin{theorem}[Mason--Stothers in characteristic zero]\\label{thm:mason}\nLet $k$ be an algebraically closed field of characteristic zero. Suppose $A,B,C\\in k[X]$ are nonzero, pairwise coprime, satisfy $A+B=C$, and are not all constant. Then\n\\[\n \\max\\{\\deg A,\\deg B,\\deg C\\}\\le n_0(ABC)-1.\n\\]\n\\end{theorem}\nThis is the polynomial $abc$ theorem. Snyder gives a short primary proof in a form that also makes the derivative hypothesis explicit~\\cite{Snyder2000}. The proof below spells out the cancellation that will motivate the arithmetic Wronskian.","\\begin{proof}\nSet $W=A'B-AB'$. It is nonzero. Indeed, if it vanished, then $(A/B)'=0$, and characteristic zero would force the rational function $A/B$ to be constant. Coprimality would then force both $A$ and $B$ to be constant, and so also $C$.","If $\\alpha$ is a root of $A$ of multiplicity $e$, then $A'$ is divisible by $(X-\\alpha)^{e-1}$ and $AB'$ by $(X-\\alpha)^e$. Thus $(X-\\alpha)^{e-1}$ divides $W$. The same argument applies to roots of $B$.","Since $C=A+B$,\n\\[\n W=A'C-AC'.\n\\]\nAt a root of $C$ of multiplicity $e$, this expression is divisible by $(X-\\alpha)^{e-1}$. The three root sets are disjoint. Therefore\n\\[\n \\frac{ABC}{\\operatorname{rad}(ABC)}\\mid W.\n\\]\nTaking degrees gives\n\\[\n \\deg A+\\deg B+\\deg C-n_0(ABC)\\le\\deg W\\le\\deg A+\\deg B-1.\n\\]\nHence $\\deg C\\le n_0(ABC)-1$. To bound $A$, apply the same argument to the relation $C+(-B)=A$. To bound $B$, use $C+(-A)=B$. Coprimality and the radical root count are unchanged by these permutations and signs.\n\\end{proof}","A+B$,\n\\[\n W=A'C-AC'.\n\\]\nAt a root of $C$ of multiplicity $e$, this expression is divisible by $(X-\\alpha)^{e-1}$. The three root sets are disjoint. Therefore\n\\[\n \\frac{ABC}{\\operatorname{rad}(ABC)}\\mid W.\n\\]\nTaking degrees gives\n\\[\n \\deg A+\\deg B+\\deg C-n_0(ABC)\\le\\deg W\\le\\deg A+\\deg B-1.\n\\]\nHence $\\deg C\\le n_0(ABC)-1$. To bound $A$, apply the same argument to the relation $C+(-B)=A$. To bound $B$, use $C+(-A)=B$. Coprimality and the radical root count are unchanged by these permutations and signs.\n\\end{proof}","W=A'C-AC'.\n\\]\nAt a root of $C$ of multiplicity $e$, this expression is divisible by $(X-\\alpha)^{e-1}$. The three root sets are disjoint. Therefore\n\\[\n \\frac{ABC}{\\operatorname{rad}(ABC)}\\mid W.\n\\]\nTaking degrees gives\n\\[\n \\deg A+\\deg B+\\deg C-n_0(ABC)\\le\\deg W\\le\\deg A+\\deg B-1.\n\\]\nHence $\\deg C\\le n_0(ABC)-1$. To bound $A$, apply the same argument to the relation $C+(-B)=A$. To bound $B$, use $C+(-A)=B$. Coprimality and the radical root count are unchanged by these permutations and signs.\n\\end{proof}","\\subsection{A sharp example}\nTake $A=X^n$, $B=1-X^n$, $C=1$ for $n\\ge1$. The root sets are disjoint. There is one distinct root of $A$ and $n$ distinct roots of $B$, so $n_0(ABC)=n+1$. The maximal degree is $n$, attaining the theorem's bound. In this example $W=nX^{n-1}$, and the repeated-root divisor is exactly $X^{n-1}$ up to a nonzero scalar.","The proof needs an actual nonzero derivative. In positive characteristic, $X^p+(1-X^p)=1$ has every derivative zero, so the characteristic-zero statement cannot simply be transferred. The correct positive-characteristic forms include a condition excluding common Frobenius powers.","\\subsection{Where the integer argument stops}\nAn ordinary integer has no distinguished derivative with both an exact Leibniz rule and an automatic relation $D(a)+D(b)=D(c)$. Arithmetic derivatives provide a family of Leibniz maps. Additivity at one triple then imposes a lattice constraint on the prime-coordinate values.","The polynomial theorem gets two benefits simultaneously. Repeated roots force a large divisor of $W$, while degree controls $W$ by a sharp expression one less than the input degree sum. For integers, a prime-coordinate Wronskian also has a large divisor. Its absolute size depends on the selected derivative packet. Choosing an additive transverse packet of uniformly small height is a separate arithmetic problem.","\\clearpage\n\\section{Consequences and the scale of the missing estimate}\\label{sec:consequences}\n\\subsection{An asymptotic Fermat consequence}\nThe following implication is elementary and illustrates how repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant.","\\clearpage\n\\section{Consequences and the scale of the missing estimate}\\label{sec:consequences}\n\\subsection{An asymptotic Fermat consequence}\nThe following implication is elementary and illustrates how repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an","ion{Consequences and the scale of the missing estimate}\\label{sec:consequences}\n\\subsection{An asymptotic Fermat consequence}\nThe following implication is elementary and illustrates how repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theo…","implication is elementary and illustrates how repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","rates how repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","repeated powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","powers amplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","mplify the conjecture.\n\\begin{theorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","eorem}[Conditional large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","al large-exponent Fermat statement]\nAssume Conjecture~\\ref{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","{conj:abc}. There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","There is an integer $n_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","_0$ such that, for every integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","y integer $n>n_0$, the equation\n\\[\n x^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","^n+y^n=z^n\n\\]\nhas no positive integer solution.\n\\end{theorem}\n\\begin{proof}\nDivide a putative solution by its common gcd. The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","The resulting $x,y,z$ are pairwise coprime. Apply $abc$ to $a=x^n$, $b=y^n$, $c=z^n$ with the fixed exponent $\\eps=1$. Then\n\\[\n z^n\\le K_1\\rad(xyz)^2\\le K_1(xyz)^2<K_1z^6,\n\\]\nbecause $x,y<z$. As $z\\ge2$, this implies $2^{n-6}<K_1$ whenever $n>6$. Any fixed integer larger than $6+\\log_2\\max(1,K_1)$ excludes all such exponents.\n\\end{proof}\nThe theorem does not give Fermat's Last Theorem for every $n\\ge3$. It supplies only a sufficiently-large-exponent implication, with a threshold depending on a conjectural constant. Fermat's Last Theorem is already an unconditional theorem. Its proof and the subsequent modularity theorem for elliptic curves form established arithmetic context~\\cite{Wiles,TaylorWiles,BCDT}.","Using a smaller fixed $\\eps$ sharpens the displayed exponent six toward three. It still leaves a constant-dependent threshold. Letting $\\eps$ vary with $n$ without controlling $K_\\eps$ would not be a valid way to remove the remaining finite exponents.","\\subsection{Elliptic curves and the modified Szpiro comparison}\nThe marked Frey curve\n\\[\n E:y^2=x(x-a)(x+b)\n\\]\ntranslates the additive triple into an elliptic curve. Its discriminant sees repeated depth, while its conductor sees prime support, with local model corrections at the small primes. This makes elliptic height estimates a natural companion to $abc$.","The precise equivalent statement uses more than the discriminant alone. For an elliptic curve over $\\Q$, let $N$ be its conductor and let $c_4,c_6$ be the invariants of a global minimal integral Weierstrass equation. The modified Szpiro assertion is\n\\[\n \\max\\{|c_4|^3,c_6^2\\}\\le C_\\eps N^{6+\\eps}\n\\]\nfor every $\\eps>0$, with a constant uniform in the elliptic curve. Oesterl\\'e identifies this form as equivalent to the full $abc$ conjecture~\\cite{Oesterle1988}.","tement uses more than the discriminant alone. For an elliptic curve over $\\Q$, let $N$ be its conductor and let $c_4,c_6$ be the invariants of a global minimal integral Weierstrass equation. The modified Szpiro assertion is\n\\[\n \\max\\{|c_4|^3,c_6^2\\}\\le C_\\eps N^{6+\\eps}\n\\]\nfor every $\\eps>0$, with a constant uniform in the elliptic curve. Oesterl\\'e identifies this form as equivalent to the full $abc$ conjecture~\\cite{Oesterle1988}.","This equivalence is cited as an external theorem. We do not derive its full small-prime and minimal-model accounting here. The ordinary statement involving only $|\\Delta_{\\min}|$ should not be silently substituted for the max-invariant formulation. A proposed project route through the Frey curve must preserve the marked integral model and the normalization of its differential, rather than only its rational isogeny class.","can carry nearly all of its finite order globally.\n\\item A logarithmic character can record the residual coherence.\n\\item A formal or analytic determinant can have large contact and be nonzero.\n\\end{enumerate}\nNone of these statements bounds the archimedean size, denominators, or arithmetic degree of a single realization that retains all the contact. The point where those costs are compared with $R_aR_b$ is the point at which the conjecture's quantitative content enters.","\\subsection{The research question to carry forward}\nThe direct output-sector target is\n\\[\n M_c\\le K_\\eps R_aR_bR^\\eps.\n\\]\nIt asks for a uniform explanation of output repetition in terms of input support. The coherence route factors $M_c$ exactly and asks for a bound on the residual product $WC$. The determinant route seeks a nonzero arithmetic witness whose divisibility and global height imply the same inequality.","These are explicit interfaces, not estimates already established. A new local formula is progress within its own scope. A new global gain must reduce the exponent or comparison cost on the correctly normalized defect $c/R$.","\\section{Signed arithmetic COS and the sparse spectral leaf}\\label{sec:sparse}\n\\subsection{What the finite assignment signal records}\nThe spectral branch of the project asks which prime assignments realize a prescribed residue. Its exact object is a finite distribution. Fix a finite abelian group $G$ and $s$ eligible variable support primes. At the $j$th prime choose one of three permitted contributions, represented by $0,\\gamma_j,-\\gamma_j$ in additive group notation. The sum of the chosen contributions is the assignment residue.","ic COS and the sparse spectral leaf}\\label{sec:sparse}\n\\subsection{What the finite assignment signal records}\nThe spectral branch of the project asks which prime assignments realize a prescribed residue. Its exact object is a finite distribution. Fix a finite abelian group $G$ and $s$ eligible variable support primes. At the $j$th prime choose one of three permitted contributions, represented by $0,\\gamma_j,-\\gamma_j$ in additive group notation. The sum of the chosen contributions is the assignment residue.","Let $N(x)$ count assignments producing $x\\in G$. Repeated contributions remain separate assignments, so\n\\[\n M=\\sum_{x\\in G}N(x)=3^s,\\qquad \\mu(x)=N(x)/M.\n\\]\nFor a character $\\chi\\in\\widehat G$, define\n\\[\n \\widehat\\mu(\\chi)=\\sum_x\\mu(x)\\chi(x)\n =\\prod_{j=1}^s\\frac{1+\\chi(\\gamma_j)+\\chi(-\\gamma_j)}3.\n\\]\nOn a cyclic channel $G=\\Z/n\\Z$, this becomes\n\\[\n m(k)=\\prod_{j=1}^s\\frac{1+2\\cos(2\\pi k\\gamma_j/n)}3.\n\\]\nThis is the discrete cosine or COS expression used in the project. It is an exact characteristic function. Individual cosine factors may be negative. No positivity of all Fourier samples is assumed.","Character orthogonality gives the target count\n\\[\n N(h)=\\frac M{|G|}\\sum_{\\chi\\in\\widehat G}\\widehat\\mu(\\chi)\\overline{\\chi(h)}.\n\\]\nThe identity is finite and exact. It solves a representation question after the arithmetic channel and admissible contributions have been specified. It does not by itself bound the height of an integer triple that generated that channel.","\\subsection{Sparsity forces transverse energy}\nLet $E=\\{x:N(x)>0\\}$ and $r=|E|$. Always $r\\le M$. Parseval yields\n\\[\n \\sum_{\\chi\\in\\widehat G}|\\widehat\\mu(\\chi)|^2\n =\\frac{|G|}{M^2}\\sum_xN(x)^2.\n\\]\nCauchy--Schwarz on the occupied support gives $\\sum_xN(x)^2\\ge M^2/r$. Hence\n\\begin{proposition}[Sparse energy barrier]\\label{prop:sparse-energy}\n\\[\n \\sum_{\\chi\\ne1}|\\widehat\\mu(\\chi)|^2\n \\ge\\frac{|G|}{r}-1\\ge\\frac{|G|}{M}-1.\n\\]\n\\end{proposition}\nWhen few assignment atoms lie in a large group, substantial nontrivial Fourier energy is unavoidable. A failure of uniform mixing is therefore a structural feature of the signal, rather than evidence that the arithmetic assignment is invalid.","\\subsection{The hypothetical exception regime is sparse}\nFor a primitive triple put $Q=abc/R$. Suppose $c>R^{1+\\eps}$ for a fixed $\\eps>0$. Since $a,b$ are positive with sum $c$, one has $ab\\ge c-1$. Thus, for large $c$,\n\\[\n Q=\\frac{abc}{R}\\ge\\frac{c(c-1)}R>\\frac12R^{1+2\\eps}.\n\\]\nThe unit group modulo $Q$ has order $\\varphi(Q)$. For every $\\delta>0$,\n\\[\n \\frac Q{\\varphi(Q)}\\ll_\\delta R^\\delta,\n \\qquad 3^{\\omega(R)}\\ll_\\delta R^\\delta.\n\\]\nBoth elementary bounds follow by separating a finite set of small primes. For all sufficiently large primes, $(1-1/p)^{-1}\\le p^\\delta$ and $3\\le p^\\delta$. Every prime dividing $Q$ also divides $R$.","are positive with sum $c$, one has $ab\\ge c-1$. Thus, for large $c$,\n\\[\n Q=\\frac{abc}{R}\\ge\\frac{c(c-1)}R>\\frac12R^{1+2\\eps}.\n\\]\nThe unit group modulo $Q$ has order $\\varphi(Q)$. For every $\\delta>0$,\n\\[\n \\frac Q{\\varphi(Q)}\\ll_\\delta R^\\delta,\n \\qquad 3^{\\omega(R)}\\ll_\\delta R^\\delta.\n\\]\nBoth elementary bounds follow by separating a finite set of small primes. For all sufficiently large primes, $(1-1/p)^{-1}\\le p^\\delta$ and $3\\le p^\\delta$. Every prime dividing $Q$ also divides $R$.","has $ab\\ge c-1$. Thus, for large $c$,\n\\[\n Q=\\frac{abc}{R}\\ge\\frac{c(c-1)}R>\\frac12R^{1+2\\eps}.\n\\]\nThe unit group modulo $Q$ has order $\\varphi(Q)$. For every $\\delta>0$,\n\\[\n \\frac Q{\\varphi(Q)}\\ll_\\delta R^\\delta,\n \\qquad 3^{\\omega(R)}\\ll_\\delta R^\\delta.\n\\]\nBoth elementary bounds follow by separating a finite set of small primes. For all sufficiently large primes, $(1-1/p)^{-1}\\le p^\\delta$ and $3\\le p^\\delta$. Every prime dividing $Q$ also divides $R$.","\\[\n Q=\\frac{abc}{R}\\ge\\frac{c(c-1)}R>\\frac12R^{1+2\\eps}.\n\\]\nThe unit group modulo $Q$ has order $\\varphi(Q)$. For every $\\delta>0$,\n\\[\n \\frac Q{\\varphi(Q)}\\ll_\\delta R^\\delta,\n \\qquad 3^{\\omega(R)}\\ll_\\delta R^\\delta.\n\\]\nBoth elementary bounds follow by separating a finite set of small primes. For all sufficiently large primes, $(1-1/p)^{-1}\\le p^\\delta$ and $3\\le p^\\delta$. Every prime dividing $Q$ also divides $R$.","R}\\ge\\frac{c(c-1)}R>\\frac12R^{1+2\\eps}.\n\\]\nThe unit group modulo $Q$ has order $\\varphi(Q)$. For every $\\delta>0$,\n\\[\n \\frac Q{\\varphi(Q)}\\ll_\\delta R^\\delta,\n \\qquad 3^{\\omega(R)}\\ll_\\delta R^\\delta.\n\\]\nBoth elementary bounds follow by separating a finite set of small primes. For all sufficiently large primes, $(1-1/p)^{-1}\\le p^\\delta$ and $3\\le p^\\delta$. Every prime dividing $Q$ also divides $R$.","If the assignment model has $M=3^s$ with $s\\le\\omega(R)$, choose $\\delta=\\eps/2$ in both estimates. Then\n\\[\n \\frac{\\varphi(Q)}M\\gg_\\eps R^{1+\\eps}.\n\\]\nConsequently a sufficiently large hypothetical exponent-gap exception, represented in this joint unit-group model, lies on the sparse side of the energy barrier. Uniform spectral mixing cannot be the universal finishing argument for this regime.","\\subsection{Exact innovation rank in a cyclic channel}\nLet $G=\\Z/n\\Z$, $\\alpha_x=e^{2\\pi ix/n}$ and\n\\[\n m(k)=\\sum_{x\\in E}\\frac{N(x)}M\\alpha_x^k.\n\\]\nForm the shift matrix $S_d=(m(k+j))_{0\\le k<n,\\,0\\le j\\le d}$. It factors as a Fourier-column matrix times a Vandermonde matrix on the distinct occupied nodes. The Fourier columns are independent, and all weights $N(x)/M$ are nonzero. Therefore\n\\[\n \\boxed{\\operatorname{rank}S_d=\\min(r,d+1).}\n\\]\nBefore degree $r$, a small singular value is a conditioning phenomenon, not an exact reduction in the number of occupied residues.","G=\\Z/n\\Z$, $\\alpha_x=e^{2\\pi ix/n}$ and\n\\[\n m(k)=\\sum_{x\\in E}\\frac{N(x)}M\\alpha_x^k.\n\\]\nForm the shift matrix $S_d=(m(k+j))_{0\\le k<n,\\,0\\le j\\le d}$. It factors as a Fourier-column matrix times a Vandermonde matrix on the distinct occupied nodes. The Fourier columns are independent, and all weights $N(x)/M$ are nonzero. Therefore\n\\[\n \\boxed{\\operatorname{rank}S_d=\\min(r,d+1).}\n\\]\nBefore degree $r$, a small singular value is a conditioning phenomenon, not an exact reduction in the number of occupied residues.","\\begin{theorem}[Exact cyclic Prony reconstruction]\\label{thm:prony}\nThe distribution on a cyclic channel with $r$ occupied residues is determined by the $2r$ exact consecutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","lic Prony reconstruction]\\label{thm:prony}\nThe distribution on a cyclic channel with $r$ occupied residues is determined by the $2r$ exact consecutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","thm:prony}\nThe distribution on a cyclic channel with $r$ occupied residues is determined by the $2r$ exact consecutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","on a cyclic channel with $r$ occupied residues is determined by the $2r$ exact consecutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","sidues is determined by the $2r$ exact consecutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","cutive samples $m(0),\\ldots,m(2r-1)$.\n\\end{theorem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","orem}\n\\begin{proof}\nWrite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","rite\n\\[\n A(z)=\\prod_{x\\in E}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","}(z-\\alpha_x)=z^r+\\sum_{j=0}^{r-1}a_jz^j.\n\\]\nFor every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","For every $k$,\n\\[\n \\sum_{j=0}^{r-1}a_jm(k+j)=-m(k+r).\n\\]\nThe coefficient matrix for $0\\le k<r$ is the Hankel matrix $H=(m(i+j))_{0\\le i,j<r}$. With $V=(\\alpha_x^i)_{i,x}$ and $D=\\operatorname{diag}(N(x)/M)$, one has $H=VDV^{\\mathsf T}$. The square Vandermonde $V$ is invertible and every diagonal entry of $D$ is nonzero. Thus $H$ is invertible, and the $2r$ samples determine the monic annihilator $A$.","Its distinct roots identify the occupied nodes. The first $r$ sample equations then give an invertible Vandermonde system for their weights. Multiplication by the known $M$ recovers every multiplicity.\n\\end{proof}\nThis is an exact reconstruction theorem. It does not promise numerical stability from approximate samples. Nor does a subpolynomial number of samples by itself imply a subpolynomial arithmetic height for the reconstructed assignment.","\\subsection{Polynomial target filters and exact count certificates}\nFor a polynomial $P(z)=\\sum_{j=0}^dc_jz^j$, define the collision quadratic form\n\\[\n \\mathcal E_d(P)=\\frac n{M^2}\\sum_{x\\in E}N(x)^2|P(\\alpha_x)|^2.\n\\]\nIt can be written as a positive semidefinite Gram form in the coefficients of $P$. Let\n\\[\n \\lambda_d(h)=\\min\\{\\mathcal E_d(P):\\deg P\\le d,\\ P(\\alpha_h)=1\\}.\n\\]\nFor every such filter, the target term is present with value $(n/M^2)N(h)^2$, so\n\\[\n N(h)\\le\\frac M{\\sqrt n}\\sqrt{\\lambda_d(h)}.\n\\]\nIf the right-hand side is rigorously less than one, the integer count is zero. An approximate decimal below one is not a certificate without a proved error enclosure.","exact count certificates}\nFor a polynomial $P(z)=\\sum_{j=0}^dc_jz^j$, define the collision quadratic form\n\\[\n \\mathcal E_d(P)=\\frac n{M^2}\\sum_{x\\in E}N(x)^2|P(\\alpha_x)|^2.\n\\]\nIt can be written as a positive semidefinite Gram form in the coefficients of $P$. Let\n\\[\n \\lambda_d(h)=\\min\\{\\mathcal E_d(P):\\deg P\\le d,\\ P(\\alpha_h)=1\\}.\n\\]\nFor every such filter, the target term is present with value $(n/M^2)N(h)^2$, so\n\\[\n N(h)\\le\\frac M{\\sqrt n}\\sqrt{\\lambda_d(h)}.\n\\]\nIf the right-hand side is rigorously less than one, the integer count is zero. An approximate decimal below one is not a certificate without a proved error enclosure.","ates}\nFor a polynomial $P(z)=\\sum_{j=0}^dc_jz^j$, define the collision quadratic form\n\\[\n \\mathcal E_d(P)=\\frac n{M^2}\\sum_{x\\in E}N(x)^2|P(\\alpha_x)|^2.\n\\]\nIt can be written as a positive semidefinite Gram form in the coefficients of $P$. Let\n\\[\n \\lambda_d(h)=\\min\\{\\mathcal E_d(P):\\deg P\\le d,\\ P(\\alpha_h)=1\\}.\n\\]\nFor every such filter, the target term is present with value $(n/M^2)N(h)^2$, so\n\\[\n N(h)\\le\\frac M{\\sqrt n}\\sqrt{\\lambda_d(h)}.\n\\]\nIf the right-hand side is rigorously less than one, the integer count is zero. An approximate decimal below one is not a certificate without a proved error enclosure.","If $h$ is absent, a degree-$r$ polynomial vanishes at every occupied node and takes value one at $\\alpha_h$, so $\\lambda_r(h)=0$. If $h$ is occupied, its degree-$(r-1)$ Lagrange polynomial isolates exactly that node. Thus\n\\[\n \\lambda_{r-1}(h)=\\frac n{M^2}N(h)^2.\n\\]\nAt the innovation degree the count bound is exact. The remaining computational issue is finding the needed filter with controlled exact coefficients and reliable arithmetic, rather than hoping for an unexplained asymptotic gain.","\\subsection{A Galois coordinate with a controlled denominator}\nFor an occupied target define\n\\[\n \\Delta_h(E)=\\prod_{x\\in E\\setminus\\{h\\}}(\\alpha_h-\\alpha_x).\n\\]\nThis is a nonzero algebraic integer in $\\Q(\\zeta_n)$. For every unit $u$ modulo $n$, its conjugate is $\\Delta_{uh}(uE)$. The absolute norm is a nonzero integer. Hence\n\\[\n \\prod_{u\\in(\\Z/n\\Z)^\\times}|\\Delta_{uh}(uE)|\\ge1.\n\\]\nAt least one factor is at least one. In that coordinate, the Lagrange numerator has coefficient $\\ell^1$ norm at most $2^{r-1}$, and dividing by the target denominator preserves that upper bound.","get define\n\\[\n \\Delta_h(E)=\\prod_{x\\in E\\setminus\\{h\\}}(\\alpha_h-\\alpha_x).\n\\]\nThis is a nonzero algebraic integer in $\\Q(\\zeta_n)$. For every unit $u$ modulo $n$, its conjugate is $\\Delta_{uh}(uE)$. The absolute norm is a nonzero integer. Hence\n\\[\n \\prod_{u\\in(\\Z/n\\Z)^\\times}|\\Delta_{uh}(uE)|\\ge1.\n\\]\nAt least one factor is at least one. In that coordinate, the Lagrange numerator has coefficient $\\ell^1$ norm at most $2^{r-1}$, and dividing by the target denominator preserves that upper bound.","h(E)=\\prod_{x\\in E\\setminus\\{h\\}}(\\alpha_h-\\alpha_x).\n\\]\nThis is a nonzero algebraic integer in $\\Q(\\zeta_n)$. For every unit $u$ modulo $n$, its conjugate is $\\Delta_{uh}(uE)$. The absolute norm is a nonzero integer. Hence\n\\[\n \\prod_{u\\in(\\Z/n\\Z)^\\times}|\\Delta_{uh}(uE)|\\ge1.\n\\]\nAt least one factor is at least one. In that coordinate, the Lagrange numerator has coefficient $\\ell^1$ norm at most $2^{r-1}$, and dividing by the target denominator preserves that upper bound.","This removes a catastrophically small denominator in at least one complex coordinate. It does not supply a uniform radical-scale bound for every conjugate or for the arithmetic height. Even if $r=R^{o(1)}$, the quantity $2^r$ need not be $R^{o(1)}$.","\\subsection{What the spectral branch leaves open}\nThe valid endpoint is exact compression and extraction of a sparse arithmetic leaf. A further height theorem must exploit the arithmetic of that leaf, such as its simultaneous logarithmic, norm-form or $S$-unit constraints. The bounded representation count, exact innovation rank and finite Fourier identities do not prove the required inequality on $c/R$.","subsection{What the spectral branch leaves open}\nThe valid endpoint is exact compression and extraction of a sparse arithmetic leaf. A further height theorem must exploit the arithmetic of that leaf, such as its simultaneous logarithmic, norm-form or $S$-unit constraints. The bounded representation count, exact innovation rank and finite Fourier identities do not prove the required inequality on $c/R$.","\\section{Correcting the global target before raising contact}\\label{sec:correction}\n\\subsection{A missing input factor changes the theorem}\nThe September continuation begins with a correction, rather than a larger contact claim. Set\n\\[\n R=\\rad(abc),\\qquad Q=abc/R.\n\\]\nSection~\\ref{sec:foundations} proved the exact identity $Q/(ab)=c/R$. A uniform subpolynomial bound on $Q$ alone is not equivalent to $abc$. It is false unconditionally.","ction{Correcting the global target before raising contact}\\label{sec:correction}\n\\subsection{A missing input factor changes the theorem}\nThe September continuation begins with a correction, rather than a larger contact claim. Set\n\\[\n R=\\rad(abc),\\qquad Q=abc/R.\n\\]\nSection~\\ref{sec:foundations} proved the exact identity $Q/(ab)=c/R$. A uniform subpolynomial bound on $Q$ alone is not equivalent to $abc$. It is false unconditionally.","\\begin{theorem}[Counterfamily for the unbalanced target]\\label{thm:false-target}\nFor every fixed $0\\le\\eta<1$ and $A\\ge0$, no constant $K$ makes\n\\[\n Q\\le K R^\\eta(\\log(3c))^A\n\\]\nvalid for all primitive positive triples.\n\\end{theorem}\n\\begin{proof}\nTake $a=1$, $b=2^k$, $c=2^k+1$ for integers $k\\ge1$. The triple is primitive and $c$ is odd. Thus\n\\[\n R=2\\rad(c)\\le2c\\le3\\cdot2^k,\n \\qquad Q=2^{k-1}\\frac c{\\rad(c)}\\ge2^{k-1}.\n\\]\nConsequently\n\\[\n \\frac{Q}{R^\\eta(\\log(3c))^A}\n \\ge\\frac{2^{(1-\\eta)k-1}}{3^\\eta(\\log(3(2^k+1)))^A}\\longrightarrow\\infty.\n\\]\nAn exponential with positive exponent eventually exceeds every fixed power of $k$.\n\\end{proof}","lanced target]\\label{thm:false-target}\nFor every fixed $0\\le\\eta<1$ and $A\\ge0$, no constant $K$ makes\n\\[\n Q\\le K R^\\eta(\\log(3c))^A\n\\]\nvalid for all primitive positive triples.\n\\end{theorem}\n\\begin{proof}\nTake $a=1$, $b=2^k$, $c=2^k+1$ for integers $k\\ge1$. The triple is primitive and $c$ is odd. Thus\n\\[\n R=2\\rad(c)\\le2c\\le3\\cdot2^k,\n \\qquad Q=2^{k-1}\\frac c{\\rad(c)}\\ge2^{k-1}.\n\\]\nConsequently\n\\[\n \\frac{Q}{R^\\eta(\\log(3c))^A}\n \\ge\\frac{2^{(1-\\eta)k-1}}{3^\\eta(\\log(3(2^k+1)))^A}\\longrightarrow\\infty.\n\\]\nAn exponential with positive exponent eventually exceeds every fixed power of $k$.\n\\end{proof}","em}\n\\begin{proof}\nTake $a=1$, $b=2^k$, $c=2^k+1$ for integers $k\\ge1$. The triple is primitive and $c$ is odd. Thus\n\\[\n R=2\\rad(c)\\le2c\\le3\\cdot2^k,\n \\qquad Q=2^{k-1}\\frac c{\\rad(c)}\\ge2^{k-1}.\n\\]\nConsequently\n\\[\n \\frac{Q}{R^\\eta(\\log(3c))^A}\n \\ge\\frac{2^{(1-\\eta)k-1}}{3^\\eta(\\log(3(2^k+1)))^A}\\longrightarrow\\infty.\n\\]\nAn exponential with positive exponent eventually exceeds every fixed power of $k$.\n\\end{proof}","\\subsection{A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact m","\\subsection{A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make","\\subsection{A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio","\\subsection{A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false targe","\\subsection{A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$","A necessary leading cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control …","ng cost for powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to…","or powers of Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$…","Q}\nA stronger elementary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","tary obstruction measures the exponent cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","nt cost of any all-triple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","iple bound on the unbalanced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","ced quantity.\n\\begin{theorem}[Necessary cost lower bound]\\label{thm:necessary-cost}\nFix $s>0$. Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.",". Suppose $B,A\\ge0$ and $C>0$ satisfy\n\\[\n Q^s\\le C R^B(\\log(3c))^A\n\\]\nfor every primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","primitive positive triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","triple. Then $B\\ge2s$.\n\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","\\end{theorem}\n\\begin{proof}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","of}\nFix an integer $d\\ge2$ and take $a=t^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","^d$, $b=(t+1)^d$, $c=t^d+(t+1)^d$ for $t\\ge1$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","$. The inputs are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","are coprime. As $\\rad(t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","t^d)\\le t$ and $\\rad((t+1)^d)\\le t+1$,\n\\[\n R\\le t(t+1)c,\\qquad\n Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","Q=\\frac{abc}{R}\\ge[t(t+1)]^{d-1}.\n\\]\nThe assumed inequality implies\n\\[\n [t(t+1)]^{s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","s(d-1)}\\le C[t(t+1)c]^B(\\log(3c))^A.\n\\]\nTake logarithms, divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.",", divide by $\\log t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","t$ and send $t$ to infinity. Since $\\log c/\\log t\\to d$ and $\\log\\log(3c)/\\log t\\to0$, this yields\n\\[\n 2s(d-1)\\le B(d+2).\n\\]\nIt holds for every $d$. Sending $d$ to infinity proves $B\\ge2s$.\n\\end{proof}\nNo contact multiplier can make the ratio $B/s$ tend to zero for this false target. A valid certificate must instead include $(ab)^s$ in its global size bound, or directly control $M_c$ relative to $R_aR_b$.","\\figplaceholder{09_balanced_defect}{The balanced defect is $c/R=Q/(ab)=M_c/(R_aR_b)$. Removing the input factor $ab$ changes the target into an unconditional false assertion.}","\\subsection{Two obligations that a new determinant must meet}\nThe proposed determinant program therefore has two separate obligations. It needs a nonzero contact expression, with an actual first nonzero coefficient and local divisibility. It also needs an arithmetic realization whose global size has the correct $(ab)^s$ normalization. The following local theorems solve parts of the first obligation. They leave the second as an explicit hypothesis.","\\clearpage\n\\section{The universal mixed-weight jet filter}\\label{sec:jet}\n\\subsection{One polynomial chooses the cancelling coefficients}\nLet $m\\ge0$ be an integer. Define\n\\[\n P_m(X)=X\\prod_{j=1}^m(jX-1)=\\sum_{s=1}^{m+1}c_{m,s}X^s,\n\\]\nand the formal series\n\\[\n \\Phi_m(t)=\\frac1{(m+1)!}\\sum_{s=1}^{m+1}c_{m,s}\\Li_s(t).\n\\]\nHere $\\Li_s(t)=\\sum_{k\\ge1}t^k/k^s$ is first interpreted in $\\Q[[t]]$. It also defines the usual analytic polylogarithm on $|t|<1$~\\cite{DLMFPolylog}. The label $s$ is its weight. The contact order will follow from its coefficients, rather than from that label.","t jet filter}\\label{sec:jet}\n\\subsection{One polynomial chooses the cancelling coefficients}\nLet $m\\ge0$ be an integer. Define\n\\[\n P_m(X)=X\\prod_{j=1}^m(jX-1)=\\sum_{s=1}^{m+1}c_{m,s}X^s,\n\\]\nand the formal series\n\\[\n \\Phi_m(t)=\\frac1{(m+1)!}\\sum_{s=1}^{m+1}c_{m,s}\\Li_s(t).\n\\]\nHere $\\Li_s(t)=\\sum_{k\\ge1}t^k/k^s$ is first interpreted in $\\Q[[t]]$. It also defines the usual analytic polylogarithm on $|t|<1$~\\cite{DLMFPolylog}. The label $s$ is its weight. The contact order will follow from its coefficients, rather than from that label.","on{One polynomial chooses the cancelling coefficients}\nLet $m\\ge0$ be an integer. Define\n\\[\n P_m(X)=X\\prod_{j=1}^m(jX-1)=\\sum_{s=1}^{m+1}c_{m,s}X^s,\n\\]\nand the formal series\n\\[\n \\Phi_m(t)=\\frac1{(m+1)!}\\sum_{s=1}^{m+1}c_{m,s}\\Li_s(t).\n\\]\nHere $\\Li_s(t)=\\sum_{k\\ge1}t^k/k^s$ is first interpreted in $\\Q[[t]]$. It also defines the usual analytic polylogarithm on $|t|<1$~\\cite{DLMFPolylog}. The label $s$ is its weight. The contact order will follow from its coefficients, rather than from that label.","\\begin{theorem}[Exact jet cancellation]\\label{thm:jet-contact}\n\\[\n \\Phi_m(t)=(-1)^m\\sum_{k\\ge1}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}.\n\\]\nThus\n\\[\n \\ord_0\\Phi_m=m+1,\\qquad\n [t^{m+1}]\\Phi_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation","\\begin{theorem}[Exact jet cancellation]\\label{thm:jet-contact}\n\\[\n \\Phi_m(t)=(-1)^m\\sum_{k\\ge1}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}.\n\\]\nThus\n\\[\n \\ord_0\\Phi_m=m+1,\\qquad\n [t^{m+1}]\\Phi_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","ncellation]\\label{thm:jet-contact}\n\\[\n \\Phi_m(t)=(-1)^m\\sum_{k\\ge1}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}.\n\\]\nThus\n\\[\n \\ord_0\\Phi_m=m+1,\\qquad\n [t^{m+1}]\\Phi_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","1)^m\\sum_{k\\ge1}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}.\n\\]\nThus\n\\[\n \\ord_0\\Phi_m=m+1,\\qquad\n [t^{m+1}]\\Phi_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","\\ord_0\\Phi_m=m+1,\\qquad\n [t^{m+1}]\\Phi_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","_m=\\frac{(-1)^m}{(m+1)^{m+2}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","f}\nThe coefficient of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","ent of $t^k$ is $P_m(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","(1/k)/(m+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","+1)!$. Direct substitution gives\n\\[\n P_m(1/k)=\\frac1k\\prod_{j=1}^m\\left(\\frac jk-1\\right)\n =(-1)^m\\frac{(k-1)\\cdots(k-m)}{k^{m+1}}.\n\\]\nDivide by $(m+1)!$ and multiply numerator and denominator by $k$ to get the asserted binomial coefficient. It vanishes for $1\\le k\\le m$ and first becomes one at $k=m+1$. The coefficient there is nonzero.\n\\end{proof}\nFor $m=0$, this is $\\Phi_0=\\Li_1$. For $m=1$,\n\\[\n \\Phi_1=(\\Li_2-\\Li_1)/2=-t^2/8-t^3/9-\\cdots.\n\\]\nThe sign is part of the exact formula. Changing the weight alone would not produce the cancellation of the linear coefficient.","\\subsection{Small coefficient norm has a limited meaning}\nThe coefficients of $P_m$ alternate in sign. Its absolute-coefficient polynomial evaluated at one gives\n\\[\n \\sum_s|c_{m,s}|=\\prod_{j=1}^m(j+1)=(m+1)!.\n\\]\nAlso $c_{m,1}=(-1)^m$, so the integer coefficient vector is primitive. After the displayed normalization, its coefficient $\\ell^1$ norm is one.","These are exact statements about a vector of rational coefficients. The number of participating weights grows with $m$. A norm of one for the coefficient vector does not bound the denominator, covolume or arithmetic height of a period realization. No such realization is built into the definition of $\\Phi_m$.","\\subsection{The depleted local series}\nFor a prime $p$, delete the terms whose index is divisible by $p$:\n\\[\n \\Phi_{m,p}(t)=(-1)^m\\sum_{p\\nmid k}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}\\in\\Z_p[[t]].\n\\]\nIt converges when $v_p(t)>0$.\n\\begin{theorem}[Exact local jet valuation]\nIf $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\Phi_{m,p}(t))=\n \\begin{cases}\n (m+1)e,&p\\nmid m+1,","\\subsection{The depleted local series}\nFor a prime $p$, delete the terms whose index is divisible by $p$:\n\\[\n \\Phi_{m,p}(t)=(-1)^m\\sum_{p\\nmid k}\\binom{k}{m+1}\\frac{t^k}{k^{m+2}}\\in\\Z_p[[t]].\n\\]\nIt converges when $v_p(t)>0$.\n\\begin{theorem}[Exact local jet valuation]\nIf $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\Phi_{m,p}(t))=\n \\begin{cases}\n (m+1)e,&p\\nmid m+1,\\\\\n (m+2)e,&p\\mid m+1.\n \\end{cases}\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $p\\nmid m+1$, the first nonzero coefficient remains the unit $(-1)^m/(m+1)^{m+2}$. Its term has valuation $(m+1)e$. Every later coefficient is $p$-integral, so every later term has strictly larger valuation.","(m+2)e,&p\\mid m+1.\n \\end{cases}\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $p\\nmid m+1$, the first nonzero coefficient remains the unit $(-1)^m/(m+1)^{m+2}$. Its term has valuation $(m+1)e$. Every later coefficient is $p$-integral, so every later term has strictly larger valuation.","If $p\\mid m+1$, that first index is deleted. The next index is $m+2$, which is a $p$-unit. Its coefficient is $(-1)^m/(m+2)^{m+1}$, also a unit. All later terms again have strictly larger valuation. The ultrametric inequality with a unique least-valuation term proves both equalities.\n\\end{proof}","\\subsection{An exact differential identity}\nLet $D=t\\,\\dd/\\dd t$ be the Euler operator. Termwise formal differentiation is valid. Applying $D^{m+2}$ multiplies the coefficient at $t^k$ by $k^{m+2}$, so\n\\[\n D^{m+2}\\Phi_m=(-1)^m\\sum_{k\\ge1}\\binom{k}{m+1}t^k\n =(-1)^m\\frac{t^{m+1}}{(1-t)^{m+2}}.\n\\]\nThe final equality is the elementary generating function for the binomial coefficients.","This universal formal identity and the prime-dependent depleted valuations can be studied simultaneously. They do not identify the different $\\Phi_{m,p}$ as local values of one algebraic number. A product formula cannot be applied merely because these functions arose from one coefficient pattern.","\\clearpage\n\\section{A nonzero Hankel contact determinant}\\label{sec:hankel}\n\\subsection{Independent moment directions replace one filter}\nFor an integer $r\\ge1$, define\n\\[\n \\mathcal H_r(t)=\\det\\bigl(\\Li_{i+j+1}(t)\\bigr)_{0\\le i,j<r}.\n\\]\nIt uses weights one through $2r-1$. The expression is a formal series over $\\Q$ and an analytic function for $|t|<1$. No algebraicity at a rational argument is assumed.","For the first two ranks,\n\\[\n \\mathcal H_1=\\Li_1,\\qquad\n \\mathcal H_2=\\Li_1\\Li_3-\\Li_2^2.\n\\]\nThe rank-two determinant compares genuinely different moment directions. Its order-three contact is not inferred from the presence of weight three.","\\begin{theorem}[Cauchy--Binet expansion]\\label{thm:hankel-expansion}\n\\[\n \\mathcal H_r(t)=\\sum_{1\\le k_1<\\cdots<k_r}\n t^{k_1+\\cdots+k_r}\n \\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The","\\begin{theorem}[Cauchy--Binet expansion]\\label{thm:hankel-expansion}\n\\[\n \\mathcal H_r(t)=\\sum_{1\\le k_1<\\cdots<k_r}\n t^{k_1+\\cdots+k_r}\n \\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly","\\begin{theorem}[Cauchy--Binet expansion]\\label{thm:hankel-expansion}\n\\[\n \\mathcal H_r(t)=\\sum_{1\\le k_1<\\cdots<k_r}\n t^{k_1+\\cdots+k_r}\n \\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","}[Cauchy--Binet expansion]\\label{thm:hankel-expansion}\n\\[\n \\mathcal H_r(t)=\\sum_{1\\le k_1<\\cdots<k_r}\n t^{k_1+\\cdots+k_r}\n \\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","k_1<\\cdots<k_r}\n t^{k_1+\\cdots+k_r}\n \\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","{i<j}(k_j-k_i)^2}{\\prod_{i=1}^r k_i^{2r-1}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","orem}\n\\begin{proof}\nWrite $v_k=(1,k^{-1},\\ldots,k^{-(r-1)})^{\\mathsf T}$. The moment matrix is\n\\[\n \\sum_{k\\ge1}\\frac{t^k}{k}v_kv_k^{\\mathsf T}.\n\\]\nFor a finite sum, Cauchy--Binet expands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","xpands its determinant over $r$-element index subsets. The determinant of the chosen columns is a Vandermonde in $k_i^{-1}$. Squaring it and including the factors $t^{k_i}/k_i$ gives the stated coefficient. In detail,\n\\[\n \\prod_{i<j}(k_j^{-1}-k_i^{-1})^2\n =\\frac{\\prod_{i<j}(k_j-k_i)^2}{\\prod_i k_i^{2(r-1)}}.\n\\]\nPassing to the formal infinite sum is coefficientwise legitimate. At a fixed power of $t$, only finitely many index subsets occur.\n\\end{proof}\nEvery coefficient in this expansion is positive. The formal cancellation implicit in the determinant is now resolved into a manifestly nonzero expansion.","\\begin{theorem}[Exact contact and first coefficient]\\label{thm:hankel-contact}\nPut $s_r=r(r+1)/2$. Then\n\\[\n \\ord_0\\mathcal H_r=s_r,\\qquad\n [t^{s_r}]\\mathcal H_r=\\prod_{k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","contact and first coefficient]\\label{thm:hankel-contact}\nPut $s_r=r(r+1)/2$. Then\n\\[\n \\ord_0\\mathcal H_r=s_r,\\qquad\n [t^{s_r}]\\mathcal H_r=\\prod_{k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.",":hankel-contact}\nPut $s_r=r(r+1)/2$. Then\n\\[\n \\ord_0\\mathcal H_r=s_r,\\qquad\n [t^{s_r}]\\mathcal H_r=\\prod_{k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","$. Then\n\\[\n \\ord_0\\mathcal H_r=s_r,\\qquad\n [t^{s_r}]\\mathcal H_r=\\prod_{k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","[t^{s_r}]\\mathcal H_r=\\prod_{k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","k=1}^r k^{-(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","(2k-1)}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","\\end{theorem}\n\\begin{proof}\nThe unique increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","increasing $r$-tuple of positive indices with least sum is $(1,\\ldots,r)$, of sum $s_r$. Its coefficient is nonzero. Its Vandermonde numerator is $\\prod_{j=1}^{r-1}(j!)^2$. Dividing this by $\\prod_{k=1}^r k^{2r-1}$ leaves exponent $2k-1$ in the denominator at each $k$, giving the formula.\n\\end{proof}\nFor $r=2$ the first coefficient is $1/8$. For $r=3$ it is $1/(2^3 3^5)$. The contact orders are $1,3,6,10,\\ldots$, the triangular numbers.","\\subsection{Real positivity and a bound uniform in rank}\n\\begin{theorem}[Positive real determinant]\\label{thm:hankel-real}\nFor every $r\\ge1$ and every real $0<t<1$,\n\\[\n 0<\\mathcal H_r(t)\\le e^{1/4}[-\\log(1-t)].\n\\]\n\\end{theorem}\n\\begin{proof}\nThe Cauchy--Binet expansion consists of positive terms and contains the nonzero term $(1,\\ldots,r)$, so positivity follows. Equivalently, the real moment matrix is positive definite: a nonzero polynomial of degree below $r$ cannot vanish at all the distinct nodes $k^{-1}$.","{theorem}[Positive real determinant]\\label{thm:hankel-real}\nFor every $r\\ge1$ and every real $0<t<1$,\n\\[\n 0<\\mathcal H_r(t)\\le e^{1/4}[-\\log(1-t)].\n\\]\n\\end{theorem}\n\\begin{proof}\nThe Cauchy--Binet expansion consists of positive terms and contains the nonzero term $(1,\\ldots,r)$, so positivity follows. Equivalently, the real moment matrix is positive definite: a nonzero polynomial of degree below $r$ cannot vanish at all the distinct nodes $k^{-1}$.","hm:hankel-real}\nFor every $r\\ge1$ and every real $0<t<1$,\n\\[\n 0<\\mathcal H_r(t)\\le e^{1/4}[-\\log(1-t)].\n\\]\n\\end{theorem}\n\\begin{proof}\nThe Cauchy--Binet expansion consists of positive terms and contains the nonzero term $(1,\\ldots,r)$, so positivity follows. Equivalently, the real moment matrix is positive definite: a nonzero polynomial of degree below $r$ cannot vanish at all the distinct nodes $k^{-1}$.","Hadamard's inequality for the positive matrix gives\n\\[\n \\mathcal H_r(t)\\le\\prod_{i=0}^{r-1}\\Li_{2i+1}(t)\n \\le[-\\log(1-t)]\\prod_{i=1}^{r-1}\\zeta(2i+1).\n\\]\nFor the higher weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","ives\n\\[\n \\mathcal H_r(t)\\le\\prod_{i=0}^{r-1}\\Li_{2i+1}(t)\n \\le[-\\log(1-t)]\\prod_{i=1}^{r-1}\\zeta(2i+1).\n\\]\nFor the higher weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","(t)\\le\\prod_{i=0}^{r-1}\\Li_{2i+1}(t)\n \\le[-\\log(1-t)]\\prod_{i=1}^{r-1}\\zeta(2i+1).\n\\]\nFor the higher weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.",")\n \\le[-\\log(1-t)]\\prod_{i=1}^{r-1}\\zeta(2i+1).\n\\]\nFor the higher weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","For the higher weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","er weights, $t^k\\le1$ bounds the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","s the series by its zeta value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","a value. Since all summands are positive,\n\\[\n \\sum_{i\\ge1}(\\zeta(2i+1)-1)\n =\\sum_{k\\ge2}\\frac1{k(k^2-1)}=\\frac14.\n\\]\nThe last sum telescopes using\n\\[\n \\frac1{k(k^2-1)}=\\frac12\\left(\\frac1{k(k-1)}-\\frac1{k(k+1)}\\right).\n\\]\nFinally $\\log(1+x)\\le x$ for $x\\ge0$ bounds the zeta product by $e^{1/4}$, uniformly in its length.\n\\end{proof}\nAt $t=a/c$ and $t=b/c$, the bounds become $e^{1/4}\\log(c/b)$ and $e^{1/4}\\log(c/a)$. Thus these real period values are nonzero and have a simple logarithmic envelope. This is valuable analytic control. It is not an all-conjugate height theorem for an algebraic scalar.","\\subsection{The third complex chart has a branch boundary}\nFor fixed rank and a consistent upper or lower boundary branch at $x=c/a>1$, the classical integer-weight inversion formula expresses each $\\Li_j(x)$ as a fixed degree-$j$ polynomial in a chosen $\\log(-x)$ plus a sign times $\\Li_j(1/x)$~\\cite{DLMFPolylog}. The latter is bounded by $\\zeta(j)$ for $j\\ge2$. For $j=1$, use $\\Li_1(x)=-\\log(1-x)$ and $x-1=b/a$.","tion{The third complex chart has a branch boundary}\nFor fixed rank and a consistent upper or lower boundary branch at $x=c/a>1$, the classical integer-weight inversion formula expresses each $\\Li_j(x)$ as a fixed degree-$j$ polynomial in a chosen $\\log(-x)$ plus a sign times $\\Li_j(1/x)$~\\cite{DLMFPolylog}. The latter is bounded by $\\zeta(j)$ for $j\\ge2$. For $j=1$, use $\\Li_1(x)=-\\log(1-x)$ and $x-1=b/a$.","Since $|\\log(b/a)|\\le\\log c$ and $\\log(c/a)\\le\\log c$, every entry has a bound by a constant times $(1+\\log c)^j$. A determinant term has total weight\n\\[\n \\sum_{i=0}^{r-1}(i+\\pi(i)+1)=r^2.\n\\]\nTherefore\n\\[\n |\\mathcal H_r(c/a)|\\le C_r(1+\\log c)^{r^2}\n\\]\nfor each of those consistent branches. The constant may depend on $r$. No nonvanishing assertion at every argument in this third chart follows from the positive expansion on $0<t<1$.","$ and $\\log(c/a)\\le\\log c$, every entry has a bound by a constant times $(1+\\log c)^j$. A determinant term has total weight\n\\[\n \\sum_{i=0}^{r-1}(i+\\pi(i)+1)=r^2.\n\\]\nTherefore\n\\[\n |\\mathcal H_r(c/a)|\\le C_r(1+\\log c)^{r^2}\n\\]\nfor each of those consistent branches. The constant may depend on $r$. No nonvanishing assertion at every argument in this third chart follows from the positive expansion on $0<t<1$.","\\clearpage\n\\section{Exact local determinant valuation and denominator budget}\\label{sec:padic}\n\\subsection{Depletion gives an integral local matrix}\nFix a prime $p$ and an integer $r\\ge1$. Define\n\\[\n F_{s,p}(t)=\\sum_{p\\nmid k}\\frac{t^k}{k^s},\\qquad\n \\mathcal H_{r,p}(t)=\\det(F_{i+j+1,p}(t))_{0\\le i,j<r}.\n\\]\nThese series converge at every $t\\in\\Q_p$ with $v_p(t)>0$. They are integral power series. The Cauchy--Binet expansion from Section~\\ref{sec:hankel} now ranges only over increasing $p$-unit indices.","dic}\n\\subsection{Depletion gives an integral local matrix}\nFix a prime $p$ and an integer $r\\ge1$. Define\n\\[\n F_{s,p}(t)=\\sum_{p\\nmid k}\\frac{t^k}{k^s},\\qquad\n \\mathcal H_{r,p}(t)=\\det(F_{i+j+1,p}(t))_{0\\le i,j<r}.\n\\]\nThese series converge at every $t\\in\\Q_p$ with $v_p(t)>0$. They are integral power series. The Cauchy--Binet expansion from Section~\\ref{sec:hankel} now ranges only over increasing $p$-unit indices.","Let $u_1<\\cdots<u_r$ be the first $r$ positive integers not divisible by $p$. Then\n\\[\n u_j=j+\\left\\lfloor\\frac{j-1}{p-1}\\right\\rfloor,\n \\qquad S_{p,r}=\\sum_{j=1}^r u_j.\n\\]\nFor $h\\ge1$, put $s_h=(p-1)p^{h-1}$ and write\n\\[\n r=q_hs_h+b_h,\\qquad 0\\le b_h<s_h.\n\\]\nDefine the finite sum\n\\[\n E_{p,r}=\\sum_{h\\ge1}\\left(s_h\\binom{q_h}{2}+b_hq_h\\right).\n\\]\nThe terms are zero once $s_h\\ge r$.","\\subsection{The residue-class energy minimum}\nFor $r$ distinct $p$-unit integers $k_1,\\ldots,k_r$, the valuation of their Vandermonde product is\n\\[\n \\sum_{i<j}v_p(k_j-k_i)\n =\\sum_{h\\ge1}\\#\\{i<j:k_i\\equiv k_j\\pmod{p^h}\\}.\n\\]\nThere are $s_h$ unit residue classes modulo $p^h$. If their occupation counts are $n_1,\\ldots,n_{s_h}$, the number of coincident pairs is $\\sum_j\\binom{n_j}{2}$.","For a fixed sum $r$, this quantity is minimized when the counts differ by at most one. If $n_i\\ge n_j+2$, moving one item from class $i$ to class $j$ lowers the pair count by $n_i-n_j-1>0$. At the balanced minimum, $b_h$ classes have count $q_h+1$ and the others count $q_h$. The minimum is therefore\n\\[\n (s_h-b_h)\\binom{q_h}{2}+b_h\\binom{q_h+1}{2}\n =s_h\\binom{q_h}{2}+b_hq_h.\n\\]\nThe first $r$ positive $p$-units realize this minimum for every $h$ simultaneously. Ordered unit residues repeat in blocks modulo $p^h$, and each full block contains each unit class exactly once. A prefix adds at most one occurrence to each class. Hence their Vandermonde valuation is exactly $E_{p,r}$.","a fixed sum $r$, this quantity is minimized when the counts differ by at most one. If $n_i\\ge n_j+2$, moving one item from class $i$ to class $j$ lowers the pair count by $n_i-n_j-1>0$. At the balanced minimum, $b_h$ classes have count $q_h+1$ and the others count $q_h$. The minimum is therefore\n\\[\n (s_h-b_h)\\binom{q_h}{2}+b_h\\binom{q_h+1}{2}\n =s_h\\binom{q_h}{2}+b_hq_h.\n\\]\nThe first $r$ positive $p$-units realize this minimum for every $h$ simultaneously. Ordered unit residues repeat in blocks modulo $p^h$, and each full block contains each unit class exactly once. A prefix adds at most one occurrence to each class. Hence their Vandermonde valuation is exactly $E_{p,r}$.","ed when the counts differ by at most one. If $n_i\\ge n_j+2$, moving one item from class $i$ to class $j$ lowers the pair count by $n_i-n_j-1>0$. At the balanced minimum, $b_h$ classes have count $q_h+1$ and the others count $q_h$. The minimum is therefore\n\\[\n (s_h-b_h)\\binom{q_h}{2}+b_h\\binom{q_h+1}{2}\n =s_h\\binom{q_h}{2}+b_hq_h.\n\\]\nThe first $r$ positive $p$-units realize this minimum for every $h$ simultaneously. Ordered unit residues repeat in blocks modulo $p^h$, and each full block contains each unit class exactly once. A prefix adds at most one occurrence to each class. Hence their Vandermonde valuation is exactly $E_{p,r}$.","g one item from class $i$ to class $j$ lowers the pair count by $n_i-n_j-1>0$. At the balanced minimum, $b_h$ classes have count $q_h+1$ and the others count $q_h$. The minimum is therefore\n\\[\n (s_h-b_h)\\binom{q_h}{2}+b_h\\binom{q_h+1}{2}\n =s_h\\binom{q_h}{2}+b_hq_h.\n\\]\nThe first $r$ positive $p$-units realize this minimum for every $h$ simultaneously. Ordered unit residues repeat in blocks modulo $p^h$, and each full block contains each unit class exactly once. A prefix adds at most one occurrence to each class. Hence their Vandermonde valuation is exactly $E_{p,r}$.",". The minimum is therefore\n\\[\n (s_h-b_h)\\binom{q_h}{2}+b_h\\binom{q_h+1}{2}\n =s_h\\binom{q_h}{2}+b_hq_h.\n\\]\nThe first $r$ positive $p$-units realize this minimum for every $h$ simultaneously. Ordered unit residues repeat in blocks modulo $p^h$, and each full block contains each unit class exactly once. A prefix adds at most one occurrence to each class. Hence their Vandermonde valuation is exactly $E_{p,r}$.","\\begin{theorem}[Exact depleted valuation]\\label{thm:depleted-valuation}\nIf $v_p(t)=e\\ge1$, then\n\\[\n \\boxed{v_p(\\mathcal H_{r,p}(t))=eS_{p,r}+2E_{p,r}.}\n\\]\nIn particular, the depleted determinant is nonzero at every such argument.\n\\end{theorem}\n\\begin{proof}\nThe denominators in the depleted expansion are $p$-units. Every $r$-element index subset has Vandermonde valuation at least $E_{p,r}$. The subset $(u_1,\\ldots,u_r)$ uniquely minimizes the sum of its indices. Its term has valuation $eS_{p,r}+2E_{p,r}$. Every other subset has a strictly larger index sum and no smaller Vandermonde valuation, so its term has strictly larger valuation.","n{theorem}[Exact depleted valuation]\\label{thm:depleted-valuation}\nIf $v_p(t)=e\\ge1$, then\n\\[\n \\boxed{v_p(\\mathcal H_{r,p}(t))=eS_{p,r}+2E_{p,r}.}\n\\]\nIn particular, the depleted determinant is nonzero at every such argument.\n\\end{theorem}\n\\begin{proof}\nThe denominators in the depleted expansion are $p$-units. Every $r$-element index subset has Vandermonde valuation at least $E_{p,r}$. The subset $(u_1,\\ldots,u_r)$ uniquely minimizes the sum of its indices. Its term has valuation $eS_{p,r}+2E_{p,r}$. Every other subset has a strictly larger index sum and no smaller Vandermonde valuation, so its term has strictly larger valuation.","then\n\\[\n \\boxed{v_p(\\mathcal H_{r,p}(t))=eS_{p,r}+2E_{p,r}.}\n\\]\nIn particular, the depleted determinant is nonzero at every such argument.\n\\end{theorem}\n\\begin{proof}\nThe denominators in the depleted expansion are $p$-units. Every $r$-element index subset has Vandermonde valuation at least $E_{p,r}$. The subset $(u_1,\\ldots,u_r)$ uniquely minimizes the sum of its indices. Its term has valuation $eS_{p,r}+2E_{p,r}$. Every other subset has a strictly larger index sum and no smaller Vandermonde valuation, so its term has strictly larger valuation.","The expansion converges $p$-adically. A unique term of minimum valuation cannot be cancelled by the remaining terms. Its valuation is therefore the valuation of the whole determinant.\n\\end{proof}\nFor rank two,\n\\[\n v_p(\\mathcal H_{2,p}(t))=\n \\begin{cases}3e,&p\\ne2,","The expansion converges $p$-adically. A unique term of minimum valuation cannot be cancelled by the remaining terms. Its valuation is therefore the valuation of the whole determinant.\n\\end{proof}\nFor rank two,\n\\[\n v_p(\\mathcal H_{2,p}(t))=\n \\begin{cases}3e,&p\\ne2,\\\\4e+2,&p=2.\\end{cases}\n\\]\nAt two the first unit indices are $1,3$, of sum four, and their difference contributes one valuation before squaring. For $p>r$ in arbitrary rank, the first $r$ indices are $1,\\ldots,r$ and all their differences are units. Thus the general formula reduces to $es_r$.","4e+2,&p=2.\\end{cases}\n\\]\nAt two the first unit indices are $1,3$, of sum four, and their difference contributes one valuation before squaring. For $p>r$ in arbitrary rank, the first $r$ indices are $1,\\ldots,r$ and all their differences are units. Thus the general formula reduces to $es_r$.","\\subsection{The universal undepleted expression at good primes}\nThe raw $\\mathcal H_r$ is one universal expression in ordinary polylogarithms. Its local power-series interpretation converges when $v_p(t)>0$, but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[","\\subsection{The universal undepleted expression at good primes}\nThe raw $\\mathcal H_r$ is one universal expression in ordinary polylogarithms. Its local power-series interpretation converges when $v_p(t)>0$, but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1","\\subsection{The universal undepleted expression at good primes}\nThe raw $\\mathcal H_r$ is one universal expression in ordinary polylogarithms. Its local power-series interpretation converges when $v_p(t)>0$, but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive inc","\\subsection{The universal undepleted expression at good primes}\nThe raw $\\mathcal H_r$ is one universal expression in ordinary polylogarithms. Its local power-series interpretation converges when $v_p(t)>0$, but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","en $v_p(t)>0$, but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.",", but its coefficients can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","can have $p$-denominators. For fixed rank, the possible denominator loss grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","grows only logarithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","rithmically with each index, whereas its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","as its numerator power of $t$ gives linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","ves linearly growing valuation. Thus the terms tend to zero and the finite Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","te Cauchy--Binet expansions pass to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","ss to the convergent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","ent local limit.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","t.\n\\begin{theorem}[Exact good-prime valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","valuation]\\label{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","bel{thm:good-valuation}\nIf $p>3r-1$ and $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","d $v_p(t)=e\\ge1$, then\n\\[\n v_p(\\mathcal H_r(t))=e\\,r(r+1)/2.\n\\]\n\\end{theorem}\n\\begin{proof}\nWrite $a_r=2r-1$ and $s_r=r(r+1)/2$. A term indexed by $k_1<\\cdots<k_r$ has valuation\n\\[\n e\\sum_i k_i+2\\sum_{i<j}v_p(k_j-k_i)-a_r\\sum_i v_p(k_i).\n\\]\nFor every $h\\ge1$, the hypothesis gives $p^h-r>a_rh$. At $h=1$ this is $p-r>a_r$. Successive increments of $p^h$ exceed $a_r$, proving the inequality for all $h$.","If $h_i=v_p(k_i)>0$, then $k_i-i\\ge p^{h_i}-r>a_rh_i$. If $h_i=0$, then $k_i-i\\ge0$ and there is no denominator loss. Thus every subset other than $(1,\\ldots,r)$ has valuation strictly above $es_r$. The first subset has a unit coefficient because its indices and differences are smaller than $p$. Its term has valuation $es_r$, and a unique minimum again proves equality.\n\\end{proof}\nThe strict prime threshold is retained. A good-prime equality is stronger than the all-prime lower bound below.","\\subsection{An explicit fixed correction at the remaining primes}\nPut $a_r=2r-1$. For each prime $p$ and $1\\le i\\le r$, define\n\\[\n C_{r,p,i}=\\max\\!\\left(0,\\sup_{h\\ge1}\\left\\{\n a_rh-\\left(p^h\\left\\lceil\\frac i{p^h}\\right\\rceil-i\\right)\\right\\}\\right).\n\\]\nThis is a finite integer maximum because $p^h$ eventually grows faster than $h$. Put $C_{r,p}=\\sum_{i=1}^rC_{r,p,i}$. The good-prime inequality just proved implies $C_{r,p}=0$ for $p>3r-1$. Hence\n\\[\n B_r=\\prod_{p\\le3r-1}p^{C_{r,p}}\n\\]\nis a fixed positive integer depending only on the rank.","icit fixed correction at the remaining primes}\nPut $a_r=2r-1$. For each prime $p$ and $1\\le i\\le r$, define\n\\[\n C_{r,p,i}=\\max\\!\\left(0,\\sup_{h\\ge1}\\left\\{\n a_rh-\\left(p^h\\left\\lceil\\frac i{p^h}\\right\\rceil-i\\right)\\right\\}\\right).\n\\]\nThis is a finite integer maximum because $p^h$ eventually grows faster than $h$. Put $C_{r,p}=\\sum_{i=1}^rC_{r,p,i}$. The good-prime inequality just proved implies $C_{r,p}=0$ for $p>3r-1$. Hence\n\\[\n B_r=\\prod_{p\\le3r-1}p^{C_{r,p}}\n\\]\nis a fixed positive integer depending only on the rank.","ction at the remaining primes}\nPut $a_r=2r-1$. For each prime $p$ and $1\\le i\\le r$, define\n\\[\n C_{r,p,i}=\\max\\!\\left(0,\\sup_{h\\ge1}\\left\\{\n a_rh-\\left(p^h\\left\\lceil\\frac i{p^h}\\right\\rceil-i\\right)\\right\\}\\right).\n\\]\nThis is a finite integer maximum because $p^h$ eventually grows faster than $h$. Put $C_{r,p}=\\sum_{i=1}^rC_{r,p,i}$. The good-prime inequality just proved implies $C_{r,p}=0$ for $p>3r-1$. Hence\n\\[\n B_r=\\prod_{p\\le3r-1}p^{C_{r,p}}\n\\]\nis a fixed positive integer depending only on the rank.","ining primes}\nPut $a_r=2r-1$. For each prime $p$ and $1\\le i\\le r$, define\n\\[\n C_{r,p,i}=\\max\\!\\left(0,\\sup_{h\\ge1}\\left\\{\n a_rh-\\left(p^h\\left\\lceil\\frac i{p^h}\\right\\rceil-i\\right)\\right\\}\\right).\n\\]\nThis is a finite integer maximum because $p^h$ eventually grows faster than $h$. Put $C_{r,p}=\\sum_{i=1}^rC_{r,p,i}$. The good-prime inequality just proved implies $C_{r,p}=0$ for $p>3r-1$. Hence\n\\[\n B_r=\\prod_{p\\le3r-1}p^{C_{r,p}}\n\\]\nis a fixed positive integer depending only on the rank.","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplic","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ rest","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they a","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The ot","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The form","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not","\\begin{theorem}[All-prime contact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","ontact lower bound]\\label{thm:budget}\nFor every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","or every prime $p$ and $v_p(t)=e\\ge1$,\n\\[\n \\boxed{v_p(B_r\\mathcal H_r(t))\\ge es_r.}\n\\]\nHere $v_p(0)=+\\infty$. This theorem alone does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","one does not establish nonvanishing at the small primes.\n\\end{theorem}\n\\begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","begin{proof}\nFor an ordered index $k_i\\ge i$ with $h_i=v_p(k_i)>0$, the smallest multiple of $p^{h_i}$ at least $i$ is $p^{h_i}\\lceil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","eil i/p^{h_i}\\rceil$. Therefore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","fore\n\\[\n a_rh_i-(k_i-i)\\le C_{r,p,i}.\n\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","\\]\nFor $h_i=0$, the same bound holds because its left side is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","is nonpositive. The numerator Vandermonde only improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","improves valuation. Since $e\\ge1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","1$, every term in the convergent expansion has valuation at least\n\\[\n es_r+\\sum_i[(k_i-i)-a_rh_i]\\ge es_r-C_{r,p}.\n\\]\nThe ultrametric inequality gives the same bound for the sum. Multiplication by $B_r$ restores the missing $C_{r,p}$.\n\\end{proof}\nFor rank two, $B_2=2^7\\cdot3^3=3456$ is a valid budget. For instance, at $p=2$ the two component maxima, in index order $i=1,2$, are three and four. At $p=3$ they are one and two. The other primes contribute zero. The formula need not be minimal.","\\subsection{The all-place comparison is still absent}\nThese results give exact depleted nonvanishing, universal good-prime equality and universal all-prime lower contact. They do not identify the real value, the various $p$-adic values, or different sector arguments as the places of one nonzero algebraic number. An arithmetic realization and its norm maps remain separate requirements.","\\clearpage\n\\section{A rational truncation that fails the output contact test}\\label{sec:truncation}\n\\subsection{A genuine algebraic candidate can be tested exactly}\nFix integers $r\\ge1$ and $K\\ge r$. Replace each polylogarithm by its finite series through index $K$, obtaining a rational polynomial $\\mathcal H_{r,K}(t)$. Its Cauchy--Binet expansion has the same positive subset coefficients, restricted to $1\\le k_i\\le K$.","ge\n\\section{A rational truncation that fails the output contact test}\\label{sec:truncation}\n\\subsection{A genuine algebraic candidate can be tested exactly}\nFix integers $r\\ge1$ and $K\\ge r$. Replace each polylogarithm by its finite series through index $K$, obtaining a rational polynomial $\\mathcal H_{r,K}(t)$. Its Cauchy--Binet expansion has the same positive subset coefficients, restricted to $1\\le k_i\\le K$.","Let\n\\[\n L_K=\\operatorname{lcm}(1,\\ldots,K),\\quad s_r=r(r+1)/2,\n \\quad d=rK-r(r-1)/2.\n\\]\nThen\n\\[\n A_{r,K}(t):=L_K^{r^2}\\mathcal H_{r,K}(t)\\in\\Z[t],\n \\qquad \\ord_0A_{r,K}=s_r,\\qquad \\deg A_{r,K}=d.\n\\]\nTo see integrality, scale row $i$ by $L_K^{i+1}$ and column $j$ by $L_K^j$. Each entry becomes\n\\[\n L_K^{i+j+1}\\sum_{k=1}^K\\frac{t^k}{k^{i+j+1}}\\in\\Z[t].\n\\]\nThe total determinant scaling exponent is $r^2$. The unique smallest subset is $(1,\\ldots,r)$ and the unique largest is $(K-r+1,\\ldots,K)$, proving the order and degree.","dots,K),\\quad s_r=r(r+1)/2,\n \\quad d=rK-r(r-1)/2.\n\\]\nThen\n\\[\n A_{r,K}(t):=L_K^{r^2}\\mathcal H_{r,K}(t)\\in\\Z[t],\n \\qquad \\ord_0A_{r,K}=s_r,\\qquad \\deg A_{r,K}=d.\n\\]\nTo see integrality, scale row $i$ by $L_K^{i+1}$ and column $j$ by $L_K^j$. Each entry becomes\n\\[\n L_K^{i+j+1}\\sum_{k=1}^K\\frac{t^k}{k^{i+j+1}}\\in\\Z[t].\n\\]\nThe total determinant scaling exponent is $r^2$. The unique smallest subset is $(1,\\ldots,r)$ and the unique largest is $(K-r+1,\\ldots,K)$, proving the order and degree.","(r-1)/2.\n\\]\nThen\n\\[\n A_{r,K}(t):=L_K^{r^2}\\mathcal H_{r,K}(t)\\in\\Z[t],\n \\qquad \\ord_0A_{r,K}=s_r,\\qquad \\deg A_{r,K}=d.\n\\]\nTo see integrality, scale row $i$ by $L_K^{i+1}$ and column $j$ by $L_K^j$. Each entry becomes\n\\[\n L_K^{i+j+1}\\sum_{k=1}^K\\frac{t^k}{k^{i+j+1}}\\in\\Z[t].\n\\]\nThe total determinant scaling exponent is $r^2$. The unique smallest subset is $(1,\\ldots,r)$ and the unique largest is $(K-r+1,\\ldots,K)$, proving the order and degree.","For a reduced rational argument $t=a/c$, an obvious integral scalar is\n\\[\n W_{r,K}(a,c)=c^dA_{r,K}(a/c).\n\\]\nIt is nonzero when $0<a<c$, since the truncated positive expansion is nonzero. Thus it is a genuine nonzero arithmetic candidate. The question is whether it retains the required output-prime contact after clearing denominators.","\\begin{theorem}[Loss of output contact]\\label{thm:truncation-failure}\nIf $p>K$, $p\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding","\\begin{theorem}[Loss of output contact]\\label{thm:truncation-failure}\nIf $p>K$, $p\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every poss","\\begin{theorem}[Loss of output contact]\\label{thm:truncation-failure}\nIf $p>K$, $p\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant r","\\begin{theorem}[Loss of output contact]\\label{thm:truncation-failure}\nIf $p>K$, $p\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant realization.","m}[Loss of output contact]\\label{thm:truncation-failure}\nIf $p>K$, $p\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant realization.","\\mid c$ and $\\gcd(a,c)=1$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant realization.","$, then\n\\[\n p\\nmid W_{r,K}(a,c).\n\\]\n\\end{theorem}\n\\begin{proof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant realization.","oof}\nExpand $W$ as a homogeneous polynomial in $a,c$. Modulo $p$, every term below the top degree in $a$ retains a positive power of $c$ and vanishes. The unique top subset is $(K-r+1,\\ldots,K)$. Its indices, nonzero differences and $L_K$ are all $p$-units, so its scaled coefficient is a unit. The remaining factor $a^d$ is a unit because $p\\nmid a$. The surviving term is therefore nonzero modulo $p$.\n\\end{proof}\nIf an output prime above the cutoff has repeated depth, this witness has no positive valuation there at all. The failed candidate does not pass the local output gate. It is not a theorem excluding every possible determinant realization.","\\subsection{Rank two exposes the loss in one line}\nThe first two truncated rank-two expressions are\n\\[\n \\mathcal H_{2,2}(t)=t^3/8,\n\\]\n\\[\n \\mathcal H_{2,3}(t)=\\frac{t^3}{8}+\\frac{4t^4}{27}+\\frac{t^5}{216}\n =\\frac{t^3(27+32t+t^2)}{216}.\n\\]\nClearing the displayed denominator and the projective denominator in the second gives\n\\[\n W(a,c)=a^3(27c^2+32ac+a^2).\n\\]\nFor every $p>3$ dividing $c$, this is congruent to $a^5$ modulo $p$. Hence it is a unit at that output prime.","{Rank two exposes the loss in one line}\nThe first two truncated rank-two expressions are\n\\[\n \\mathcal H_{2,2}(t)=t^3/8,\n\\]\n\\[\n \\mathcal H_{2,3}(t)=\\frac{t^3}{8}+\\frac{4t^4}{27}+\\frac{t^5}{216}\n =\\frac{t^3(27+32t+t^2)}{216}.\n\\]\nClearing the displayed denominator and the projective denominator in the second gives\n\\[\n W(a,c)=a^3(27c^2+32ac+a^2).\n\\]\nFor every $p>3$ dividing $c$, this is congruent to $a^5$ modulo $p$. Hence it is a unit at that output prime.","oses the loss in one line}\nThe first two truncated rank-two expressions are\n\\[\n \\mathcal H_{2,2}(t)=t^3/8,\n\\]\n\\[\n \\mathcal H_{2,3}(t)=\\frac{t^3}{8}+\\frac{4t^4}{27}+\\frac{t^5}{216}\n =\\frac{t^3(27+32t+t^2)}{216}.\n\\]\nClearing the displayed denominator and the projective denominator in the second gives\n\\[\n W(a,c)=a^3(27c^2+32ac+a^2).\n\\]\nFor every $p>3$ dividing $c$, this is congruent to $a^5$ modulo $p$. Hence it is a unit at that output prime.","The original local series had contact near zero. The algebraic scalar evaluates at $a/c$ and clears powers of $c$ dictated by its top degree. At primes dividing $c$, it is the top term that survives. Long truncation does not automatically solve this mismatch. A valid construction must prove the contact at the actual local chart and retain it through the exact normalization map.","\\section{The correctly normalized global bridge}\\label{sec:bridge}\n\\subsection{A sufficient certificate with every hypothesis visible}\nThe jet block supplies formal and local expressions. The next theorem states exactly how a suitable global arithmetic witness would imply $abc$. Its hypotheses have not been established for the present determinant candidate.","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$.","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\f","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\e","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta","\\begin{theorem}[Conditional integral witness criterion]\\label{thm:global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2…",":global-certificate}\nLet $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power …","Let $\\mathcal R$ be an unbounded set of positive integer ranks, and put $s_r=r(r+1)/2$. Suppose that for each $r\\in\\mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positi…","mathcal R$ there are constants $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D…","s $C_r>0$, $A_r\\ge0$ and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\…","and $E_r\\ge0$, independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\del…","independent of the primitive triple, such that every primitive positive triple admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chose…","admits a nonzero integer $W_r$ with\n\\[\n Q^{s_r}\\mid W_r,\n \\qquad |W_r|\\le C_r(ab)^{s_r}R^{E_r}(\\log(3c))^{A_r}.\n\\]\nIf $E_r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb…","r/s_r\\to0$ along $\\mathcal R$, then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ …","then Conjecture~\\ref{conj:abc} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly o…","c} follows.\n\\end{theorem}\n\\begin{proof}\nNonzero integer divisibility gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each…","ty gives $Q^{s_r}\\le|W_r|$. Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be…","Divide by $(ab)^{s_r}$ and use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A ra…","nd use $Q/(ab)=c/R$:\n\\[\n (c/R)^{s_r}\\le C_rR^{E_r}(\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the t…","\\log(3c))^{A_r}.\n\\]\nTaking positive roots yields\n\\[\n c\\le D_rR^{1+\\eta_r}(\\log(3c))^{a_r},\n\\quad D_r=C_r^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not me…","^{1/s_r},\\quad \\eta_r=E_r/s_r,\\quad a_r=A_r/s_r.\n\\]\nFix the target $\\eps>0$. Choose one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","one rank with $\\eta_r<\\eps$. Next choose\n\\[\n 0<\\delta<\\frac{\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","\\eps-\\eta_r}{1+\\eps}.\n\\]\nFor the fixed exponent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","onent $a_r$, there is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","re is a constant $L_{r,\\delta}$ such that $(\\log(3c))^{a_r}\\le L_{r,\\delta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","ta}c^\\delta$ for all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","all $c\\ge2$. The elementary fact follows because a fixed logarithmic power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","c power divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","divided by any positive power of $c$ tends to zero, and the remainder is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","is bounded on a finite initial interval. Thus\n\\[\n c^{1-\\delta}\\le D_rL_{r,\\delta}R^{1+\\eta_r}.\n\\]\nRaise to $1/(1-\\delta)$. The chosen inequality for $\\delta$ gives\n\\[\n \\frac{1+\\eta_r}{1-\\delta}<1+\\eps.\n\\]\nAbsorb the fixed constants into $K_\\eps$ and use $R\\ge2$ to obtain the desired bound. No prior power bound relating $c$ to $R$ was assumed.\n\\end{proof}\nOnly one rank is selected for each target $\\eps$. Its constants may be arbitrarily large but must be uniform in the triple. A rank that depends on the triple without uniform constant control would not meet the theorem.","arithmetic object whose local realization is the chosen determinant.\n\\item A nonzero algebraic scalar, or a nonzero section of a rigorously metrized determinant line.\n\\item The full local divisibility in the correct sector charts, with every comparison factor included.\n\\item Bounds on every finite denominator and every archimedean conjugate or norm.\n\\item A global size estimate with the required $(ab)^{s_r}$ normalization and vanishing relative radical cost.\n\\end{enumerate}\nReal positivity proves nonvanishing of a real analytic value. It does not prove that value algebraic. The depleted exact valuations belong to prime-dependent functions. The raw determinant supplies a common expression but not an authenticated product formula across its analytic realizations. The rational truncation constructs an algebraic number but fails the output-prime contact gate.","A product formula applies to the places of one nonzero algebraic number. It does not apply to an arbitrary list of formally related complex and $p$-adic periods. A determinant-line alternative would likewise need its own proved arithmetic-degree identity and metric comparisons.","\\subsection{The coherence route has a parallel endpoint}\nThe inherited exact factorization is\n\\[\n M_c=\\epsilon_2KWC,\\qquad\\epsilon_2\\le2,\\quad K\\le\\log_2c.\n\\]\nHere the factor denoted $W$ is the local logarithmic defect, distinct from the integer witness $W_r$ in Theorem~\\ref{thm:global-certificate}. The letters have separate domains.","If for every $\\eta>0$ one proves\n\\[\n WC\\le C_\\eta R_aR_bR^\\eta,\n\\]\nthen $c\\le C'_{\\eta}R^{1+\\eta}\\log(3c)$. The same absorption argument gives $abc$. Conversely $WC\\le M_c$, so the all-$\\eps$ output-sector $abc$ inequality immediately supplies this bound. The uniform combined $WC$ estimate is therefore the exact endpoint. A bound on $C$ alone suffices only under independent control of $W$, for example in a stated class where $W=1$.","es\n\\[\n WC\\le C_\\eta R_aR_bR^\\eta,\n\\]\nthen $c\\le C'_{\\eta}R^{1+\\eta}\\log(3c)$. The same absorption argument gives $abc$. Conversely $WC\\le M_c$, so the all-$\\eps$ output-sector $abc$ inequality immediately supplies this bound. The uniform combined $WC$ estimate is therefore the exact endpoint. A bound on $C$ alone suffices only under independent control of $W$, for example in a stated class where $W=1$.","\\subsection{What a useful next rank-two result would be}\nThe strongest immediate target is an explicit arithmetic realization of\n\\[\n \\mathcal H_2(t)=\\Li_1(t)\\Li_3(t)-\\Li_2(t)^2,\n\\]\nwith every local comparison and global norm recorded. A motivic or iterated-integral construction is a possible research direction, not an input assumed here. A different nonzero algebraic determinant could also be useful if it passes the same gates.","A new block should demonstrate a quantitative gain on $c/R$ or $M_c/(R_aR_b)$. Another local valuation, formal transformation or finite successful example is evidence within its declared scope. It does not establish that global gain.","\\midrule\nCoefficient cancellation & $m=0,\\ldots,30$, indices $k=1,\\ldots,80$","Finite Cauchy--Binet & Ranks $1,\\ldots,5$, $K=r,\\ldots,r+4$, arguments $1/3,2/3,-1/2$","Residue energy minimum & Primes $2,3,5,7$, ranks through six, subsets of the first eleven unit indices","Depleted valuations & Primes through seventeen in the selected seven-prime list, ranks through eight, $e=1,2,3$, three unit choices","The unchanged canonical mathematical audit reproduced all seven archived output groups. Its phase Smith check covered 152,095 nondegenerate primitive triples with $a\\le b$ and $c\\le1000$. It also checked 195 polyphase cases, 2,499 seeded Wronskian packets, 5,001 divisor-packet examples, 391 Kummer examples and 254 diagonal Walsh packets. The Wronskian packets are sampling, and the Walsh model is an algebraic diagonal model, rather than an automorphic construction.","The nine canonical implementation groups also reproduced their archived values, after two disclosed filesystem path adaptations. The mathematical code was unchanged. At $3+125=128$, the weighted derivative minimum is $\\Theta=6$, while the exponent-free moving-search minimum is $\\Phi=8$. They optimize different packets. The latter has obstruction multiple one and exact master gap one. At $1+4374=4375$, the moving-search minimum is 62 and its exact master gap is $248/243$.","hived values, after two disclosed filesystem path adaptations. The mathematical code was unchanged. At $3+125=128$, the weighted derivative minimum is $\\Theta=6$, while the exponent-free moving-search minimum is $\\Phi=8$. They optimize different packets. The latter has obstruction multiple one and exact master gap one. At $1+4374=4375$, the moving-search minimum is 62 and its exact master gap is $248/243$.","\\subsection{An independent primitive census}\nA new exact enumeration covered every primitive positive triple with $a\\le b$ and $2\\le c\\le10000$. It contains 15,198,743 triples. A separate Euler totient sieve gives the same count. Exactly 121 satisfy $c>R$, with membership decided by the integer inequality rather than a floating quality threshold.","Within that declared domain, the largest displayed quality is approximately $1.567887$ at $1+4374=4375$, with $R=210$. The example $2+109\\cdot3^{10}=23^5$ lies outside the census. This computation supplies neither a uniform conjectural constant nor a conclusion about unsearched triples.","The strict good-prime boundary was also tested with justified infinite-series tails. For rank two and $p=5=3r-1$, the raw determinant at $t=10$ has valuation four, while its depleted version has valuation three. For $p=7$ and $t=14$, both have valuation three. The modular precision levels were 6, 8, 12 and 16. The tail bounds in the computation report authenticate these named valuations. Thus the strict condition $p>3r-1$ cannot be weakened to equality in the universal exact-valuation theorem.","ime boundary was also tested with justified infinite-series tails. For rank two and $p=5=3r-1$, the raw determinant at $t=10$ has valuation four, while its depleted version has valuation three. For $p=7$ and $t=14$, both have valuation three. The modular precision levels were 6, 8, 12 and 16. The tail bounds in the computation report authenticate these named valuations. Thus the strict condition $p>3r-1$ cannot be weakened to equality in the universal exact-valuation theorem.","At a small prime, a raw lower bound can even coexist with zero. For rank one at $p=2,t=2$, the standard convergent logarithm identity gives $\\Li_1(2)=-\\log_2(-1)=0$. Finite zero residues support this identity but do not prove it by themselves. The depleted moment at that argument has valuation one. This is why the all-prime budget theorem explicitly uses $v_p(0)=+\\infty$ and does not promise raw nonvanishing at every small prime.","These scripts do not formalize convergence, general infinite Hankel determinants, positivity, the all-rank valuation theorem, the residue-class minimum proof or the global $abc$ criterion. The Gram identities alone have no positivity hypotheses. The integer remainder identity is a polynomial equality, rather than a completed prime-divisibility theorem. The formal scope is finite algebra, distinct from the written analytic and arithmetic proofs.","d witness-preservation theorem and a monotone observable, a hull containment yields the scalar observable bound. The source witness for an actual IUT instantiation is still missing. A finite Boolean model demonstrates that the earlier abstract fields do not logically imply the missing containment. That is an independence result inside the defined interface, rather than a counterexample to IUT or $abc$.","\\subsection{Failures are part of the evidence}\nThe historical jet verifier first encountered a float at the rank-one empty product. The exact identity test detected it, and the accepted historical implementation corrected it. The latest replay uses the corrected exact source. The false unbalanced $Q$ target and the failed rational witness are mathematical corrections retained as explicit theorems.","The prime contact and determinant calculations are derived in this text. Pasten's arithmetic derivative work supplies an established base for that branch. Its small-derivative conjecture and its weak $abc$ equivalences are not silently identified with the stronger all-$\\eps$ packet target stated here. The Wiles-defect and motivic references explain relevant established frameworks. They do not construct the project-specific persistence-to-Hecke or integral period transfer maps.","\\bibitem{SerreAbelian1989}\nJ.-P. Serre,\n\\emph{Abelian $\\ell$-adic representations and elliptic curves},\nwith the collaboration of W. Kuyk and J. Labute,\nAddison--Wesley, 1989, reprint of the 1968 original.\nChapter IV, Appendix A.1.2--A.1.3; Special Preface, p.~ix.","\\bibitem{StewartYu2001}\nC. L. Stewart and K. R. Yu,\n``On the $abc$ conjecture, II,''\n\\emph{Duke Mathematical Journal} 108, no.~1 (2001), 169--181.\n\\href{https://doi.org/10.1215/S0012-7094-01-10815-6}{doi:10.1215/S0012-7094-01-10815-6}.","\\bibitem{VojtaABC}\nP. Vojta,\n``A more general $ABC$ conjecture,''\n\\emph{International Mathematics Research Notices} 1998, no.~21, 1103--1116.\n\\href{https://doi.org/10.1155/S1073792898000658}{doi:10.1155/S1073792898000658}.","\\bibitem{ScholzeStix2018}\nP. Scholze and J. Stix,\n``Why $abc$ is still a conjecture,'' August 2018 manuscript.\n\\url{https://www.math.uni-bonn.de/people/scholze/WhyABCisStillaConjecture.pdf}.","\\bibitem{BCDT}\nC. Breuil, B. Conrad, F. Diamond, and R. Taylor,\n``On the modularity of elliptic curves over $\\mathbb Q$: wild 3-adic exercises,''\n\\emph{Journal of the American Mathematical Society} 14, no.~4 (2001), 843--939.\n\\href{https://doi.org/10.1090/S0894-0347-01-00370-8}{doi:10.1090/S0894-0347-01-00370-8}.","\\bibitem{ZagierPolylog}\nD. Zagier,\n``Polylogarithms, Dedekind zeta functions, and the algebraic $K$-theory of fields,''\nin \\emph{Arithmetic Algebraic Geometry}, Progress in Mathematics 89,\nBirkh\\\"auser, 1990, 391--430.\n\\href{https://doi.org/10.1007/978-1-4612-0457-2_19}{doi:10.1007/978-1-4612-0457-2\\_19}.","\\bibitem{SUnit1988}\nJ.-H. Evertse, K. Gy\\H{o}ry, C. L. Stewart, and R. Tijdeman,\n``On $S$-unit equations in two unknowns,''\n\\emph{Inventiones Mathematicae} 92 (1988), 461--477.\n\\href{https://doi.org/10.1007/BF01393743}{doi:10.1007/BF01393743}.","\\subsection{The prime-coordinate module}\nLet\n\\[\n a+b=c,\n \\qquad\n \\gcd(a,b)=1.\n\\]\nThe three integers are pairwise coprime. Write\n\\[\n S=S_a\\sqcup S_b\\sqcup S_c\n =\\{p:p\\mid abc\\}.\n\\]\nThe first object of the calculus is the free prime-packet module\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-","\\subsection{The prime-coordinate module}\nLet\n\\[\n a+b=c,\n \\qquad\n \\gcd(a,b)=1.\n\\]\nThe three integers are pairwise coprime. Write\n\\[\n S=S_a\\sqcup S_b\\sqcup S_c\n =\\{p:p\\mid abc\\}.\n\\]\nThe first object of the calculus is the free prime-packet module\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","nate module}\nLet\n\\[\n a+b=c,\n \\qquad\n \\gcd(a,b)=1.\n\\]\nThe three integers are pairwise coprime. Write\n\\[\n S=S_a\\sqcup S_b\\sqcup S_c\n =\\{p:p\\mid abc\\}.\n\\]\nThe first object of the calculus is the free prime-packet module\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","e three integers are pairwise coprime. Write\n\\[\n S=S_a\\sqcup S_b\\sqcup S_c\n =\\{p:p\\mid abc\\}.\n\\]\nThe first object of the calculus is the free prime-packet module\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","ct of the calculus is the free prime-packet module\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","le\n\\[\n \\Omega_S=\\bigoplus_{p\\in S}\\Z\\xi_p.\n\\]\nFirst define the global free module over all primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","primes, and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]",", and restrict it to $\\Omega_S$ for $S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","$S$-supported integers. For such a positive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","sitive integer $n$, define the universal multiplicative differential\n\\[\n \\boxed{\n \\mathrm dn\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p\\,\\xi_p.\n }\n\\]\nEvery coefficient is an integer because $p\\mid n$. The map obeys the exact Leibniz law\n\\[\n \\mathrm d(mn)=m\\,\\mathrm dn+n\\,\\mathrm dm.\n\\]\nThe verification is local. For each prime $p$,\n\\[\n v_p(mn)=v_p(m)+v_p(n),\n\\]\nso the $p$-coordinate of the right side is\n\\[\n mn\\frac{v_p(m)+v_p(n)}p.\n\\]","A scalar packet is an integer vector\n\\[\n x=(x_p)_{p\\in S}\\in\\Z^S.\n\\]\nPairing with $x$ gives the arithmetic derivative\n\\[\n \\boxed{\n \\mathrm d_x(n)\n =\n n\\sum_{p\\mid n}\\frac{v_p(n)}p x_p.\n }\n\\]\nThe Leibniz rule survives scalarization.","Arithmetic derivatives and their connection with $abc$ have an established literature. The present construction is organized here in prime-packet coordinates and later refined into filtered shell objects~\\cite{PastenDerivatives,VojtaABC}.","\\subsection{The additive row}\nThe condition\n\\[\n \\mathrm d_x(a)+\\mathrm d_x(b)=\\mathrm d_x(c)\n\\]\nis one integer equation. Define\n\\[\n \\lambda_p=\n \\begin{cases}\n \\dfrac{a v_p(a)}p,&p\\mid a,","\\subsection{The additive row}\nThe condition\n\\[\n \\mathrm d_x(a)+\\mathrm d_x(b)=\\mathrm d_x(c)\n\\]\nis one integer equation. Define\n\\[\n \\lambda_p=\n \\begin{cases}\n \\dfrac{a v_p(a)}p,&p\\mid a,\\\\[6pt]\n \\dfrac{b v_p(b)}p,&p\\mid b,\\\\[6pt]\n -\\dfrac{c v_p(c)}p,&p\\mid c.\n \\end{cases}\n\\]\nThen\n\\[\n \\boxed{\\lambda\\cdot x=0.}\n\\]\nThe additive derivative lattice is\n\\[\n \\boxed{\n \\mathcal T_{a,b,c}\n =\n \\ker(\\lambda)\\cap\\Z^S.\n }\n\\]\nSince the row is nonzero, its rank is $|S|-1$. The problem is not existence of additive derivatives. The problem is finding an additive derivative that remains transverse to the dependence direction introduced by $a$ and $b$.","[6pt]\n -\\dfrac{c v_p(c)}p,&p\\mid c.\n \\end{cases}\n\\]\nThen\n\\[\n \\boxed{\\lambda\\cdot x=0.}\n\\]\nThe additive derivative lattice is\n\\[\n \\boxed{\n \\mathcal T_{a,b,c}\n =\n \\ker(\\lambda)\\cap\\Z^S.\n }\n\\]\nSince the row is nonzero, its rank is $|S|-1$. The problem is not existence of additive derivatives. The problem is finding an additive derivative that remains transverse to the dependence direction introduced by $a$ and $b$.","]\n -\\dfrac{c v_p(c)}p,&p\\mid c.\n \\end{cases}\n\\]\nThen\n\\[\n \\boxed{\\lambda\\cdot x=0.}\n\\]\nThe additive derivative lattice is\n\\[\n \\boxed{\n \\mathcal T_{a,b,c}\n =\n \\ker(\\lambda)\\cap\\Z^S.\n }\n\\]\nSince the row is nonzero, its rank is $|S|-1$. The problem is not existence of additive derivatives. The problem is finding an additive derivative that remains transverse to the dependence direction introduced by $a$ and $b$.","ac{c v_p(c)}p,&p\\mid c.\n \\end{cases}\n\\]\nThen\n\\[\n \\boxed{\\lambda\\cdot x=0.}\n\\]\nThe additive derivative lattice is\n\\[\n \\boxed{\n \\mathcal T_{a,b,c}\n =\n \\ker(\\lambda)\\cap\\Z^S.\n }\n\\]\nSince the row is nonzero, its rank is $|S|-1$. The problem is not existence of additive derivatives. The problem is finding an additive derivative that remains transverse to the dependence direction introduced by $a$ and $b$.","\\subsection{The arithmetic Wronskian}\nDefine\n\\[\n \\boxed{\n W_x(a,b)\n =a\\,\\mathrm d_x(b)-b\\,\\mathrm d_x(a).\n }\n\\]\nWriting\n\\[\n \\mu_p=\n \\begin{cases}\n -\\dfrac{v_p(a)}p,&p\\mid a,","\\subsection{The arithmetic Wronskian}\nDefine\n\\[\n \\boxed{\n W_x(a,b)\n =a\\,\\mathrm d_x(b)-b\\,\\mathrm d_x(a).\n }\n\\]\nWriting\n\\[\n \\mu_p=\n \\begin{cases}\n -\\dfrac{v_p(a)}p,&p\\mid a,\\\\[6pt]\n \\dfrac{v_p(b)}p,&p\\mid b,\\\\[6pt]\n 0,&p\\mid c,\n \\end{cases}\n\\]\nwe obtain\n\\[\n W_x(a,b)=ab\\,\\mu\\cdot x.\n\\]\nAn additive packet is \\emph{transverse} when $W_x(a,b)\\neq0$. It is degenerate when the two input terms remain derivative-proportional.","[6pt]\n 0,&p\\mid c,\n \\end{cases}\n\\]\nwe obtain\n\\[\n W_x(a,b)=ab\\,\\mu\\cdot x.\n\\]\nAn additive packet is \\emph{transverse} when $W_x(a,b)\\neq0$. It is degenerate when the two input terms remain derivative-proportional.","Over $\\R$, the additive projection is\n\\[\n P_{\\mathrm{add}}\n =\n I-\n \\frac{\\lambda\\lambda^{\\mathsf T}}{\\norm{\\lambda}_2^2}.\n\\]\nThe obstruction direction inside the additive hyperplane is\n\\[\n \\mu_\\perp=P_{\\mathrm{add}}\\mu.\n\\]\nWhen $\\mu_\\perp\\neq0$, the rank-one obstruction projector is\n\\[\n P_{\\mathrm{obs}}\n =\n \\frac{\\mu_\\perp\\mu_\\perp^{\\mathsf T}}\n {\\norm{\\mu_\\perp}_2^2}.\n\\]\nThis is the first appearance of the projector-resolved pattern. A large ambient prime space is decomposed into one equation direction, one obstruction direction, and the remaining null packets.","\\subsection{Radical divisibility}\nSet\n\\[\n R=\\rad(abc).\n\\]\nThe central first-order theorem is an exact divisibility law.","\\begin{theorem}[Wronskian divisibility]\nFor every additive packet $x\\in\\mathcal T_{a,b,c}$,\n\\[\n \\boxed{\n \\frac{abc}{R}\\mid W_x(a,b).\n }\n\\]\n\\end{theorem}","\\begin{proof}\nLet $p^e\\parallel a$. Every term of $\\mathrm d_x(a)$ is divisible by $p^{e-1}$. Since $p\\nmid b$, the term $b\\,\\mathrm d_x(a)$ is divisible by $p^{e-1}$. Also $p^e\\mid a$, so $a\\,\\mathrm d_x(b)$ is divisible by $p^e$, hence by $p^{e-1}$. Therefore\n\\[\n p^{e-1}\\mid W_x(a,b).\n\\]\nThe same argument applies to primes dividing $b$.","Now let $p^e\\parallel c$. Additivity gives\n\\[\n \\mathrm d_x(c)=\\mathrm d_x(a)+\\mathrm d_x(b).\n\\]\nUsing $c=a+b$,\n\\[\n \\begin{aligned}\n W_x(a,b)\n &=a\\,\\mathrm d_x(b)-b\\,\\mathrm d_x(a)","Now let $p^e\\parallel c$. Additivity gives\n\\[\n \\mathrm d_x(c)=\\mathrm d_x(a)+\\mathrm d_x(b).\n\\]\nUsing $c=a+b$,\n\\[\n \\begin{aligned}\n W_x(a,b)\n &=a\\,\\mathrm d_x(b)-b\\,\\mathrm d_x(a)\\\\\n &=a\\bigl(\\mathrm d_x(c)-\\mathrm d_x(a)\\bigr)-b\\,\\mathrm d_x(a)\\\\\n &=a\\,\\mathrm d_x(c)-c\\,\\mathrm d_x(a).\n \\end{aligned}\n\\]\nThe first term is divisible by $p^{e-1}$ because $\\mathrm d_x(c)$ is. The second is divisible by $p^e$. Hence $p^{e-1}\\mid W_x(a,b)$.","&=a\\,\\mathrm d_x(c)-c\\,\\mathrm d_x(a).\n \\end{aligned}\n\\]\nThe first term is divisible by $p^{e-1}$ because $\\mathrm d_x(c)$ is. The second is divisible by $p^e$. Hence $p^{e-1}\\mid W_x(a,b)$.","The prime supports of $a$, $b$, and $c$ are disjoint. Multiplying all local divisibilities gives\n\\[\n \\prod_{p^e\\parallel abc}p^{e-1}\n =\n \\frac{abc}{R}\n \\mid W_x(a,b).\n\\]\n\\end{proof}","For a transverse packet,\n\\[\n \\boxed{\n |W_x(a,b)|\\ge\\frac{abc}{R}.\n }\n\\]\nThis is the integer analogue of the repeated-root divisor appearing in the polynomial Mason--Stothers theorem.","\\subsection{The master height inequality}\nFor every positive $n$,\n\\[\n \\begin{aligned}\n |\\mathrm d_x(n)|\n &\\le\n n\\norm{x}_\\infty\n \\sum_{p\\mid n}\\frac{v_p(n)}p.\n \\end{aligned}\n\\]\nSince $p\\ge2$,\n\\[\n \\frac1p\\le\\frac1{2\\log2}\\log p.\n\\]\nTherefore\n\\[\n \\sum_{p\\mid n}\\frac{v_p(n)}p\n \\le\n \\frac{\\log n}{2\\log2},\n\\]\nand\n\\[\n |\\mathrm d_x(n)|\n \\le\n \\frac{n\\log n}{2\\log2}\\norm{x}_\\infty.\n\\]\nConsequently,\n\\[\n \\begin{aligned}\n |W_x(a,b)|\n &\\le\n a|\\mathrm d_x(b)|+b|\\mathrm d_x(a)|","\\subsection{The master height inequality}\nFor every positive $n$,\n\\[\n \\begin{aligned}\n |\\mathrm d_x(n)|\n &\\le\n n\\norm{x}_\\infty\n \\sum_{p\\mid n}\\frac{v_p(n)}p.\n \\end{aligned}\n\\]\nSince $p\\ge2$,\n\\[\n \\frac1p\\le\\frac1{2\\log2}\\log p.\n\\]\nTherefore\n\\[\n \\sum_{p\\mid n}\\frac{v_p(n)}p\n \\le\n \\frac{\\log n}{2\\log2},\n\\]\nand\n\\[\n |\\mathrm d_x(n)|\n \\le\n \\frac{n\\log n}{2\\log2}\\norm{x}_\\infty.\n\\]\nConsequently,\n\\[\n \\begin{aligned}\n |W_x(a,b)|\n &\\le\n a|\\mathrm d_x(b)|+b|\\mathrm d_x(a)|\\\\\n &\\le\n \\frac{ab}{\\log2}\\norm{x}_\\infty\\log c.\n \\end{aligned}\n\\]\nCombining this with radical divisibility gives\n\\[\n \\boxed{\n c\n \\le\n \\frac{R}{\\log2}\\norm{x}_\\infty\\log c.\n }\n\\]","&\\le\n \\frac{ab}{\\log2}\\norm{x}_\\infty\\log c.\n \\end{aligned}\n\\]\nCombining this with radical divisibility gives\n\\[\n \\boxed{\n c\n \\le\n \\frac{R}{\\log2}\\norm{x}_\\infty\\log c.\n }\n\\]","Except for $(1,1,2)$, define the shortest transverse packet\n\\[\n \\boxed{\n \\Theta(a,b,c)\n =\n \\min\\left\\{\n \\norm{x}_\\infty:\n x\\in\\mathcal T_{a,b,c},\n W_x(a,b)\\neq0\n \\right\\}.\n }\n\\]\nThen\n\\[\n c\n \\le\n \\frac{R}{\\log2}\\Theta(a,b,c)\\log c.\n\\]\nThe exceptional triple $(1,1,2)$ has zero input derivatives and is omitted from $\\Theta$. It is absorbed into the constant in the conjecture. A subpolynomial estimate\n\\[\n \\Theta(a,b,c)=c^{o(1)}\n\\]\nwould imply the strong $abc$ inequality. Later sections explain why this target is too close to the conjecture to be reached by generic lattice geometry alone.","ept for $(1,1,2)$, define the shortest transverse packet\n\\[\n \\boxed{\n \\Theta(a,b,c)\n =\n \\min\\left\\{\n \\norm{x}_\\infty:\n x\\in\\mathcal T_{a,b,c},\n W_x(a,b)\\neq0\n \\right\\}.\n }\n\\]\nThen\n\\[\n c\n \\le\n \\frac{R}{\\log2}\\Theta(a,b,c)\\log c.\n\\]\nThe exceptional triple $(1,1,2)$ has zero input derivatives and is omitted from $\\Theta$. It is absorbed into the constant in the conjecture. A subpolynomial estimate\n\\[\n \\Theta(a,b,c)=c^{o(1)}\n\\]\nwould imply the strong $abc$ inequality. Later sections explain why this target is too close to the conjecture to be reached by generic lattice geometry alone.","cket\n\\[\n \\boxed{\n \\Theta(a,b,c)\n =\n \\min\\left\\{\n \\norm{x}_\\infty:\n x\\in\\mathcal T_{a,b,c},\n W_x(a,b)\\neq0\n \\right\\}.\n }\n\\]\nThen\n\\[\n c\n \\le\n \\frac{R}{\\log2}\\Theta(a,b,c)\\log c.\n\\]\nThe exceptional triple $(1,1,2)$ has zero input derivatives and is omitted from $\\Theta$. It is absorbed into the constant in the conjecture. A subpolynomial estimate\n\\[\n \\Theta(a,b,c)=c^{o(1)}\n\\]\nwould imply the strong $abc$ inequality. Later sections explain why this target is too close to the conjecture to be reached by generic lattice geometry alone.","\\subsection{Three-sector compression}\nFor the empty unit sector, set $g_1=0$, $t_1=0$, and $\\rho_1(0)=0$. Any nonzero target in that sector is infeasible. Within a nonempty $a$-sector, define the integer row\n\\[\n \\lambda^{(a)}\n =\n \\left(\\frac{a v_p(a)}p\\right)_{p\\mid a}\n\\]\nand let\n\\[\n g_a=\\gcd\\left\\{\\frac{a v_p(a)}p:p\\mid a\\right\\}.\n\\]\nDefine $g_b$ and $g_c$ similarly. A one-row Smith normal form in each sector produces a unimodular coordinate system in which only one aggregate channel survives. The additive equation becomes\n\\[\n \\boxed{\n g_at_a+g_bt_b-g_ct_c=0.\n }\n\\]\nThe Wronskian becomes\n\\[\n \\boxed{\n a g_bt_b-b g_at_a\\neq0.\n }\n\\]\nAll remaining sector coordinates are lift gauges. They matter only because they determine the cost of realizing an aggregate channel by a","\\subsection{Three-sector compression}\nFor the empty unit sector, set $g_1=0$, $t_1=0$, and $\\rho_1(0)=0$. Any nonzero target in that sector is infeasible. Within a nonempty $a$-sector, define the integer row\n\\[\n \\lambda^{(a)}\n =\n \\left(\\frac{a v_p(a)}p\\right)_{p\\mid a}\n\\]\nand let\n\\[\n g_a=\\gcd\\left\\{\\frac{a v_p(a)}p:p\\mid a\\right\\}.\n\\]\nDefine $g_b$ and $g_c$ similarly. A one-row Smith normal form in each sector produces a unimodular coordinate system in which only one aggregate channel survives. The additive equation becomes\n\\[\n \\boxed{\n g_at_a+g_bt_b-g_ct_c=0.\n }\n\\]\nThe Wronskian becomes\n\\[\n \\boxed{\n a g_bt_b-b g_at_a\\neq0.\n }\n\\]\nAll remaining sector coordinates are lift gauges. They matter only because they determine the cost of realizing an aggregate channel by an integra","\\subsection{Three-sector compression}\nFor the empty unit sector, set $g_1=0$, $t_1=0$, and $\\rho_1(0)=0$. Any nonzero target in that sector is infeasible. Within a nonempty $a$-sector, define the integer row\n\\[\n \\lambda^{(a)}\n =\n \\left(\\frac{a v_p(a)}p\\right)_{p\\mid a}\n\\]\nand let\n\\[\n g_a=\\gcd\\left\\{\\frac{a v_p(a)}p:p\\mid a\\right\\}.\n\\]\nDefine $g_b$ and $g_c$ similarly. A one-row Smith normal form in each sector produces a unimodular coordinate system in which only one aggregate channel survives. The additive equation becomes\n\\[\n \\boxed{\n g_at_a+g_bt_b-g_ct_c=0.\n }\n\\]\nThe Wronskian becomes\n\\[\n \\boxed{\n a g_bt_b-b g_at_a\\neq0.\n }\n\\]\nAll remaining sector coordinates are lift gauges. They matter only because they determine the cost of realizing an aggregate channel by an integral prime packet.","sector is infeasible. Within a nonempty $a$-sector, define the integer row\n\\[\n \\lambda^{(a)}\n =\n \\left(\\frac{a v_p(a)}p\\right)_{p\\mid a}\n\\]\nand let\n\\[\n g_a=\\gcd\\left\\{\\frac{a v_p(a)}p:p\\mid a\\right\\}.\n\\]\nDefine $g_b$ and $g_c$ similarly. A one-row Smith normal form in each sector produces a unimodular coordinate system in which only one aggregate channel survives. The additive equation becomes\n\\[\n \\boxed{\n g_at_a+g_bt_b-g_ct_c=0.\n }\n\\]\nThe Wronskian becomes\n\\[\n \\boxed{\n a g_bt_b-b g_at_a\\neq0.\n }\n\\]\nAll remaining sector coordinates are lift gauges. They matter only because they determine the cost of realizing an aggregate channel by an integral prime packet.","nteger row\n\\[\n \\lambda^{(a)}\n =\n \\left(\\frac{a v_p(a)}p\\right)_{p\\mid a}\n\\]\nand let\n\\[\n g_a=\\gcd\\left\\{\\frac{a v_p(a)}p:p\\mid a\\right\\}.\n\\]\nDefine $g_b$ and $g_c$ similarly. A one-row Smith normal form in each sector produces a unimodular coordinate system in which only one aggregate channel survives. The additive equation becomes\n\\[\n \\boxed{\n g_at_a+g_bt_b-g_ct_c=0.\n }\n\\]\nThe Wronskian becomes\n\\[\n \\boxed{\n a g_bt_b-b g_at_a\\neq0.\n }\n\\]\nAll remaining sector coordinates are lift gauges. They matter only because they determine the cost of realizing an aggregate channel by an integral prime packet.","For each term define\n\\[\n \\rho_a(t)\n =\n \\min\\left\\{\n \\norm{x^{(a)}}_\\infty:\n \\lambda^{(a)}\\cdot x^{(a)}=g_at\n \\right\\},\n\\]\nwith analogous functions for $b$ and $c$. Then\n\\[\n \\Theta(a,b,c)\n =\n \\min\n \\max\\{\\rho_a(t_a),\\rho_b(t_b),\\rho_c(t_c)\\}\n\\]\nover aggregate solutions satisfying additivity and transversality.","\\subsection{Worked example: \\texorpdfstring{$3+125=128$}{3+125=128}}\nThe prime packets are $3$, $5$, and $2$. The additive row is\n\\[\n \\lambda=(1,75,-448).\n\\]\nA shortest transverse packet is\n\\[\n x=(-2,6,1).\n\\]\nIt gives\n\\[\n \\mathrm d_x(3)=-2,\n \\qquad\n \\mathrm d_x(125)=450,\n \\qquad\n \\mathrm d_x(128)=448.\n\\]\nThus additivity holds. The Wronskian is\n\\[\n W_x(3,125)\n =3(450)-125(-2)\n =1600.\n\\]\nThe radical is\n\\[\n R=2\\cdot3\\cdot5=30,\n\\]\nand\n\\[\n \\frac{3\\cdot125\\cdot128}{30}=1600.\n\\]\nHence the radical lower bound is attained exactly.","ction{Worked example: \\texorpdfstring{$3+125=128$}{3+125=128}}\nThe prime packets are $3$, $5$, and $2$. The additive row is\n\\[\n \\lambda=(1,75,-448).\n\\]\nA shortest transverse packet is\n\\[\n x=(-2,6,1).\n\\]\nIt gives\n\\[\n \\mathrm d_x(3)=-2,\n \\qquad\n \\mathrm d_x(125)=450,\n \\qquad\n \\mathrm d_x(128)=448.\n\\]\nThus additivity holds. The Wronskian is\n\\[\n W_x(3,125)\n =3(450)-125(-2)\n =1600.\n\\]\nThe radical is\n\\[\n R=2\\cdot3\\cdot5=30,\n\\]\nand\n\\[\n \\frac{3\\cdot125\\cdot128}{30}=1600.\n\\]\nHence the radical lower bound is attained exactly.","string{$3+125=128$}{3+125=128}}\nThe prime packets are $3$, $5$, and $2$. The additive row is\n\\[\n \\lambda=(1,75,-448).\n\\]\nA shortest transverse packet is\n\\[\n x=(-2,6,1).\n\\]\nIt gives\n\\[\n \\mathrm d_x(3)=-2,\n \\qquad\n \\mathrm d_x(125)=450,\n \\qquad\n \\mathrm d_x(128)=448.\n\\]\nThus additivity holds. The Wronskian is\n\\[\n W_x(3,125)\n =3(450)-125(-2)\n =1600.\n\\]\nThe radical is\n\\[\n R=2\\cdot3\\cdot5=30,\n\\]\nand\n\\[\n \\frac{3\\cdot125\\cdot128}{30}=1600.\n\\]\nHence the radical lower bound is attained exactly.","The minimality is elementary. At height at most five, $|x_3+75x_5|\\le380<448$, so additivity forces $x_2=0$. Then $x_3=-75x_5$ and the same height restriction force both coordinates to vanish. Thus no transverse packet has height below six, and the displayed packet proves $\\Theta=6$.","\\subsection{Replacing one exponent by a valuation word}\nThe coefficient $v_p(n)$ compresses a run of valuation shells into one scalar. For a prime power $p^e$, define the valuation word\n\\[\n \\boxed{\n V_e(T)=1+T+\\cdots+T^{e-1}.\n }\n\\]\nSet $V_0(T)=0$ for an absent prime. It obeys the concatenation identity\n\\[\n V_{e+f}(T)=V_e(T)+T^eV_f(T).\n\\]\nThe refined partial derivative\n\\[\n \\partial_{p,T}(n)\n =\n \\frac npV_{v_p(n)}(T)\n\\]\ntherefore satisfies the twisted product rule\n\\[\n \\boxed{\n \\partial_{p,T}(mn)\n =\n n\\partial_{p,T}(m)\n +\n mT^{v_p(m)}\\partial_{p,T}(n).\n }\n\\]\nAt $T=1$, the ordinary arithmetic partial derivative is recovered.","nt $v_p(n)$ compresses a run of valuation shells into one scalar. For a prime power $p^e$, define the valuation word\n\\[\n \\boxed{\n V_e(T)=1+T+\\cdots+T^{e-1}.\n }\n\\]\nSet $V_0(T)=0$ for an absent prime. It obeys the concatenation identity\n\\[\n V_{e+f}(T)=V_e(T)+T^eV_f(T).\n\\]\nThe refined partial derivative\n\\[\n \\partial_{p,T}(n)\n =\n \\frac npV_{v_p(n)}(T)\n\\]\ntherefore satisfies the twisted product rule\n\\[\n \\boxed{\n \\partial_{p,T}(mn)\n =\n n\\partial_{p,T}(m)\n +\n mT^{v_p(m)}\\partial_{p,T}(n).\n }\n\\]\nAt $T=1$, the ordinary arithmetic partial derivative is recovered.","the valuation word\n\\[\n \\boxed{\n V_e(T)=1+T+\\cdots+T^{e-1}.\n }\n\\]\nSet $V_0(T)=0$ for an absent prime. It obeys the concatenation identity\n\\[\n V_{e+f}(T)=V_e(T)+T^eV_f(T).\n\\]\nThe refined partial derivative\n\\[\n \\partial_{p,T}(n)\n =\n \\frac npV_{v_p(n)}(T)\n\\]\ntherefore satisfies the twisted product rule\n\\[\n \\boxed{\n \\partial_{p,T}(mn)\n =\n n\\partial_{p,T}(m)\n +\n mT^{v_p(m)}\\partial_{p,T}(n).\n }\n\\]\nAt $T=1$, the ordinary arithmetic partial derivative is recovered.","\\subsection{Finite polyphase rings}\nFix an integer $N\\ge2$ and work in\n\\[\n A_N=\\Z[T]/(T^N-1).\n\\]\nThe residue of $V_e(T)$ records the number of valuation shells in each congruence class modulo $N$. Multiplication by $V_e$ is represented by a circulant matrix $C_{e,N}$.","\\begin{theorem}[Polyphase Smith normal form]\nIf $\\gcd(e,N)=1$, then\n\\[\n \\boxed{\n \\SNF(C_{e,N})\n =\n \\operatorname{diag}(1,\\ldots,1,e).\n }\n\\]\n\\end{theorem}","\\begin{proof}\nLet\n\\[\n I_N=\\ker\\bigl(A_N\\xrightarrow{\\,T\\mapsto1\\,}\\Z\\bigr)\n\\]\nbe the augmentation ideal. It is a free abelian group of rank $N-1$, and multiplication by $V_e$ preserves it. After extension to $\\C$, the eigenvalues on $I_N\\otimes\\C$ are\n\\[\n V_e(\\zeta)\n =\n \\frac{1-\\zeta^e}{1-\\zeta},\n \\qquad\n \\zeta^N=1,\n \\quad\n \\zeta\\neq1.\n\\]\nBecause $\\gcd(e,N)=1$, the map $\\zeta\\mapsto\\zeta^e$ permutes the nontrivial $N$th roots. Therefore\n\\[\n \\det\\bigl(V_e\\mid I_N\\bigr)\n =\n \\prod_{\\substack{\\zeta^N=1","[\n I_N=\\ker\\bigl(A_N\\xrightarrow{\\,T\\mapsto1\\,}\\Z\\bigr)\n\\]\nbe the augmentation ideal. It is a free abelian group of rank $N-1$, and multiplication by $V_e$ preserves it. After extension to $\\C$, the eigenvalues on $I_N\\otimes\\C$ are\n\\[\n V_e(\\zeta)\n =\n \\frac{1-\\zeta^e}{1-\\zeta},\n \\qquad\n \\zeta^N=1,\n \\quad\n \\zeta\\neq1.\n\\]\nBecause $\\gcd(e,N)=1$, the map $\\zeta\\mapsto\\zeta^e$ permutes the nontrivial $N$th roots. Therefore\n\\[\n \\det\\bigl(V_e\\mid I_N\\bigr)\n =\n \\prod_{\\substack{\\zeta^N=1","ghtarrow{\\,T\\mapsto1\\,}\\Z\\bigr)\n\\]\nbe the augmentation ideal. It is a free abelian group of rank $N-1$, and multiplication by $V_e$ preserves it. After extension to $\\C$, the eigenvalues on $I_N\\otimes\\C$ are\n\\[\n V_e(\\zeta)\n =\n \\frac{1-\\zeta^e}{1-\\zeta},\n \\qquad\n \\zeta^N=1,\n \\quad\n \\zeta\\neq1.\n\\]\nBecause $\\gcd(e,N)=1$, the map $\\zeta\\mapsto\\zeta^e$ permutes the nontrivial $N$th roots. Therefore\n\\[\n \\det\\bigl(V_e\\mid I_N\\bigr)\n =\n \\prod_{\\substack{\\zeta^N=1\\\\\\zeta\\neq1}}\n \\frac{1-\\zeta^e}{1-\\zeta}\n =1.\n\\]\nThus multiplication by $V_e$ is an automorphism of the integral lattice $I_N$. On the quotient\n\\[\n A_N/I_N\\cong\\Z,\n\\]\nit acts as multiplication by $V_e(1)=e$. Applying the snake lemma to\n\\[\n 0\\longrightarrow I_N\\longrightarrow A_N\\longrightarrow\\Z\\longrightarrow0\n\\]\nshows that the full map is injective and has cokernel $\\Z/e\\Z$. Hence its Smith normal form is\n\\[\n \\operatorname{diag}(1,\\ldots,1,e).\n\\]\n\\end{proof}","zeta\\neq1}}\n \\frac{1-\\zeta^e}{1-\\zeta}\n =1.\n\\]\nThus multiplication by $V_e$ is an automorphism of the integral lattice $I_N$. On the quotient\n\\[\n A_N/I_N\\cong\\Z,\n\\]\nit acts as multiplication by $V_e(1)=e$. Applying the snake lemma to\n\\[\n 0\\longrightarrow I_N\\longrightarrow A_N\\longrightarrow\\Z\\longrightarrow0\n\\]\nshows that the full map is injective and has cokernel $\\Z/e\\Z$. Hence its Smith normal form is\n\\[\n \\operatorname{diag}(1,\\ldots,1,e).\n\\]\n\\end{proof}","\\frac{1-\\zeta^e}{1-\\zeta}\n =1.\n\\]\nThus multiplication by $V_e$ is an automorphism of the integral lattice $I_N$. On the quotient\n\\[\n A_N/I_N\\cong\\Z,\n\\]\nit acts as multiplication by $V_e(1)=e$. Applying the snake lemma to\n\\[\n 0\\longrightarrow I_N\\longrightarrow A_N\\longrightarrow\\Z\\longrightarrow0\n\\]\nshows that the full map is injective and has cokernel $\\Z/e\\Z$. Hence its Smith normal form is\n\\[\n \\operatorname{diag}(1,\\ldots,1,e).\n\\]\n\\end{proof}","The refined block has $N-1$ integrally invertible phase channels and one aggregate multiplicity channel. Every local denominator lies in the aggregate channel.","\\subsection{Dyadic and odd-adic primary lifts}\nFactor\n\\[\n e=2^\\alpha\\prod_{\\ell\\text{ odd}}\\ell^{\\alpha_\\ell}.\n\\]\nThe cokernel of the valuation block is\n\\[\n \\Z/e\\Z\n \\cong\n \\Z/2^\\alpha\\Z\n \\times\n \\prod_{\\ell\\text{ odd}}\n \\Z/\\ell^{\\alpha_\\ell}\\Z.\n\\]\nThe primary factors specify the local lifting problems for the aggregate multiplicity channel. Its $2$-primary component must be treated $2$-adically, while the odd components admit the corresponding odd-adic lifts. The phase channels require no local denominator correction.","This Smith decomposition identifies where separate $2$-primary bookkeeping first enters. The additional one-level shift in multiplicative principal-unit coordinates at the prime $2$ comes from the local $2$-adic logarithm and is established later. It is not a consequence of the Smith theorem alone.","\\subsection{Coarse and phase derivatives}\nChoose a polyphase order $N$ coprime to every active valuation exponent. After the Smith decomposition, each refined prime packet has one coarse coordinate and integrally invertible phase coordinates in the common augmentation lattice. The additive equation separates accordingly.","The coarse equation recovers the original exponent-weighted row\n\\[\n \\sum_{p\\mid a}\\frac{av_p(a)}p t_p\n +\n \\sum_{p\\mid b}\\frac{bv_p(b)}p t_p\n -\n \\sum_{p\\mid c}\\frac{cv_p(c)}p t_p\n =0.\n\\]\nThe phase equation is exponent free:\n\\[\n \\boxed{\n \\sum_{p\\mid a}\\frac ap y_p\n +\n \\sum_{p\\mid b}\\frac bp y_p\n -\n \\sum_{p\\mid c}\\frac cp y_p\n =0.\n }\n\\]\nDefine the phase derivative\n\\[\n D_y(n)=n\\sum_{p\\mid n}\\frac{y_p}{p}.\n\\]\nThe same radical divisibility argument gives\n\\[\n \\frac{abc}{R}\\mid aD_y(b)-bD_y(a).\n\\]","hted row\n\\[\n \\sum_{p\\mid a}\\frac{av_p(a)}p t_p\n +\n \\sum_{p\\mid b}\\frac{bv_p(b)}p t_p\n -\n \\sum_{p\\mid c}\\frac{cv_p(c)}p t_p\n =0.\n\\]\nThe phase equation is exponent free:\n\\[\n \\boxed{\n \\sum_{p\\mid a}\\frac ap y_p\n +\n \\sum_{p\\mid b}\\frac bp y_p\n -\n \\sum_{p\\mid c}\\frac cp y_p\n =0.\n }\n\\]\nDefine the phase derivative\n\\[\n D_y(n)=n\\sum_{p\\mid n}\\frac{y_p}{p}.\n\\]\nThe same radical divisibility argument gives\n\\[\n \\frac{abc}{R}\\mid aD_y(b)-bD_y(a).\n\\]","Let\n\\[\n \\sigma(a,b)\n =\n \\sum_{p\\mid a}\\frac1p\n +\n \\sum_{p\\mid b}\\frac1p.\n\\]\nFor primitive triples other than $(1,1,2)$, let $\\Phi(a,b,c)$ be the shortest transverse phase packet. Then\n\\[\n \\boxed{\n c\n \\le\n R\\,\\sigma(a,b)\\,\\Phi(a,b,c).\n }\n\\]\nThis can be sharper than the exponent-weighted inequality on individual triples, but finite evidence still shows positive power growth rather than a subpolynomial law.","\\subsection{First-order no-free-refinement principle}\nThe shell-refined factor\n\\[\n \\prod_{r=0}^{e-1}\n \\left(1+\\frac{T_r}{p}\\right)\n\\]\nhas first-order expansion\n\\[\n 1+\\frac1p\\sum_rT_r+O(T^2).\n\\]\nOnly the shell sum appears. Every internal phase in the kernel of shell synthesis is invisible to both first-order additivity and the first-order Wronskian.","\\begin{theorem}[First-order no-free-polyphase principle]\nLet\n\\[\n S:(T_{p,r})_{p,r}\\longmapsto\n \\left(\\sum_rT_{p,r}\\right)_p\n\\]\nbe shell synthesis. The first-order additive map and first-order Wronskian factor through $S$. Hence\n\\[\n \\Ker S\n\\]\nis contained in the kernels of both first-order maps.\n\\end{theorem}","\\subsection{Second-order phase energy}\nAt second order,\n\\[\n \\prod_{r=0}^{e-1}\n \\left(1+\\frac{T_r}{p}\\right)\n =\n 1+\n \\frac1p\\sum_rT_r+\n \\frac1{p^2}\\sum_{r<s}T_rT_s+\n O(T^3).\n\\]\nLet $\\widehat T_j$ be the discrete Fourier transform of the shell coordinates. Parseval gives\n\\[\n \\sum_{r<s}T_rT_s\n =\n \\frac12\\left[\n \\widehat T_0^2\n -\n \\frac1e\\sum_j\n \\widehat T_j\\widehat T_{-j}\n \\right].\n\\]\nFor real coordinates,\n\\[\n \\sum_{r<s}T_rT_s\n =\n \\frac12\\left[\n \\left(1-\\frac1e\\right)|\\widehat T_0|^2\n -\n \\frac1e\\sum_{j\\neq0}|\\widehat T_j|^2\n \\right].\n\\]\nThus nonzero polyphase energy first becomes visible at second order. Later formal-smoothness results show that this curvature is not itself a lifting obstruction.","on{Second-order phase energy}\nAt second order,\n\\[\n \\prod_{r=0}^{e-1}\n \\left(1+\\frac{T_r}{p}\\right)\n =\n 1+\n \\frac1p\\sum_rT_r+\n \\frac1{p^2}\\sum_{r<s}T_rT_s+\n O(T^3).\n\\]\nLet $\\widehat T_j$ be the discrete Fourier transform of the shell coordinates. Parseval gives\n\\[\n \\sum_{r<s}T_rT_s\n =\n \\frac12\\left[\n \\widehat T_0^2\n -\n \\frac1e\\sum_j\n \\widehat T_j\\widehat T_{-j}\n \\right].\n\\]\nFor real coordinates,\n\\[\n \\sum_{r<s}T_rT_s\n =\n \\frac12\\left[\n \\left(1-\\frac1e\\right)|\\widehat T_0|^2\n -\n \\frac1e\\sum_{j\\neq0}|\\widehat T_j|^2\n \\right].\n\\]\nThus nonzero polyphase energy first becomes visible at second order. Later formal-smoothness results show that this curvature is not itself a lifting obstruction.","\\subsection{Exact local lift gauges}\nLet\n\\[\n r_u=\\rad(u),\n \\qquad\n P_u=\\frac{u}{r_u}.\n\\]\nA shell packet $y=(y_{p,r})$ produces\n\\[\n D_y(u)\n =\n u\\sum_{p\\mid u}\\frac1p\n \\sum_{r=0}^{v_p(u)-1}y_{p,r}\n =P_ut_u,\n\\]\nwhere\n\\[\n t_u=\n \\sum_{p\\mid u}\n \\frac{r_u}{p}\n \\sum_r y_{p,r}.\n\\]\nBecause\n\\[\n \\gcd_{p\\mid u}\\frac{r_u}{p}=1,\n\\]\nevery integer $t_u$ is attainable when $u>1$.","The shell lift gauge is\n\\[\n \\rho_u^{\\mathrm{sh}}(t)\n =\n \\min\\left\\{\n \\norm{y}_\\infty:\n \\sum_{p\\mid u}\n \\frac{r_u}{p}\n \\sum_r y_{p,r}=t\n \\right\\}.\n\\]\nAt the aggregate level,\n\\[\n P_at_a+P_bt_b=P_ct_c,\n\\]\nand\n\\[\n W_y(a,b)\n =\n P_aP_b(r_at_b-r_bt_a).\n\\]\nThe radical divisor is attained sharply when\n\\[\n r_at_b-r_bt_a=P_c.\n\\]\nWhen $a,b>1$, all sharp aggregate solutions form a one-parameter family\n\\[\n \\begin{aligned}\n t_a(n)&=t_a^{(0)}+nr_a,","The shell lift gauge is\n\\[\n \\rho_u^{\\mathrm{sh}}(t)\n =\n \\min\\left\\{\n \\norm{y}_\\infty:\n \\sum_{p\\mid u}\n \\frac{r_u}{p}\n \\sum_r y_{p,r}=t\n \\right\\}.\n\\]\nAt the aggregate level,\n\\[\n P_at_a+P_bt_b=P_ct_c,\n\\]\nand\n\\[\n W_y(a,b)\n =\n P_aP_b(r_at_b-r_bt_a).\n\\]\nThe radical divisor is attained sharply when\n\\[\n r_at_b-r_bt_a=P_c.\n\\]\nWhen $a,b>1$, all sharp aggregate solutions form a one-parameter family\n\\[\n \\begin{aligned}\n t_a(n)&=t_a^{(0)}+nr_a,\\\\\n t_b(n)&=t_b^{(0)}+nr_b,\\\\\n t_c(n)&=t_c^{(0)}+nr_c.\n \\end{aligned}\n\\]\nThe unit-sector cases have no moving parameter. If $a=1$, then\n\\[\n (t_a,t_b,t_c)=(0,P_c,P_b),\n\\]\nand if $b=1$, then\n\\[\n (t_a,t_b,t_c)=(-P_c,0,-P_a).\n\\]\nFor obstruction multiple $k$, these fixed triples are multiplied by $k$. Thus the high-dimensional sharp search reduces either to one integer parameter plus three local lift gauges, or to one fixed aggregate triple in a unit-sector case.","t_c(n)&=t_c^{(0)}+nr_c.\n \\end{aligned}\n\\]\nThe unit-sector cases have no moving parameter. If $a=1$, then\n\\[\n (t_a,t_b,t_c)=(0,P_c,P_b),\n\\]\nand if $b=1$, then\n\\[\n (t_a,t_b,t_c)=(-P_c,0,-P_a).\n\\]\nFor obstruction multiple $k$, these fixed triples are multiplied by $k$. Thus the high-dimensional sharp search reduces either to one integer parameter plus three local lift gauges, or to one fixed aggregate triple in a unit-sector case.","\\subsection{A joint exponent-free two-row complex}\nThe Smith obstruction is attached to the exponent-free phase, or shell-synthesis, rows rather than to the exponent-weighted derivative row. For every prime $p\\mid abc$, define\n\\[\n \\Lambda_p=\n \\begin{cases}\n a/p,&p\\mid a,","\\subsection{A joint exponent-free two-row complex}\nThe Smith obstruction is attached to the exponent-free phase, or shell-synthesis, rows rather than to the exponent-weighted derivative row. For every prime $p\\mid abc$, define\n\\[\n \\Lambda_p=\n \\begin{cases}\n a/p,&p\\mid a,\\\\\n b/p,&p\\mid b,\\\\\n -c/p,&p\\mid c,\n \\end{cases}\n \\qquad\n \\mathsf w_p=\n \\begin{cases}\n -ab/p,&p\\mid a,\\\\\n ab/p,&p\\mid b,\\\\\n 0,&p\\mid c.\n \\end{cases}\n\\]\nConsider\n\\[\n \\mathsf M_{a,b,c}\n =\n \\begin{pmatrix}\n \\Lambda","0,&p\\mid c.\n \\end{cases}\n\\]\nConsider\n\\[\n \\mathsf M_{a,b,c}\n =\n \\begin{pmatrix}\n \\Lambda\\\\\n \\mathsf w\n \\end{pmatrix}:\n \\Z^S\\longrightarrow\\Z^2.\n\\]\nThe first row is the exponent-free additive phase row. The second is its integer Wronskian row. Set\n\\[\n Q=\\frac{abc}{R}.\n\\]","\\mathsf w\n \\end{pmatrix}:\n \\Z^S\\longrightarrow\\Z^2.\n\\]\nThe first row is the exponent-free additive phase row. The second is its integer Wronskian row. Set\n\\[\n Q=\\frac{abc}{R}.\n\\]","\\begin{theorem}[Smith obstruction theorem]\\label{thm:phase-smith}\nLet $a+b=c$ be primitive, with at least two of $a,b,c$ greater than one. Then\n\\[\n \\boxed{\n \\SNF(\\mathsf M_{a,b,c})\n =\n \\operatorname{diag}(1,Q).\n }\n\\]\nConsequently,\n\\[\n \\boxed{\n \\operatorname{coker}(\\mathsf M_{a,b,c})_{\\mathrm{tors}}\n \\cong\n \\Z/Q\\Z.\n }\n\\]\nThe exceptional triple $(1,1,2)$ has only one nonempty prime sector, so the matrix has rank one and is excluded from the two-row statement.\n\\end{theorem}","\\begin{proof}\nThe gcd of the entries of $\\Lambda$ is one. Indeed, every prime dividing one sector is absent from every other nonempty sector, and an entry from another sector is not divisible by it. Hence the first Smith factor is one.","Rows restricted to one sector are proportional, so same-sector $2\\times2$ minors vanish. If $p$ and $q$ lie in two different nonempty sectors, the corresponding minor is\n\\[\n \\pm\\frac{abc}{pq}\n =\n \\pm Q\\frac{R}{pq}.\n\\]\nEvery nonzero minor is therefore divisible by $Q$. Conversely, for each prime $\\ell\\mid R$, choose $p=\\ell$ and choose $q$ in a different nonempty sector. Since $R$ is squarefree, $R/(pq)$ is not divisible by $\\ell$. Thus the gcd of the reduced minors $R/(pq)$ is one. The gcd of all nonzero minors is exactly $Q$. The determinantal-divisor description of Smith normal form now gives the result.\n\\end{proof}","r are proportional, so same-sector $2\\times2$ minors vanish. If $p$ and $q$ lie in two different nonempty sectors, the corresponding minor is\n\\[\n \\pm\\frac{abc}{pq}\n =\n \\pm Q\\frac{R}{pq}.\n\\]\nEvery nonzero minor is therefore divisible by $Q$. Conversely, for each prime $\\ell\\mid R$, choose $p=\\ell$ and choose $q$ in a different nonempty sector. Since $R$ is squarefree, $R/(pq)$ is not divisible by $\\ell$. Thus the gcd of the reduced minors $R/(pq)$ is one. The gcd of all nonzero minors is exactly $Q$. The determinantal-divisor description of Smith normal form now gives the result.\n\\end{proof}","2$ minors vanish. If $p$ and $q$ lie in two different nonempty sectors, the corresponding minor is\n\\[\n \\pm\\frac{abc}{pq}\n =\n \\pm Q\\frac{R}{pq}.\n\\]\nEvery nonzero minor is therefore divisible by $Q$. Conversely, for each prime $\\ell\\mid R$, choose $p=\\ell$ and choose $q$ in a different nonempty sector. Since $R$ is squarefree, $R/(pq)$ is not divisible by $\\ell$. Thus the gcd of the reduced minors $R/(pq)$ is one. The gcd of all nonzero minors is exactly $Q$. The determinantal-divisor description of Smith normal form now gives the result.\n\\end{proof}","mpty sectors, the corresponding minor is\n\\[\n \\pm\\frac{abc}{pq}\n =\n \\pm Q\\frac{R}{pq}.\n\\]\nEvery nonzero minor is therefore divisible by $Q$. Conversely, for each prime $\\ell\\mid R$, choose $p=\\ell$ and choose $q$ in a different nonempty sector. Since $R$ is squarefree, $R/(pq)$ is not divisible by $\\ell$. Thus the gcd of the reduced minors $R/(pq)$ is one. The gcd of all nonzero minors is exactly $Q$. The determinantal-divisor description of Smith normal form now gives the result.\n\\end{proof}","\\subsection{The exact lattice extension}\nThroughout the remaining height and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The diffic","\\subsection{The exact lattice extension}\nThroughout the remaining height and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section,","\\subsection{The exact lattice extension}\nThroughout the remaining height and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","act lattice extension}\nThroughout the remaining height and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","the remaining height and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","ght and exact-sequence constructions, exclude $(1,1,2)$. Define\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","ne\n\\[\n L=\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","\\ker(\\Lambda)\\cap\\Z^S\n\\]\nand\n\\[\n L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","L_0=L\\cap\\ker(\\mathsf w).\n\\]\nBecause $\\Lambda$ is primitive, choose $t\\in\\Z^S$ with $\\Lambda(t)=1$. Then\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","n\n\\[\n \\Z^S=L\\oplus\\Z t.\n\\]\nThe image of $\\mathsf M_{a,b,c}$ is generated by $(1,\\mathsf w(t))$ together with $(0,\\mathsf w(L))$. Its index in $\\Z^2$ is therefore the index of $\\mathsf w(L)$ in $\\Z$. Theorem~\\ref{thm:phase-smith} shows that this index is $Q$, so\n\\[\n \\mathsf w(L)=Q\\Z.\n\\]\nHence\n\\[\n \\omega(x)=\\frac{\\mathsf w\\cdot x}{Q}\n\\]\nis an integer-valued surjection and gives the exact sequence\n\\[\n \\boxed{\n 0\\longrightarrow L_0\n \\longrightarrow L\n \\xrightarrow{\\omega}\n \\Z\n \\longrightarrow0.\n }\n\\]\nThe extension splits algebraically. The difficult quantity is the norm of a section, not its existence.","For $k\\in\\Z$, define\n\\[\n H(k)\n =\n \\min\\left\\{\n \\norm{x}_\\infty:\n x\\in L,\n \\omega(x)=k\n \\right\\}.\n\\]\nThree heights must be distinguished:\n\\[\n H(1),\n \\qquad\n \\Phi=\\min_{k\\neq0}H(k),\n \\qquad\n \\overline H=\n \\inf_{k\\ge1}\\frac{H(k)}k.\n\\]\nThe shortest nonzero detector may live in a nonprimitive obstruction coset.","\\subsection{The stable real height}\nRelax the integral packet to a real vector in the additive hyperplane. The exact continuous width can be computed from the three sector weights. Let\n\\[\n \\sigma_u=\\sum_{p\\mid u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","to a real vector in the additive hyperplane. The exact continuous width can be computed from the three sector weights. Let\n\\[\n \\sigma_u=\\sum_{p\\mid u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","ve hyperplane. The exact continuous width can be computed from the three sector weights. Let\n\\[\n \\sigma_u=\\sum_{p\\mid u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","e sector weights. Let\n\\[\n \\sigma_u=\\sum_{p\\mid u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","_u=\\sum_{p\\mid u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","u}\\frac1p\n\\]\nand\n\\[\n \\Delta(a,b,c)\n =\n \\min\\left\\{\n \\sigma_a+\\sigma_b,\n \\frac cb(\\sigma_a+\\sigma_c),\n \\frac ca(\\sigma_b+\\sigma_c)\n \\right\\}.\n\\]\nThen\n\\[\n \\boxed{\n \\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","\\overline H\n =\n \\frac{c}{R\\Delta(a,b,c)}.\n }\n\\]\nThe discrete master inequality becomes\n\\[\n c\\le R\\Delta\\Phi.\n\\]\nSince\n\\[\n \\Phi=\\overline H\\,k_*\\Gamma\n\\]\nfor the optimal obstruction multiple $k_*$ and an integrality factor $\\Gamma\\ge1$,\n\\[\n \\boxed{\n \\Delta\\Phi\n =\n \\frac cR k_*\\Gamma.\n }\n\\]\nThis is the no-escape identity. Optimizing the same first-order invariant cannot lower the continuous floor $c/R$.","\\subsection{The full shell deformation}\nIntroduce one variable for every valuation shell and define\n\\[\n \\mathscr J_n(\\mathbf T)\n =\n \\prod_{p^e\\parallel n}\n \\prod_{r=0}^{e-1}(p+T_{p,r}).\n\\]\nAt $\\mathbf T=0$,\n\\[\n \\mathscr J_n(0)=n.\n\\]\nFor $a+b=c$, set\n\\[\n F_{a,b,c}(\\mathbf T)\n =\n \\mathscr J_a(\\mathbf T)\n +\n \\mathscr J_b(\\mathbf T)\n -\n \\mathscr J_c(\\mathbf T).\n\\]\nThe shell gradient has entries\n\\[\n \\frac ap,\n \\qquad\n \\frac bp,\n \\qquad\n -\\frac cp\n\\]\nrepeated according to valuation depth. The gcd of all entries is one.","\\begin{theorem}[Formal affine-space theorem]\nLet $d$ be the number of shell variables. After a unimodular integral change of shell coordinates, the $(\\mathbf T)$-adic completion of\n\\[\n \\Z[\\mathbf T]/(F_{a,b,c})\n\\]\nat the zero section is isomorphic to a formal power-series ring $\\Z[[U_2,\\ldots,U_d]]$.\n\\end{theorem}","\\begin{proof}\nA primitive gradient row can be completed to a unimodular matrix. In the new coordinates, one variable has linear coefficient one, which is a unit in $\\Z$. The formal implicit-function recursion therefore solves uniquely for that variable at every degree with integral coefficients, giving the stated isomorphism of completed rings.\n\\end{proof}","\\begin{corollary}[Integral jet extension]\nEvery integral first-order tangent vector to $F_{a,b,c}=0$ extends to an integral formal curve through all orders.\n\\end{corollary}","\\subsection{Aggregate products and the integral boundary}\nOver $\\Q$, on the locus where all shell factors are nonzero, product-preserving directions within a sector form a torus. For $a,b>1$, the normalized products\n\\[\n A=\\mathscr J_a/a,\\qquad B=\\mathscr J_b/b,\\qquad C=\\mathscr J_c/c\n\\]\nsatisfy $aA+bB=cC$, leaving a two-dimensional aggregate plane and internal product gauges. A unit input sector has no shell variables and leaves a one-dimensional aggregate line.","\\subsection{Obstruction cosets}\nExclude $(1,1,2)$ throughout this section. Choose a section $x_0\\in L$ with $\\omega(x_0)=1$. Then\n\\[\n L\n =\n \\bigsqcup_{k\\in\\Z}\n (kx_0+L_0).\n\\]\nFor fixed $k$, the problem is a closest-vector problem to the moving target $-kx_0$ in the lattice $L_0$. The three-sector compression makes this search much more explicit.","\\subsection{Aggregate affine line}\nWrite\n\\[\n u=P_ur_u,\n \\qquad\n r_u=\\rad(u),\n \\qquad\n P_u=u/r_u.\n\\]\nThe sharp aggregate conditions are\n\\[\n P_at_a+P_bt_b=P_ct_c\n\\]\nand\n\\[\n r_at_b-r_bt_a=kP_c.\n\\]\nSince $\\gcd(r_a,r_b)=1$, choose $s,t$ with\n\\[\n sr_a+tr_b=1.\n\\]\nOne solution is\n\\[\n t_a^{(0)}=-tkP_c,\n \\qquad\n t_b^{(0)}=skP_c.\n\\]\nEvery aggregate solution is\n\\[\n \\boxed{\n \\begin{aligned}\n t_a(k,n)&=t_a^{(0)}+nr_a,","\\subsection{Aggregate affine line}\nWrite\n\\[\n u=P_ur_u,\n \\qquad\n r_u=\\rad(u),\n \\qquad\n P_u=u/r_u.\n\\]\nThe sharp aggregate conditions are\n\\[\n P_at_a+P_bt_b=P_ct_c\n\\]\nand\n\\[\n r_at_b-r_bt_a=kP_c.\n\\]\nSince $\\gcd(r_a,r_b)=1$, choose $s,t$ with\n\\[\n sr_a+tr_b=1.\n\\]\nOne solution is\n\\[\n t_a^{(0)}=-tkP_c,\n \\qquad\n t_b^{(0)}=skP_c.\n\\]\nEvery aggregate solution is\n\\[\n \\boxed{\n \\begin{aligned}\n t_a(k,n)&=t_a^{(0)}+nr_a,\\\\\n t_b(k,n)&=t_b^{(0)}+nr_b,\\\\\n t_c(k,n)&=t_c^{(0)}+nr_c,\n \\end{aligned}\n \\qquad n\\in\\Z.\n }\n\\]\nWhen $a,b>1$, the global search in one obstruction coset is therefore one-dimensional. If $a=1$, the empty $a$-sector forces $t_a=0$ and fixes the coarse parameter, giving\n\\[\n (t_a,t_b,t_c)=(0,kP_c,kP_b).\n\\]\nIf $b=1$, the corresponding fixed triple is\n\\[\n (t_a,t_b,t_c)=(-kP_c,0,-kP_a).\n\\]\nThus a unit-sector coset has one aggregate candidate rather than a moving affine line.","t_c(k,n)&=t_c^{(0)}+nr_c,\n \\end{aligned}\n \\qquad n\\in\\Z.\n }\n\\]\nWhen $a,b>1$, the global search in one obstruction coset is therefore one-dimensional. If $a=1$, the empty $a$-sector forces $t_a=0$ and fixes the coarse parameter, giving\n\\[\n (t_a,t_b,t_c)=(0,kP_c,kP_b).\n\\]\nIf $b=1$, the corresponding fixed triple is\n\\[\n (t_a,t_b,t_c)=(-kP_c,0,-kP_a).\n\\]\nThus a unit-sector coset has one aggregate candidate rather than a moving affine line.","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be non","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer betwee","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their en","\\subsection{Local CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient","CRT normal form}\nFix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","ix one sector with primes\n\\[\n p_1,\\ldots,p_m\n\\]\nand radical\n\\[\n r=\\prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","prod_i p_i.\n\\]\nFor a nonempty sector, the local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","e local equation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","uation is\n\\[\n \\sum_i\\frac r{p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","p_i}x_i=t.\n\\]\nAn empty unit sector has no prime coordinates and is feasible exactly when $t=0$.\nModulo $p_i$, the coordinate $x_i$ is forced:\n\\[\n x_i\n \\equiv\n t\\left(\\frac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","rac r{p_i}\\right)^{-1}\n \\pmod{p_i}.\n\\]\nChoose the centered residue $\\xi_i$ and write\n\\[\n x_i=\\xi_i+p_i z_i.\n\\]\nSubstitution gives one quotient equation\n\\[\n \\sum_i z_i=q(t).\n\\]\nFor a box radius $H$, each $z_i$ lies in the interval\n\\[\n \\ell_i(H)\n =\n \\left\\lceil\\frac{-H-\\xi_i}{p_i}\\right\\rceil\n \\le z_i\\le\n \\left\\lfloor\\frac{H-\\xi_i}{p_i}\\right\\rfloor\n =u_i(H).\n\\]\nThe local lift exists exactly when\n\\[\n \\boxed{\n \\ell_i(H)\\le u_i(H)\\ \\text{for every }i,\\qquad\n \\sum_i\\ell_i(H)\\le q(t)\\le\\sum_i u_i(H).\n }\n\\]\nEvery coordinate interval must be nonempty. Once that condition holds, the sum of the integer intervals contains every integer between the sums of their endpoints. This proves the criterion. Summing endpoints alone is insufficient when one interval is empty.","\\subsection{Moving-interval theorem}\nUsing the complete local criterion, including the individual nonemptiness guards, for fixed $k$ and $H$ the three sectors produce three integer intervals for the coarse parameter $n$. A packet of height at most $H$ exists if and only if those intervals share an integer.","\\begin{theorem}[Moving-interval theorem]\nFor fixed obstruction multiple $k$ and radius $H$,\n\\[\n H(k)\\le H\n\\]\nif and only if\n\\[\n I_a(H,k)\n \\cap\n I_b(H,k)\n \\cap\n I_c(H,k)\n \\cap\\Z\n \\neq\\varnothing.\n\\]\n\\end{theorem}","The first feasible $H$ gives the exact coset height. The preceding radius is an explicit no-smaller-box certificate.","\\subsection{Finite global stopping}\nThe continuous lower bound gives\n\\[\n H(k)\\ge k\\overline H.\n\\]\nIf the best exact height found so far is $B$, then no coset with\n\\[\n k\\overline H\\ge B\n\\]\ncan improve it. Therefore only finitely many obstruction multiples must be searched:\n\\[\n \\boxed{\n 1\\le k\\le\n \\left\\lfloor\\frac{B-1}{\\overline H}\\right\\rfloor.\n }\n\\]\nEvery unsearched coset is excluded by a proved lower bound.","\\subsection{Phase and coarse separation}\nThe local target has the form\n\\[\n t_u(k,n)=\\alpha_uk+r_un.\n\\]\nFor a local prime $p\\mid r_u$,\n\\[\n t_u(k,n)\\equiv\\alpha_uk\\pmod p.\n\\]\nThus the obstruction multiple $k$ determines the finite phase packet. The coarse parameter $n$ changes only the quotient load. The exact search is a moving polyphase mechanism rather than a generic mixed-integer program.","\\subsection{Arithmetic correction and arbitrary precision}\nSome large high-quality triples exceed signed $64$-bit range, and intermediate quantities such as\n\\[\n ab,\n \\qquad\n \\frac{abc}{R},\n \\qquad\n w\\cdot x\n\\]\ncan overflow even when the input terms individually fit. Every public arithmetic entry point must therefore promote integer-like inputs to arbitrary-precision integers before multiplication. Floating-point values are display-only. All logical comparisons and stopping bounds are exact rational or integer computations.","Several reported ratios are mathematically at least one:\n\\[\n \\frac{\\Phi}{\\overline H}\\ge1,\n \\qquad\n \\frac{\\Phi}{k_*\\overline H}\\ge1,\n \\qquad\n \\frac{H(1)}{\\Phi}\\ge1.\n\\]\nValues above one are not probability errors.","\\subsection{What the exact search proved}\nThe moving search found nonprimitive optima on many finite examples. It also found increasingly deep phase resonances. This showed that a fixed cutoff in $k$ is unjustified. More importantly, the no-escape identity proved that even a complete exact search of the first-order invariant cannot imply $abc$ independently. The search became a diagnostic and classification engine rather than the final proof mechanism.","\\subsection{Prime support as bandwidth and valuation as persistence}\nFor\n\\[\n n=\\prod_p p^{e_p},\n\\]\ndefine the prime-shell signal\n\\[\n \\Psi_n(p,r)\n =\n \\sqrt{\\log p}\\,\\mathbf 1_{0\\le r<e_p}.\n\\]\nIts total energy is\n\\[\n \\norm{\\Psi_n}_2^2\n =\n \\sum_p e_p\\log p\n =\n \\boxed{\\log n}.\n\\]\nLet $P_{\\mathrm{act}}$ retain only the first shell in each active prime channel. Then\n\\[\n \\norm{P_{\\mathrm{act}}\\Psi_n}_2^2\n =\n \\sum_{p\\mid n}\\log p\n =\n \\boxed{\\log\\rad(n)}.\n\\]\nThus the radical is activated bandwidth, while the difference\n\\[\n \\log n-\\log\\rad(n)\n\\]\nis repeated persistence depth.","ell signal\n\\[\n \\Psi_n(p,r)\n =\n \\sqrt{\\log p}\\,\\mathbf 1_{0\\le r<e_p}.\n\\]\nIts total energy is\n\\[\n \\norm{\\Psi_n}_2^2\n =\n \\sum_p e_p\\log p\n =\n \\boxed{\\log n}.\n\\]\nLet $P_{\\mathrm{act}}$ retain only the first shell in each active prime channel. Then\n\\[\n \\norm{P_{\\mathrm{act}}\\Psi_n}_2^2\n =\n \\sum_{p\\mid n}\\log p\n =\n \\boxed{\\log\\rad(n)}.\n\\]\nThus the radical is activated bandwidth, while the difference\n\\[\n \\log n-\\log\\rad(n)\n\\]\nis repeated persistence depth.","For a primitive triple, the direct-sum signal\n\\[\n \\Psi_{a,b,c}\n =\n \\Psi_a\\oplus\\Psi_b\\oplus\\Psi_c\n\\]\nhas disjoint prime support in the three sectors. The standard quality becomes the exact energy ratio\n\\[\n \\boxed{\n q(a,b,c)\n =\n \\frac{\\log c}{\\log\\rad(abc)}\n =\n \\frac{\\norm{P_c\\Psi_{a,b,c}}_2^2}\n {\\norm{P_{\\mathrm{act}}\\Psi_{a,b,c}}_2^2}.\n }\n\\]\nHigh quality means large output persistence relative to activated global bandwidth.","\\begin{proposition}[Unitary no-gain principle]\nLet $U$ be unitary on the shell coordinate of every active prime channel. Then\n\\[\n \\norm{U\\Psi_n}_2=\\norm{\\Psi_n}_2.\n\\]\nTransport the activation projector to $UP_{\\mathrm{act}}U^{-1}$. Its projected energy then also agrees exactly with the original activation energy. The same holds after exact synthesis. An arbitrary unitary transform need not preserve the energy under a fixed, untransported first-shell projection.\n\\end{proposition}","\\subsection{The three-spike \\texorpdfstring{$p$-adic}{p-adic} coherence signal}\nRepresent the additive triple by\n\\[\n \\nu_{a,b,c}=\\delta_0+\\delta_a+\\delta_c.\n\\]\nThe pairwise differences are $a$, $b$, and $c$. At a prime dividing exactly one of these integers, exactly one pair of spikes is $p$-adically close. The depth of that edge is the corresponding valuation.","The triple therefore determines a prime-indexed coherence triangle. The radical counts the active triangles. The valuations count how long a pair remains coupled through the $p$-adic refinement tree. This picture anticipates the later cross-prime logarithmic network.","\\subsection{The Frey curve}\nAttach the marked Frey curve\n\\[\n \\boxed{\n E_{a,b,c}:y^2=x(x-a)(x+b).\n }\n\\]\nThe roots $0$, $a$, and $-b$ retain the three differences $a$, $b$, and $c$. The basic invariants are\n\\[\n \\Delta_E=16a^2b^2c^2\n\\]\nand\n\\[\n c_4(E)=16(a^2+ab+b^2).\n\\]\nAt every odd prime $p\\mid abc$, the curve has multiplicative reduction. The conductor records that the channel is active, while the minimal discriminant records the full depth.","In signal language, the Frey curve is a nonlinear encoder that converts:\n\\[\n \\begin{aligned}\n \\text{prime support}&\\longrightarrow\\text{conductor},\\\\\n \\text{valuation depth}&\\longrightarrow\\text{minimal discriminant depth},\\\\\n \\text{additive relation}&\\longrightarrow\\text{elliptic geometry}.\n \\end{aligned}\n\\]\nThis is precisely the kind of nonlinear transform that a unitary shell system could not provide.","\\subsection{The automorphic eigenchannel}\nModularity associates a weight-two newform $f_E$ to the rational isogeny class of $E$. Its Hecke coefficients\n\\[\n a_\\ell(E)\n\\]\nform a global frequency signal. The conductor is the automorphic bandwidth. The newform, however, remembers the rational isogeny class rather than the marked integral representative.","\\subsection{Dyadic Landen descent}\nLet $u>v>0$ be coprime odd integers. Consider\n\\[\n T_0=(v^2,u^2-v^2,u^2)\n\\]\nand\n\\[\n T_1=\n \\left(\n uv,\n \\left(\\frac{u-v}{2}\\right)^2,\n \\left(\\frac{u+v}{2}\\right)^2\n \\right).\n\\]\nBoth are primitive triples. Their radicals are equal:\n\\[\n \\rad(T_0)=\\rad(T_1)=\\rad\\bigl(uv(u-v)(u+v)\\bigr).\n\\]\nThe second output is strictly smaller:\n\\[\n \\left(\\frac{u+v}{2}\\right)^2<u^2.\n\\]\nHence the second quality is smaller.","The Legendre parameter of the parent is\n\\[\n \\lambda=t^2,\n \\qquad\n t=v/u.\n\\]\nThe degree-two Landen transformation gives\n\\[\n \\boxed{\n \\lambda\n \\longmapsto\n \\frac{4\\sqrt\\lambda}{(1+\\sqrt\\lambda)^2}.\n }\n\\]\nThis is exactly the child ratio $a_1/c_1$. The transformation preserves the automorphic channel up to the corresponding two-isogeny while lowering the integral height. It proves that the integral representative inside an isogeny class is relevant to $abc$.","^2,\n \\qquad\n t=v/u.\n\\]\nThe degree-two Landen transformation gives\n\\[\n \\boxed{\n \\lambda\n \\longmapsto\n \\frac{4\\sqrt\\lambda}{(1+\\sqrt\\lambda)^2}.\n }\n\\]\nThis is exactly the child ratio $a_1/c_1$. The transformation preserves the automorphic channel up to the corresponding two-isogeny while lowering the integral height. It proves that the integral representative inside an isogeny class is relevant to $abc$.","\\subsection{Exact automorphic analysis--synthesis identity}\nLet\n\\[\n \\phi_E:X_0(N_E)\\longrightarrow E\n\\]\nbe a nonconstant modular parametrization, and let\n\\[\n \\phi_E^*\\omega_E\n =\n 2\\pi i\\,c_\\phi f_E(z)\\,dz.\n\\]\nHere $c_\\phi$ is the differential multiplier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree,","\\subsection{Exact automorphic analysis--synthesis identity}\nLet\n\\[\n \\phi_E:X_0(N_E)\\longrightarrow E\n\\]\nbe a nonconstant modular parametrization, and let\n\\[\n \\phi_E^*\\omega_E\n =\n 2\\pi i\\,c_\\phi f_E(z)\\,dz.\n\\]\nHere $c_\\phi$ is the differential multiplier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","automorphic analysis--synthesis identity}\nLet\n\\[\n \\phi_E:X_0(N_E)\\longrightarrow E\n\\]\nbe a nonconstant modular parametrization, and let\n\\[\n \\phi_E^*\\omega_E\n =\n 2\\pi i\\,c_\\phi f_E(z)\\,dz.\n\\]\nHere $c_\\phi$ is the differential multiplier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","parametrization, and let\n\\[\n \\phi_E^*\\omega_E\n =\n 2\\pi i\\,c_\\phi f_E(z)\\,dz.\n\\]\nHere $c_\\phi$ is the differential multiplier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","l multiplier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","lier of this chosen parametrization and differential. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","al. The marked Frey curve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","ve need not be the optimal representative of its isogeny class. Define the differential area\n\\[\n A(E)\n =\n \\frac{i}{2}\n \\int_{E(\\C)}\n \\omega_E\\wedge\\overline{\\omega_E}.\n\\]\nPulling the area form back gives\n\\[\n \\boxed{\n \\deg(\\phi_E)A(E)\n =\n 4\\pi^2 c_\\phi^2\n \\norm{f_E}_{\\mathrm{Pet}}^2.\n }\n\\]\nFor the Frey model, scaling $x=cX$, $y=c^{3/2}Y$ gives an area factor proportional to $1/c$. Thus the height $c$ can be written as a shape-corrected synthesis gain involving modular degree, Manin normalization, and Petersson energy.","\\subsection{Exact-order divisor packets}\nLet\n\\[\n n=\\prod_p p^{e_p}.\n\\]\nFor every $d\\ge1$, define\n\\[\n \\boxed{\n B_d(n)\n =\n \\prod_{\\substack{p\\mid n","\\subsection{Exact-order divisor packets}\nLet\n\\[\n n=\\prod_p p^{e_p}.\n\\]\nFor every $d\\ge1$, define\n\\[\n \\boxed{\n B_d(n)\n =\n \\prod_{\\substack{p\\mid n\\\\d\\mid e_p}}p.\n }\n\\]\nSince\n\\[\n \\sum_{d\\mid e}\\varphi(d)=e,\n\\]\nwe have the perfect reconstruction identity\n\\[\n \\boxed{\n n\n =\n \\prod_{d\\ge1}\n B_d(n)^{\\varphi(d)}.\n }\n\\]\nThe first packet is\n\\[\n B_1(n)=\\rad(n).\n\\]\nThe powerful part is\n\\[\n \\boxed{\n \\frac{n}{\\rad(n)}\n =\n \\prod_{d\\ge2}\n B_d(n)^{\\varphi(d)}.\n }\n\\]\nThis is a multiplicative polyphase decomposition indexed by exact character orders.","d\\mid e_p}}p.\n }\n\\]\nSince\n\\[\n \\sum_{d\\mid e}\\varphi(d)=e,\n\\]\nwe have the perfect reconstruction identity\n\\[\n \\boxed{\n n\n =\n \\prod_{d\\ge1}\n B_d(n)^{\\varphi(d)}.\n }\n\\]\nThe first packet is\n\\[\n B_1(n)=\\rad(n).\n\\]\nThe powerful part is\n\\[\n \\boxed{\n \\frac{n}{\\rad(n)}\n =\n \\prod_{d\\ge2}\n B_d(n)^{\\varphi(d)}.\n }\n\\]\nThis is a multiplicative polyphase decomposition indexed by exact character orders.","\\subsection{Quality as input-output persistence imbalance}\nFor a primitive triple,\n\\[\n q-1\n =\n \\frac{\n \\log(c/R_c)-\\log(R_aR_b)\n }{\n \\log R\n }.\n\\]\nUsing the divisor packets,\n\\[\n \\boxed{\n q-1\n =\n \\frac{\n \\sum_{d\\ge2}\\varphi(d)\\log B_d(c)\n -\n \\log(R_aR_b)\n }{\\log R}.\n }\n\\]\nA high-quality triple has more repeated-depth persistence in the output than the input radicals supply as activated bandwidth.","Strong $abc$ is equivalent to\n\\[\n \\prod_{d\\ge2}\n B_d(c)^{\\varphi(d)}\n \\le\n C_\\eps R_aR_bR^\\eps.\n\\]\nThe decomposition makes the target transparent, but not easier by itself. A global channel must explain why the exact-order packets cannot all be large simultaneously.","\\subsection{Local Tate-depth interpretation}\nAt an odd prime $p^e\\parallel c$, the Frey discriminant has local depth $2e$. The divisors of $e$ organize the levels at which prime-power coefficient systems can become insensitive to the local extension. The exact-order divisor packet therefore has a local representation-theoretic interpretation, although local disappearance does not automatically produce a global low-level eigenform.","\\subsection{Atkin--Lehner Walsh packets}\nLet\n\\[\n W_1,\\ldots,W_r\n\\]\nbe commuting Atkin--Lehner involutions. For a sign vector\n\\[\n \\varepsilon\\in\\{\\pm1\\}^r,\n\\]\ndefine\n\\[\n A_\\varepsilon\n =\n \\prod_{j=1}^{r}(I+\\varepsilon_jW_j).\n\\]\nThen\n\\[\n \\boxed{\n A_\\varepsilon^2=2^rA_\\varepsilon,\n }\n\\]\n\\[\n A_\\varepsilon A_{\\varepsilon'}=0\n \\quad(\\varepsilon\\neq\\varepsilon'),\n\\]\nand\n\\[\n \\boxed{\n \\sum_\\varepsilon A_\\varepsilon=2^rI.\n }\n\\]\nThe unnormalized analysis packets are integral. The common synthesis denominator is $2^r$, which is subpolynomial in a squarefree conductor. Local sign separation is therefore not the large persistence carrier.","For a sign vector\n\\[\n \\varepsilon\\in\\{\\pm1\\}^r,\n\\]\ndefine\n\\[\n A_\\varepsilon\n =\n \\prod_{j=1}^{r}(I+\\varepsilon_jW_j).\n\\]\nThen\n\\[\n \\boxed{\n A_\\varepsilon^2=2^rA_\\varepsilon,\n }\n\\]\n\\[\n A_\\varepsilon A_{\\varepsilon'}=0\n \\quad(\\varepsilon\\neq\\varepsilon'),\n\\]\nand\n\\[\n \\boxed{\n \\sum_\\varepsilon A_\\varepsilon=2^rI.\n }\n\\]\nThe unnormalized analysis packets are integral. The common synthesis denominator is $2^r$, which is subpolynomial in a squarefree conductor. Local sign separation is therefore not the large persistence carrier.","\\subsection{The same-sign separation problem}\nAfter fixing the local signs, several eigenforms can remain congruent inside the same packet. Let $\\Lambda$ be an integral Hecke lattice and let $e_f$ be the rational projector onto the Frey eigenline. Define\n\\[\n \\eta_f\n =\n \\min\\{m\\ge1:m e_f(\\Lambda)\\subseteq\\Lambda\\}.\n\\]\nThis is the integral eigenchannel separation denominator. It is an arithmetic analogue of leakage in a filter bank. The difficulty is not rational spectral isolation. It is isolation inside the integral lattice.","\\subsection{Projector denominator as a congruence exponent}\nLet $V=U\\oplus W$ be a rational decomposition and let $\\Lambda\\subset V$ be an integral lattice. Let $e_U$ be the rational projector onto $U$. Define\n\\[\n \\Lambda_U=\\Lambda\\cap U,\n \\qquad\n \\Lambda_W=\\Lambda\\cap W,\n\\]\nand\n\\[\n C_U=\\Lambda/(\\Lambda_U+\\Lambda_W).\n\\]","\\begin{theorem}[Integral projector theorem]\nThe least positive integer $\\eta_U$ satisfying\n\\[\n \\eta_U e_U(\\Lambda)\\subseteq\\Lambda\n\\]\nis the exponent of the finite quotient $C_U$.\n\\end{theorem}","\\begin{proof}\nFor $x=u+w\\in\\Lambda$, the condition $m e_U(x)=mu\\in\\Lambda_U$ is equivalent to\n\\[\n mx=mu+mw\\in\\Lambda_U+\\Lambda_W.\n\\]\nThus $m$ clears every projector denominator exactly when it annihilates every class in $C_U$.\n\\end{proof}","For a primitive rank-one vector $v\\in\\Z^n$, the orthogonal projector denominator is\n\\[\n \\langle v,v\\rangle,\n\\]\nand the congruence quotient is cyclic of that order.","\\subsection{Why one cyclic quotient is not enough for split persistence}\nThe split persistence module\n\\[\n \\mathcal P_c\n \\cong\n \\bigoplus_{p^e\\parallel c}\n (\\Z/p\\Z)^{e-1}\n\\]\nhas large $p$-rank. The $p$-torsion of a cyclic target has rank at most one. Therefore every map from $\\mathcal P_c$ to a cyclic group has a forced kernel of order at least\n\\[\n \\prod_{p^e\\parallel c}p^{\\max(e-2,0)}.\n\\]\nThis is a true obstruction to the split source. It motivates the Hecke cotangent module\n\\[\n \\mathfrak p/\\mathfrak p^2,\n\\]\nwhich can retain multiple independent congruence directions before cyclic compression.","persistence module\n\\[\n \\mathcal P_c\n \\cong\n \\bigoplus_{p^e\\parallel c}\n (\\Z/p\\Z)^{e-1}\n\\]\nhas large $p$-rank. The $p$-torsion of a cyclic target has rank at most one. Therefore every map from $\\mathcal P_c$ to a cyclic group has a forced kernel of order at least\n\\[\n \\prod_{p^e\\parallel c}p^{\\max(e-2,0)}.\n\\]\nThis is a true obstruction to the split source. It motivates the Hecke cotangent module\n\\[\n \\mathfrak p/\\mathfrak p^2,\n\\]\nwhich can retain multiple independent congruence directions before cyclic compression.","\\subsection{A toy Hecke order}\nConsider\n\\[\n \\mathbb T(\\mathbf m)\n =\n \\{(x_0,x_1,\\ldots,x_r)\\in\\Z^{r+1}:x_i\\equiv x_0\\pmod{m_i}\\}.\n\\]\nFor the target coordinate $x_0$, the cotangent module is\n\\[\n \\boxed{\n \\mathfrak p/\\mathfrak p^2\n \\cong\n \\bigoplus_{i=1}^{r}\\Z/m_i\\Z.\n }\n\\]\nThe cyclic congruence quotient is\n\\[\n \\boxed{\n \\Z/\\lcm(m_1,\\ldots,m_r)\\Z.\n }\n\\]\nThe quotient of cotangent order by congruence order is\n\\[\n \\frac{\\prod_i m_i}{\\lcm(m_i)}.\n\\]\nThis is a toy Wiles-defect factor. It measures the multiplicity discarded by cyclic compression.","on{A toy Hecke order}\nConsider\n\\[\n \\mathbb T(\\mathbf m)\n =\n \\{(x_0,x_1,\\ldots,x_r)\\in\\Z^{r+1}:x_i\\equiv x_0\\pmod{m_i}\\}.\n\\]\nFor the target coordinate $x_0$, the cotangent module is\n\\[\n \\boxed{\n \\mathfrak p/\\mathfrak p^2\n \\cong\n \\bigoplus_{i=1}^{r}\\Z/m_i\\Z.\n }\n\\]\nThe cyclic congruence quotient is\n\\[\n \\boxed{\n \\Z/\\lcm(m_1,\\ldots,m_r)\\Z.\n }\n\\]\nThe quotient of cotangent order by congruence order is\n\\[\n \\frac{\\prod_i m_i}{\\lcm(m_i)}.\n\\]\nThis is a toy Wiles-defect factor. It measures the multiplicity discarded by cyclic compression.","\\subsection{Complete-intersection control case}\nLet $F\\in\\Z[X]$ be monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-","\\subsection{Complete-intersection control case}\nLet $F\\in\\Z[X]$ be monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than","\\subsection{Complete-intersection control case}\nLet $F\\in\\Z[X]$ be monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","intersection control case}\nLet $F\\in\\Z[X]$ be monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","}\nLet $F\\in\\Z[X]$ be monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","monic and squarefree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","efree over $\\Q$, with an integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","n integer simple root $a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","$a_0$. Put\n\\[\n \\mathbb T=\\Z[X]/(F)\n\\]\nand let $\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","$\\lambda:\\mathbb T\\to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","to\\Z$ evaluate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","ate at $a_0$. The ring is finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","finite free over $\\Z$ and reduced over $\\Q$. Then the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","hen the congruence quotient $\\Z/\\lambda(\\operatorname{Ann}I)$ and cotangent module both have order\n\\[\n |F'(a_0)|.\n\\]\nTo see this directly, write $F=(X-a_0)G$. The kernel $I$ is generated by $X-a_0$. Modulo $I^2$, the relation is $G(a_0)(X-a_0)=0$, so $I/I^2\\cong\\Z/F'(a_0)\\Z$. The annihilator of $I$ is generated by $G$ modulo $F$, whose evaluation is $G(a_0)=F'(a_0)$. Thus the congruence ideal is $F'(a_0)\\Z$ and the defect is one. Extra multiplicity is therefore tied to non-complete-intersection structure rather than being a normalization artifact.","\\subsection{Two prime indices}\nThe persistence is indexed by conductor primes $q\\mid abc$. Congruence depth is indexed by coefficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full ba","\\subsection{Two prime indices}\nThe persistence is indexed by conductor primes $q\\mid abc$. Congruence depth is indexed by coefficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later","\\subsection{Two prime indices}\nThe persistence is indexed by conductor primes $q\\mid abc$. Congruence depth is indexed by coefficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient","\\subsection{Two prime indices}\nThe persistence is indexed by conductor primes $q\\mid abc$. Congruence depth is indexed by coefficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel","\\subsection{Two prime indices}\nThe persistence is indexed by conductor primes $q\\mid abc$. Congruence depth is indexed by coefficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel.","fficient primes $\\ell$. The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel.",". The natural local signal is a matrix\n\\[\n D_f(\\ell,q).\n\\]\nLet $E/\\Q_q$ have split multiplicative reduction with Tate parameter $q_E$, and let $\\ell\\ne q$. Scalarity of the full local Galois representation on $E[\\ell^n]$ forces\n\\[\n n\\le\\min\\{v_\\ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel.","ell(q-1),v_\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel.","\\ell(v_q(q_E))\\}.\n\\]\nIn a Tate basis, the cyclotomic graded diagonal and the inertia Kummer extension respectively force $\\ell^n\\mid(q-1)$ and $\\ell^n\\mid v_q(q_E)$~\\cite{SerreAbelian1989}. This is a necessary condition, rather than a converse or a condition on Frobenius alone. These off-diagonal channels carry tame congruence information. They do not naturally reproduce the full base-$q$ factor $q^{e-1}$. This mismatch motivates the later filtered contact correction and the diagonal coefficient-prime channel.","\\subsection{The three-cusp section}\nView a primitive triple as the integral section\n\\[\n \\sigma_{a,b,c}:\n \\Spec\\Z\\longrightarrow\\mathbb P^1_\\Z,\n \\qquad\n \\sigma=[a:c].\n\\]\nThe pullbacks of the boundary divisors $0$, $1$, and $\\infty$ are\n\\[\n \\operatorname{div}(a),\n \\qquad\n \\operatorname{div}(b),\n \\qquad\n \\operatorname{div}(c).\n\\]\nFor the output cusp, let\n\\[\n C=\\operatorname{div}(c),\n \\qquad\n C_{\\mathrm{red}}=\\operatorname{div}(R_c).\n\\]\nPlace the two-term contact complex in degrees $-1,0$ and define\n\\[\n \\boxed{\n \\mathbf J_c\n =\n [\\mathcal O(-C)\\longrightarrow\\mathcal O(-C_{\\mathrm{red}})].\n }\n\\]\nOn global sections,\n\\[\n \\mathbf J_c\n =\n [c\\Z\\hookrightarrow R_c\\Z]\n \\simeq\n [\\Z\\xrightarrow{c/R_c}\\Z].\n\\]","\\[\n \\sigma_{a,b,c}:\n \\Spec\\Z\\longrightarrow\\mathbb P^1_\\Z,\n \\qquad\n \\sigma=[a:c].\n\\]\nThe pullbacks of the boundary divisors $0$, $1$, and $\\infty$ are\n\\[\n \\operatorname{div}(a),\n \\qquad\n \\operatorname{div}(b),\n \\qquad\n \\operatorname{div}(c).\n\\]\nFor the output cusp, let\n\\[\n C=\\operatorname{div}(c),\n \\qquad\n C_{\\mathrm{red}}=\\operatorname{div}(R_c).\n\\]\nPlace the two-term contact complex in degrees $-1,0$ and define\n\\[\n \\boxed{\n \\mathbf J_c\n =\n [\\mathcal O(-C)\\longrightarrow\\mathcal O(-C_{\\mathrm{red}})].\n }\n\\]\nOn global sections,\n\\[\n \\mathbf J_c\n =\n [c\\Z\\hookrightarrow R_c\\Z]\n \\simeq\n [\\Z\\xrightarrow{c/R_c}\\Z].\n\\]","\\Spec\\Z\\longrightarrow\\mathbb P^1_\\Z,\n \\qquad\n \\sigma=[a:c].\n\\]\nThe pullbacks of the boundary divisors $0$, $1$, and $\\infty$ are\n\\[\n \\operatorname{div}(a),\n \\qquad\n \\operatorname{div}(b),\n \\qquad\n \\operatorname{div}(c).\n\\]\nFor the output cusp, let\n\\[\n C=\\operatorname{div}(c),\n \\qquad\n C_{\\mathrm{red}}=\\operatorname{div}(R_c).\n\\]\nPlace the two-term contact complex in degrees $-1,0$ and define\n\\[\n \\boxed{\n \\mathbf J_c\n =\n [\\mathcal O(-C)\\longrightarrow\\mathcal O(-C_{\\mathrm{red}})].\n }\n\\]\nOn global sections,\n\\[\n \\mathbf J_c\n =\n [c\\Z\\hookrightarrow R_c\\Z]\n \\simeq\n [\\Z\\xrightarrow{c/R_c}\\Z].\n\\]","d $\\infty$ are\n\\[\n \\operatorname{div}(a),\n \\qquad\n \\operatorname{div}(b),\n \\qquad\n \\operatorname{div}(c).\n\\]\nFor the output cusp, let\n\\[\n C=\\operatorname{div}(c),\n \\qquad\n C_{\\mathrm{red}}=\\operatorname{div}(R_c).\n\\]\nPlace the two-term contact complex in degrees $-1,0$ and define\n\\[\n \\boxed{\n \\mathbf J_c\n =\n [\\mathcal O(-C)\\longrightarrow\\mathcal O(-C_{\\mathrm{red}})].\n }\n\\]\nOn global sections,\n\\[\n \\mathbf J_c\n =\n [c\\Z\\hookrightarrow R_c\\Z]\n \\simeq\n [\\Z\\xrightarrow{c/R_c}\\Z].\n\\]","\\begin{theorem}[Filtered contact quotient]\n\\[\n \\boxed{\n H^0(\\mathbf J_c)\n =\n R_c\\Z/c\\Z\n \\cong\n \\Z/(c/R_c)\\Z.\n }\n\\]\n\\end{theorem}","\\subsection{Associated graded and carry extensions}\nAt $p^e\\parallel c$,\n\\[\n \\Jcal_{p,e}\n =\n p\\Z_p/p^e\\Z_p\n \\cong\n \\Z/p^{e-1}\\Z.\n\\]\nIts filtration\n\\[\n p^j\\Z_p/p^e\\Z_p\n\\]\nhas successive quotients $\\F_p$. Thus\n\\[\n \\gr\\Jcal_{p,e}\\cong(\\F_p)^{e-1}.\n\\]\nThe exact sequences\n\\[\n 0\\longrightarrow\n \\Z/p^{j-1}\\Z\n \\xrightarrow{\\times p}\n \\Z/p^j\\Z\n \\longrightarrow\n \\F_p\n \\longrightarrow0\n\\]\nare nonsplit for $j\\ge2$. The $j=1$ sequence has zero left term and is split. These carry extensions are precisely what the split persistence packet forgot.","Associated graded and carry extensions}\nAt $p^e\\parallel c$,\n\\[\n \\Jcal_{p,e}\n =\n p\\Z_p/p^e\\Z_p\n \\cong\n \\Z/p^{e-1}\\Z.\n\\]\nIts filtration\n\\[\n p^j\\Z_p/p^e\\Z_p\n\\]\nhas successive quotients $\\F_p$. Thus\n\\[\n \\gr\\Jcal_{p,e}\\cong(\\F_p)^{e-1}.\n\\]\nThe exact sequences\n\\[\n 0\\longrightarrow\n \\Z/p^{j-1}\\Z\n \\xrightarrow{\\times p}\n \\Z/p^j\\Z\n \\longrightarrow\n \\F_p\n \\longrightarrow0\n\\]\nare nonsplit for $j\\ge2$. The $j=1$ sequence has zero left term and is split. These carry extensions are precisely what the split persistence packet forgot.","\\begin{corollary}[Correction of the cyclic-rank barrier]\nThe $p$-rank obstruction applies to the associated graded source. It does not obstruct a filtered map from $\\Z/p^{e-1}\\Z$ to a cyclic target of matching depth.\n\\end{corollary}","\\subsection{Exact cyclic transducer}\nLet $m,\\eta$ be positive integers and put\n\\[\n C(m)=[\\Z\\xrightarrow{m}\\Z]\n\\]\nand\n\\[\n C(\\eta)=[\\Z\\xrightarrow{\\eta}\\Z].\n\\]\nSet $g=\\gcd(m,\\eta)$. Multiplication by $m/g$ in degree $-1$ and by $\\eta/g$ in degree $0$ defines a chain map. On cohomology,\n\\[\n \\boxed{\n |\\ker|=m/g,\n \\qquad\n |\\im|=g,\n \\qquad\n |\\coker|=\\eta/g.\n }\n\\]\nThe maximal transferable packet is the common prime-power depth.","For the output source $m=c/R_c$, the corrected algebraic defect against a cyclic target of order $\\eta_f$ is\n\\[\n \\boxed{\n \\delta_{\\mathrm{alg}}(c,f)\n =\n \\frac{c/R_c}{\\gcd(c/R_c,\\eta_f)}.\n }\n\\]\nPrime by prime,\n\\[\n \\delta_{\\mathrm{alg}}(c,f)\n =\n \\prod_{p^e\\parallel c}\n p^{\\max(e-1-v_p(\\eta_f),0)}.\n\\]\nThe problem is missing depth, not missing rank.","\\subsection{Witt realization}\nFor $e\\ge2$, the local packet has a natural additive identification\n\\[\n \\boxed{\n \\Jcal_{p,e}\n \\cong\n W_{e-1}(\\F_p).\n }\n\\]\nDivision by $p$ identifies the source with $\\Z_p/p^{e-1}\\Z_p$, and the standard additive identification $W_n(\\F_p)\\cong\\Z/p^n\\Z$ gives the result. For $e=1$, take $W_0=0$. The Verschiebung filtration matches multiplication by successive powers of $p$. This is the typed local object underlying the carry chain.","At odd $p$, the $p$-adic logarithm identifies\n\\[\n (1+p\\Z_p)/(1+p^e\\Z_p)\n \\cong\n p\\Z_p/p^e\\Z_p.\n\\]\nAt $2$, the additive contact packet remains uniform, while the multiplicative coordinate begins on $1+4\\Z_2$ and therefore carries the familiar weighted dyadic shift.","\\subsection{The filtered local-global target}\nThe intended transducer is a filtered chain\n\\[\n \\mathbf J_c\n \\longrightarrow\n \\bigoplus_{p\\mid c}\\mathbf L_p^{\\mathrm{st}}\n \\longrightarrow\n \\mathbf L_{R^{\\mathrm{glob}}}\n \\xrightarrow{R=\\mathbb T}\n \\mathbf L_{\\mathbb T_f}\n \\longrightarrow\n \\mathbf C_f.\n\\]\nIt must preserve the Bockstein carry maps. The local contact packet exists canonically. The unproved step is its transport through global deformation and Hecke complexes with small kernel and controlled archimedean size.","\\begin{theorem}[Conditional filtered-transfer criterion]\nSuppose every primitive positive triple admits a canonical map\n\\[\n H^0(\\Theta_f):\\Z/(c/R_c)\\Z\\longrightarrow H^0(\\mathbf C_f)\n\\]\nsatisfies, for every $\\eps>0$, with constants independent of the triple,\n\\[\n |\\ker H^0(\\Theta_f)|\\le C_\\eps R^\\eps\n\\]\nand\n\\[\n |H^0(\\mathbf C_f)|\\le C'_\\eps R_aR_bR^\\eps.\n\\]\nThen the strong $abc$ conjecture follows.\n\\end{theorem}","\\begin{proof}\nThe image has order\n\\[\n \\frac{c/R_c}{|\\ker|}\n\\]\nand is bounded by the target order. Therefore\n\\[\n \\frac{c/R_c}{C_\\eps R^\\eps}\n \\le\n C'_\\eps R_aR_bR^\\eps.\n\\]\nMultiplying by $R_c$ and replacing $\\eps$ by $\\eps/2$ gives the strong bound.\n\\end{proof}","\\subsection{The contact modulus}\nLet\n\\[\n c=\\prod_{i=1}^{r}p_i^{e_i},\n \\qquad\n R_c=\\prod_i p_i,\n \\qquad\n M=\\frac{c}{R_c}\n =\\prod_i p_i^{e_i-1}.\n\\]\nThe filtered output contact packet is the cyclic group\n\\[\n \\Z/M\\Z.\n\\]\nThe first global question is whether this packet can be realized by one arithmetic class without losing most of its order.","\\subsection{The global Kummer map}\nKummer theory gives\n\\[\n H^1(\\Q,\\mu_M)\n \\cong\n \\Q^\\times/(\\Q^\\times)^M.\n\\]\nDefine\n\\[\n \\boxed{\n \\kappa_c:\n \\Z/M\\Z\n \\longrightarrow\n H^1(\\Q,\\mu_M),\n \\qquad\n k\\longmapsto[c^k].\n }\n\\]\nThe valuation vector of $c$ is\n\\[\n (e_1,\\ldots,e_r)\n \\in\n (\\Z/M\\Z)^r.\n\\]","\\begin{theorem}[Exact Kummer kernel]\nLet\n\\[\n g_c=\\gcd(M,e_1,\\ldots,e_r).\n\\]\nThen\n\\[\n \\boxed{\n |\\ker\\kappa_c|=g_c,\n \\qquad\n |\\im\\kappa_c|=M/g_c.\n }\n\\]\n\\end{theorem}","\\begin{proof}\nThe class $[c]^k$ is trivial exactly when every prime valuation $ke_i$ is divisible by $M$. The least positive such $k$ is\n\\[\n \\operatorname{lcm}_i\n \\frac{M}{\\gcd(M,e_i)}\n =\n \\frac{M}{\\gcd(M,e_1,\\ldots,e_r)}.\n\\]\nThis is the order of the image. The kernel size follows.\n\\end{proof}","Because\n\\[\n g_c\\le\\min_i e_i\\le\\log_2 c,\n\\]\nwe obtain the universal near-injectivity estimate\n\\[\n \\boxed{\n |\\ker\\kappa_c|\\le\\log_2c.\n }\n\\]\nThis closes the global-occupancy problem for the contact packet up to logarithmic loss.","\\subsection{Balanced Kummer class}\nThe class of $c$ does not remember the additive relation. A better coordinate is\n\\[\n \\beta=-\\frac ba=1-\\frac ca.\n\\]\nLet\n\\[\n \\nu_q=v_q(b)-v_q(a)\n\\]\nfor input primes $q\\mid ab$. The order of $[\\beta]$ in\n\\[\n \\Q^\\times/(\\Q^\\times)^M\n\\]\nis determined by the signed valuation vector, with one possible dyadic sign correction when $M$ is even. If\n\\[\n g_{\\mathrm{val}}\n =\n \\gcd(M,\\{\\nu_q\\}),\n \\qquad\n m_0=M/g_{\\mathrm{val}},\n\\]\nthen the order is $m_0$ unless $M$ is even and $m_0$ is odd, in which case it is $2m_0$.","dditive relation. A better coordinate is\n\\[\n \\beta=-\\frac ba=1-\\frac ca.\n\\]\nLet\n\\[\n \\nu_q=v_q(b)-v_q(a)\n\\]\nfor input primes $q\\mid ab$. The order of $[\\beta]$ in\n\\[\n \\Q^\\times/(\\Q^\\times)^M\n\\]\nis determined by the signed valuation vector, with one possible dyadic sign correction when $M$ is even. If\n\\[\n g_{\\mathrm{val}}\n =\n \\gcd(M,\\{\\nu_q\\}),\n \\qquad\n m_0=M/g_{\\mathrm{val}},\n\\]\nthen the order is $m_0$ unless $M$ is even and $m_0$ is odd, in which case it is $2m_0$.","ion. A better coordinate is\n\\[\n \\beta=-\\frac ba=1-\\frac ca.\n\\]\nLet\n\\[\n \\nu_q=v_q(b)-v_q(a)\n\\]\nfor input primes $q\\mid ab$. The order of $[\\beta]$ in\n\\[\n \\Q^\\times/(\\Q^\\times)^M\n\\]\nis determined by the signed valuation vector, with one possible dyadic sign correction when $M$ is even. If\n\\[\n g_{\\mathrm{val}}\n =\n \\gcd(M,\\{\\nu_q\\}),\n \\qquad\n m_0=M/g_{\\mathrm{val}},\n\\]\nthen the order is $m_0$ unless $M$ is even and $m_0$ is odd, in which case it is $2m_0$.","ordinate is\n\\[\n \\beta=-\\frac ba=1-\\frac ca.\n\\]\nLet\n\\[\n \\nu_q=v_q(b)-v_q(a)\n\\]\nfor input primes $q\\mid ab$. The order of $[\\beta]$ in\n\\[\n \\Q^\\times/(\\Q^\\times)^M\n\\]\nis determined by the signed valuation vector, with one possible dyadic sign correction when $M$ is even. If\n\\[\n g_{\\mathrm{val}}\n =\n \\gcd(M,\\{\\nu_q\\}),\n \\qquad\n m_0=M/g_{\\mathrm{val}},\n\\]\nthen the order is $m_0$ unless $M$ is even and $m_0$ is odd, in which case it is $2m_0$.","\\frac ca.\n\\]\nLet\n\\[\n \\nu_q=v_q(b)-v_q(a)\n\\]\nfor input primes $q\\mid ab$. The order of $[\\beta]$ in\n\\[\n \\Q^\\times/(\\Q^\\times)^M\n\\]\nis determined by the signed valuation vector, with one possible dyadic sign correction when $M$ is even. If\n\\[\n g_{\\mathrm{val}}\n =\n \\gcd(M,\\{\\nu_q\\}),\n \\qquad\n m_0=M/g_{\\mathrm{val}},\n\\]\nthen the order is $m_0$ unless $M$ is even and $m_0$ is odd, in which case it is $2m_0$.","\\subsection{The Kummer--Fermat lift}\nChoose algebraic numbers $u$ and $v$ satisfying\n\\[\n u^M=\\frac ac,\n \\qquad\n v^M=\\frac bc.\n\\]\nThen\n\\[\n \\boxed{u^M+v^M=1.}\n\\]\nThus\n\\[\n P_M=[u:v:1]\n\\]\nlies on the Fermat curve\n\\[\n F_M:X^M+Y^M=Z^M.\n\\]\nThe construction is canonical up to the action of $\\mu_M^2$.","\\begin{theorem}[Exact height compression]\n\\[\n \\boxed{\n h_{\\mathcal O(1)}(P_M)=\\frac{\\log c}{M}.\n }\n\\]\n\\end{theorem}","\\begin{proof}\nThe coordinatewise $M$th-power map sends $P_M$ to $[a:b:c]$. Absolute logarithmic projective height multiplies by $M$ under this map. Since the primitive positive triple has height $\\log c$, the result follows.\n\\end{proof}","\\subsection{Fermat character polyphase}\nThe group $\\mu_M^2$ acts by\n\\[\n (\\xi,\\eta)\\cdot(X,Y)=(\\xi X,\\eta Y).\n\\]\nLet $\\sigma$ and $\\tau$ denote the two coordinate rotations. For\n\\[\n (r,s)\\in(\\Z/M\\Z)^2,\n\\]\ndefine\n\\[\n P_{r,s}\n =\n \\frac1{M^2}\n \\sum_{i,j=0}^{M-1}\n \\zeta_M^{-ri-sj}\n \\sigma^i\\tau^j.\n\\]\nCharacter orthogonality gives\n\\[\n P_{r,s}P_{r',s'}\n =\n \\delta_{r,r'}\\delta_{s,s'}P_{r,s}\n\\]\nand\n\\[\n \\boxed{\n \\sum_{r,s}P_{r,s}=I.\n }\n\\]\nThis is an exact two-dimensional character filter bank.","\\subsection{Line-bundle cost of the full Fermat model}\nFor $M\\ge4$, adjunction gives\n\\[\n K_{F_M}\\cong\\mathcal O_{F_M}(M-3).\n\\]\nFor compatible Weil-height normalizations,\n\\[\n h_{K_{F_M}}(P_M)\n =\n (M-3)h_{\\mathcal O(1)}(P_M)+O(1)\n =\n \\frac{M-3}{M}\\log c+O(1).\n\\]\nIf the metric on $K_{F_M}$ is defined by the induced tensor-power metric on $\\mathcal O(1)^{\\otimes(M-3)}$, the displayed equality holds without the $O(1)$ term. Thus the full plane-curve polarization reintroduces a factor of order $M$ into the height accounting.","{Line-bundle cost of the full Fermat model}\nFor $M\\ge4$, adjunction gives\n\\[\n K_{F_M}\\cong\\mathcal O_{F_M}(M-3).\n\\]\nFor compatible Weil-height normalizations,\n\\[\n h_{K_{F_M}}(P_M)\n =\n (M-3)h_{\\mathcal O(1)}(P_M)+O(1)\n =\n \\frac{M-3}{M}\\log c+O(1).\n\\]\nIf the metric on $K_{F_M}$ is defined by the induced tensor-power metric on $\\mathcal O(1)^{\\otimes(M-3)}$, the displayed equality holds without the $O(1)$ term. Thus the full plane-curve polarization reintroduces a factor of order $M$ into the height accounting.","M}\\cong\\mathcal O_{F_M}(M-3).\n\\]\nFor compatible Weil-height normalizations,\n\\[\n h_{K_{F_M}}(P_M)\n =\n (M-3)h_{\\mathcal O(1)}(P_M)+O(1)\n =\n \\frac{M-3}{M}\\log c+O(1).\n\\]\nIf the metric on $K_{F_M}$ is defined by the induced tensor-power metric on $\\mathcal O(1)^{\\otimes(M-3)}$, the displayed equality holds without the $O(1)$ term. Thus the full plane-curve polarization reintroduces a factor of order $M$ into the height accounting.","\\begin{proposition}[Line-bundle cost of the full plane model]\nThe Kummer lift reduces the $\\mathcal O(1)$ coordinate height by $M$, while the canonical line bundle of the complete Fermat curve has degree factor $M-3$. This rules out a height saving obtained merely by retaining the full plane-curve carrier with its induced polarization. It does not rule out a lower-complexity quotient equipped with a different arithmetic metric.\n\\end{proposition}","\\subsection{Root-distribution compression}\nLet $u^M=z$ and let $\\zeta_M$ be a primitive $M$th root of unity. For $n\\ge1$,\n\\[\n \\boxed{\n \\sum_{j=0}^{M-1}\n \\operatorname{Li}_n(\\zeta_M^ju)\n =\n M^{1-n}\\operatorname{Li}_n(u^M).\n }\n\\]\nInside the disk of convergence, this follows by summing the defining series and using the root-of-unity filter. Analytic continuation extends the identity with compatible branches.","The inverse synthesis gain at weight $n$ is $M^{n-1}$. Through weights one, two, and three, the total gain is\n\\[\n \\boxed{1\\cdot M\\cdot M^2=M^3.}\n\\]\nThe displayed coordinate rescaling has determinant $M^3$. It has not been identified here with an actual integral motivic or period-lattice map. That identification would require the missing arithmetic realization.","\\subsection{Three-point functional closure}\nSet\n\\[\n z=\\frac ac,\n \\qquad\n 1-z=\\frac bc,\n \\qquad\n \\beta=1-\\frac1z=-\\frac ba.\n\\]\nThe Rogers dilogarithm\n\\[\n L_2(z)\n =\n \\operatorname{Li}_2(z)\n +\n \\frac12\\log z\\log(1-z)\n\\]\nsatisfies\n\\[\n \\boxed{\n L_2(z)+L_2(1-z)=\\frac{\\pi^2}{6}.\n }\n\\]\nAt weight three,\n\\[\n \\boxed{\n \\begin{aligned}\n &\\operatorname{Li}_3(z)\n +\\operatorname{Li}_3(1-z)\n +\\operatorname{Li}_3(\\beta)","\\subsection{Three-point functional closure}\nSet\n\\[\n z=\\frac ac,\n \\qquad\n 1-z=\\frac bc,\n \\qquad\n \\beta=1-\\frac1z=-\\frac ba.\n\\]\nThe Rogers dilogarithm\n\\[\n L_2(z)\n =\n \\operatorname{Li}_2(z)\n +\n \\frac12\\log z\\log(1-z)\n\\]\nsatisfies\n\\[\n \\boxed{\n L_2(z)+L_2(1-z)=\\frac{\\pi^2}{6}.\n }\n\\]\nAt weight three,\n\\[\n \\boxed{\n \\begin{aligned}\n &\\operatorname{Li}_3(z)\n +\\operatorname{Li}_3(1-z)\n +\\operatorname{Li}_3(\\beta)\\\\\n &\\qquad=\n \\zeta(3)\n +\\frac16\\log^3z\n +\\frac{\\pi^2}{6}\\log z\n -\\frac12\\log^2z\\log(1-z).\n \\end{aligned}\n }\n\\]\nFor $0<z<1$, all three displayed arguments use their real branches. The dilogarithm complement identity proves the Rogers relation. The auxiliary identity\n\\[\n \\Li_2(1-1/z)+\\Li_2(1-z)=-\\tfrac12\\log^2z\n\\]\nfollows by differentiation and the limit at $z=1$. Using $\\dd\\Li_3(w)=\\Li_2(w)\\,\\dd w/w$, the derivative of the displayed trilogarithm sum is\n\\[\n \\frac{\\pi^2}{6z}-\\frac{\\log z\\log(1-z)}z+\n \\frac{\\log^2z}{2z(1-z)}.\n\\]\nIt is also the derivative of the right side. Both sides tend to $\\zeta(3)$ as $z\\to1$, proving the formula. The balanced Kummer coordinate is therefore exactly the…","&\\qquad=\n \\zeta(3)\n +\\frac16\\log^3z\n +\\frac{\\pi^2}{6}\\log z\n -\\frac12\\log^2z\\log(1-z).\n \\end{aligned}\n }\n\\]\nFor $0<z<1$, all three displayed arguments use their real branches. The dilogarithm complement identity proves the Rogers relation. The auxiliary identity\n\\[\n \\Li_2(1-1/z)+\\Li_2(1-z)=-\\tfrac12\\log^2z\n\\]\nfollows by differentiation and the limit at $z=1$. Using $\\dd\\Li_3(w)=\\Li_2(w)\\,\\dd w/w$, the derivative of the displayed trilogarithm sum is\n\\[\n \\frac{\\pi^2}{6z}-\\frac{\\log z\\log(1-z)}z+\n \\frac{\\log^2z}{2z(1-z)}.\n\\]\nIt is also the derivative of the right side. Both sides tend to $\\zeta(3)$ as $z\\to1$, proving the formula. The balanced Kummer coordinate is therefore exactly the third argument in the depth-three functional closure.","og z\n -\\frac12\\log^2z\\log(1-z).\n \\end{aligned}\n }\n\\]\nFor $0<z<1$, all three displayed arguments use their real branches. The dilogarithm complement identity proves the Rogers relation. The auxiliary identity\n\\[\n \\Li_2(1-1/z)+\\Li_2(1-z)=-\\tfrac12\\log^2z\n\\]\nfollows by differentiation and the limit at $z=1$. Using $\\dd\\Li_3(w)=\\Li_2(w)\\,\\dd w/w$, the derivative of the displayed trilogarithm sum is\n\\[\n \\frac{\\pi^2}{6z}-\\frac{\\log z\\log(1-z)}z+\n \\frac{\\log^2z}{2z(1-z)}.\n\\]\nIt is also the derivative of the right side. Both sides tend to $\\zeta(3)$ as $z\\to1$, proving the formula. The balanced Kummer coordinate is therefore exactly the third argument in the depth-three functional closure.","omplement identity proves the Rogers relation. The auxiliary identity\n\\[\n \\Li_2(1-1/z)+\\Li_2(1-z)=-\\tfrac12\\log^2z\n\\]\nfollows by differentiation and the limit at $z=1$. Using $\\dd\\Li_3(w)=\\Li_2(w)\\,\\dd w/w$, the derivative of the displayed trilogarithm sum is\n\\[\n \\frac{\\pi^2}{6z}-\\frac{\\log z\\log(1-z)}z+\n \\frac{\\log^2z}{2z(1-z)}.\n\\]\nIt is also the derivative of the right side. Both sides tend to $\\zeta(3)$ as $z\\to1$, proving the formula. The balanced Kummer coordinate is therefore exactly the third argument in the depth-three functional closure.","tity proves the Rogers relation. The auxiliary identity\n\\[\n \\Li_2(1-1/z)+\\Li_2(1-z)=-\\tfrac12\\log^2z\n\\]\nfollows by differentiation and the limit at $z=1$. Using $\\dd\\Li_3(w)=\\Li_2(w)\\,\\dd w/w$, the derivative of the displayed trilogarithm sum is\n\\[\n \\frac{\\pi^2}{6z}-\\frac{\\log z\\log(1-z)}z+\n \\frac{\\log^2z}{2z(1-z)}.\n\\]\nIt is also the derivative of the right side. Both sides tend to $\\zeta(3)$ as $z\\to1$, proving the formula. The balanced Kummer coordinate is therefore exactly the third argument in the depth-three functional closure.","A fixed path packet is\n\\[\n \\Pi_{\\le3}(z)\n =\n \\left(\n \\log z,\n \\log(1-z),\n L_2(z),\n \\operatorname{Li}_3(z),\n \\operatorname{Li}_3(1-z)\n \\right).\n\\]\nIts coordinate count is independent of $M$.","\\subsection{Archimedean envelope}\nFor $0<z<1$,\n\\[\n |\\log z|,\n |\\log(1-z)|\n \\le\\log c.\n\\]\nThe positive dilogarithm and trilogarithm values on $(0,1)$ are bounded by $\\zeta(2)$ and $\\zeta(3)$. The three-point identity gives\n\\[\n \\left|\\operatorname{Li}_3\\left(-\\frac ba\\right)\\right|\n \\le\n 3\\zeta(3)\n +\n \\frac{\\pi^2}{6}\\log c\n +\n \\frac23(\\log c)^3.\n\\]\nThus the complex packet has only cubic logarithmic growth. The surrounding motivic and regulator language is informed by the polylogarithmic complexes of Goncharov, the regulator framework discussed by Zagier, and Borel's rank theorem~\\cite{Goncharov,ZagierPolylog,Borel}.","\\subsection{Fixed-rank integral uncertainty}\nLet $A\\in M_d(\\Z)$ be nonsingular and suppose $D\\mid\\det A$. If every entry satisfies $|A_{ij}|\\le B$, then Hadamard's inequality gives\n\\[\n D\\le|\\det A|\\le d^{d/2}B^d.\n\\]\nHence\n\\[\n \\boxed{\n B\\ge\\frac{D^{1/d}}{\\sqrt d}.\n }\n\\]\nA small complex period packet does not imply a small integral realization.","\\subsection{Exact finite-place notch}\nAt an output prime $p^e\\parallel c$,\n\\[\n \\beta=1-\\frac ca.\n\\]\nSince $p\\nmid a$,\n\\[\n v_p(\\beta-1)=e.\n\\]\nFor odd $p$,\n\\[\n \\boxed{\n v_p(\\log_p\\beta)=e.\n }\n\\]\nThe corresponding weighted dyadic identity holds on the principal branch when $e\\ge2$.","Writing\n\\[\n \\beta=-\\prod_{q\\mid ab}q^{\\nu_q},\n\\]\nwe obtain the cross-prime logarithmic filter relation\n\\[\n \\boxed{\n \\log_p\\beta\n =\n \\sum_{q\\mid ab}\n \\nu_q\\log_p\\langle q\\rangle_p.\n }\n\\]\nThe same depth-three path therefore has a small complex envelope and exact deep finite notches.","\\subsection{Weight-three motivation and limitation}\nOver $\\Q$, Borel's rank theorem gives $\\operatorname{rank}K_3(\\Q)=0$ and $\\operatorname{rank}K_5(\\Q)=1$, the rational weights two and three. This makes weight three a plausible rational regulator layer. It does not assert the absence of all weight-two periods or of regulator directions over arbitrary number fields. This motivates the depth-three path packet. It does not prove that weight three suffices uniformly. A complete proof may require adaptive higher weights or a different motivic quotient.","\\subsection{Conditional motivic transfer}\nLet $K_\\beta$ be the cyclic group generated by the balanced Kummer class. Suppose that for every primitive positive triple there exists an integral finite object $\\mathcal H_3(a,b,c)$ and a canonical map\n\\[\n \\Theta_3:K_\\beta\\longrightarrow\\mathcal H_3(a,b,c)\n\\]\nsuch that, for every $\\eps>0$, constants independent of the triple give\n\\[\n |\\ker\\Theta_3|\\le C_\\eps R^\\eps\n\\]\nand\n\\[\n |\\mathcal H_3(a,b,c)|\\le C'_\\eps R_aR_bR^\\eps.\n\\]\nThen the strong $abc$ conjecture follows after absorbing the logarithmic Kummer kernel.","al motivic transfer}\nLet $K_\\beta$ be the cyclic group generated by the balanced Kummer class. Suppose that for every primitive positive triple there exists an integral finite object $\\mathcal H_3(a,b,c)$ and a canonical map\n\\[\n \\Theta_3:K_\\beta\\longrightarrow\\mathcal H_3(a,b,c)\n\\]\nsuch that, for every $\\eps>0$, constants independent of the triple give\n\\[\n |\\ker\\Theta_3|\\le C_\\eps R^\\eps\n\\]\nand\n\\[\n |\\mathcal H_3(a,b,c)|\\le C'_\\eps R_aR_bR^\\eps.\n\\]\nThen the strong $abc$ conjecture follows after absorbing the logarithmic Kummer kernel.","he balanced Kummer class. Suppose that for every primitive positive triple there exists an integral finite object $\\mathcal H_3(a,b,c)$ and a canonical map\n\\[\n \\Theta_3:K_\\beta\\longrightarrow\\mathcal H_3(a,b,c)\n\\]\nsuch that, for every $\\eps>0$, constants independent of the triple give\n\\[\n |\\ker\\Theta_3|\\le C_\\eps R^\\eps\n\\]\nand\n\\[\n |\\mathcal H_3(a,b,c)|\\le C'_\\eps R_aR_bR^\\eps.\n\\]\nThen the strong $abc$ conjecture follows after absorbing the logarithmic Kummer kernel.","\\subsection{The signed input exponent signal}\nLet\n\\[\n S_{ab}=\\{q:q\\mid ab\\}.\n\\]\nDefine\n\\[\n \\nu_q=v_q(b)-v_q(a).\n\\]\nThen\n\\[\n -\\frac ba=-\\prod_{q\\mid ab}q^{\\nu_q}.\n\\]\nFor every output prime $p\\mid c$, the same input exponent packet produces the deep principal unit\n\\[\n 1-\\frac ca.\n\\]\nThe output persistence is therefore a simultaneous cross-prime cancellation phenomenon.","\\subsection{Odd logarithmic rows}\nFor odd $p^e\\parallel c$, define the finite-depth row coefficient\n\\[\n \\ell_{p,e}(q)\n =\n \\frac1p\\log_p\\langle q\\rangle_p\n \\pmod{p^{e-1}}.\n\\]\nThe balanced relation gives\n\\[\n \\boxed{\n \\sum_{q\\mid ab}\n \\nu_q\\ell_{p,e}(q)\n \\equiv0\n \\pmod{p^{e-1}}.\n }\n\\]\nDefine the truncated valuation on $\\Z/p^{e-1}\\Z$ by\n\\[\n \\bar v_p(0)=e-1,\n \\qquad\n \\bar v_p(x)=\\min\\{e-1,v_p(\\widetilde x)\\}\n\\]\nfor any integer lift $\\widetilde x$ of a nonzero residue. Then\n\\[\n \\boxed{\n \\bar v_p(\\ell_{p,e}(q))\n =\n \\min\\{e-1,v_p(q^{p-1}-1)-1\\}.\n }\n\\]\nExceptional entrywise depth is a generalized Wieferich event.","}\nFor odd $p^e\\parallel c$, define the finite-depth row coefficient\n\\[\n \\ell_{p,e}(q)\n =\n \\frac1p\\log_p\\langle q\\rangle_p\n \\pmod{p^{e-1}}.\n\\]\nThe balanced relation gives\n\\[\n \\boxed{\n \\sum_{q\\mid ab}\n \\nu_q\\ell_{p,e}(q)\n \\equiv0\n \\pmod{p^{e-1}}.\n }\n\\]\nDefine the truncated valuation on $\\Z/p^{e-1}\\Z$ by\n\\[\n \\bar v_p(0)=e-1,\n \\qquad\n \\bar v_p(x)=\\min\\{e-1,v_p(\\widetilde x)\\}\n\\]\nfor any integer lift $\\widetilde x$ of a nonzero residue. Then\n\\[\n \\boxed{\n \\bar v_p(\\ell_{p,e}(q))\n =\n \\min\\{e-1,v_p(q^{p-1}-1)-1\\}.\n }\n\\]\nExceptional entrywise depth is a generalized Wieferich event.","te-depth row coefficient\n\\[\n \\ell_{p,e}(q)\n =\n \\frac1p\\log_p\\langle q\\rangle_p\n \\pmod{p^{e-1}}.\n\\]\nThe balanced relation gives\n\\[\n \\boxed{\n \\sum_{q\\mid ab}\n \\nu_q\\ell_{p,e}(q)\n \\equiv0\n \\pmod{p^{e-1}}.\n }\n\\]\nDefine the truncated valuation on $\\Z/p^{e-1}\\Z$ by\n\\[\n \\bar v_p(0)=e-1,\n \\qquad\n \\bar v_p(x)=\\min\\{e-1,v_p(\\widetilde x)\\}\n\\]\nfor any integer lift $\\widetilde x$ of a nonzero residue. Then\n\\[\n \\boxed{\n \\bar v_p(\\ell_{p,e}(q))\n =\n \\min\\{e-1,v_p(q^{p-1}-1)-1\\}.\n }\n\\]\nExceptional entrywise depth is a generalized Wieferich event.","\\subsection{Weighted dyadic row}\nFor $2^e\\parallel c$ with $e\\ge2$ and an odd input prime $q$, define\n\\[\n \\ell_{2,e}(q)\n =\n \\frac14\n \\log_2\\left((-1)^{(q-1)/2}q\\right)\n \\pmod{2^{e-2}}.\n\\]\nWith the analogous truncated valuation on $\\Z/2^{e-2}\\Z$,\n\\[\n \\bar v_2(\\ell_{2,e}(q))\n =\n \\min\\{e-2,v_2(q^2-1)-3\\}.\n\\]\nAt $e=1$, the output contact depth is zero and this row is omitted. A separate sign packet supplies the missing first dyadic layer. This is the weighted dyadic branch in its final logarithmic form.","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coeff","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. D","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\mi","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized glo","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\pro","\\subsection{Sequential factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qq","ial factorization of the powerful part}\nThe factorization is defined prime by prime. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{…","me. For an odd repeated output prime $p^e\\parallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign lay…","rallel c$, set\n\\[\n d_p=e-1.\n\\]\nFor a repeated dyadic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic …","dic prime $2^e\\parallel c$, remove one sign layer and set\n\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the co…","\\[\n s_2=1,\n \\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe","\\qquad\n d_2=e-2.\n\\]\nIf $2\\nmid M$, take $s_2=0$. Put $\\epsilon_2=2^{s_2}$. Let $K_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of re…","_{\\mathrm{Kum}}$ be the kernel size of the balanced Kummer class and put\n\\[\n r_p=\\min\\{d_p,v_p(K_{\\mathrm{Kum}})\\}.\n\\]\nThe balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer pre…","balanced order formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modul…","r formula implies\n\\[\n K_{\\mathrm{Kum}}\\mid M/\\epsilon_2,\n\\]\nso $K_{\\mathrm{Kum}}=\\prod_{p\\mid c}p^{r_p}$. Write $K=K_{\\mathrm{Kum}}$. Divide the input exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modulo $p^{\\chi_p}…","put exponent vector by $K$. Define $a_{p,q}$ as the original odd or weighted dyadic coefficient, now reduced modulo $p^{d_p-r_p}$. Dividing the relation vector reduces the needed relation modulus. It does not divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modulo $p^{\\chi_p}$.","divide the coefficient row by $K$. Define\n\\[\n w_p\n =\n \\min\\left\n \\{\n d_p-r_p,\n \\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modulo $p^{\\chi_p}$.","\\min_{q\\mid ab}\\bar v_p(a_{p,q})\n \\right\\},\n \\qquad\n \\chi_p=d_p-r_p-w_p.\n\\]\nThe normalized global factors are\n\\[\n K=\\prod_{p\\mid c}p^{r_p},\n \\qquad\n W=\\prod_{p\\mid c}p^{w_p},\n \\qquad\n C=\\prod_{p\\mid c}p^{\\chi_p}.\n\\]\nThe balanced Kummer order formula guarantees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modulo $p^{\\chi_p}$.","ntees that the dyadic sign layer is removed before the principal $2$-adic Kummer depth, so these exponents partition the contact depth. Hence\n\\[\n \\boxed{\n M=\\epsilon_2KWC.\n }\n\\]\nThe order of removal matters. The local logarithmic defect is measured only after the Kummer preload has been divided out. The resulting row is then divided by its common logarithmic depth, leaving a primitive row modulo $p^{\\chi_p}$.","\\subsection{The coherence character}\nSet $\\chi_p=0$ for absent or squarefree output channels. When the residual modulus is one, use the trivial row and character. After normalization, every output prime with nontrivial residual depth produces a primitive row\n\\[\n g_p\n \\in\n (\\Z/p^{\\chi_p}\\Z)^{S_{ab}}\n\\]\nsatisfying\n\\[\n g_p\\cdot\\nu_0\n \\equiv0\n \\pmod{p^{\\chi_p}},\n\\]\nwhere $\\nu_0$ is the Kummer-normalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{a","\\subsection{The coherence character}\nSet $\\chi_p=0$ for absent or squarefree output channels. When the residual modulus is one, use the trivial row and character. After normalization, every output prime with nontrivial residual depth produces a primitive row\n\\[\n g_p\n \\in\n (\\Z/p^{\\chi_p}\\Z)^{S_{ab}}\n\\]\nsatisfying\n\\[\n g_p\\cdot\\nu_0\n \\equiv0\n \\pmod{p^{\\chi_p}},\n\\]\nwhere $\\nu_0$ is the Kummer-normalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","sidual modulus is one, use the trivial row and character. After normalization, every output prime with nontrivial residual depth produces a primitive row\n\\[\n g_p\n \\in\n (\\Z/p^{\\chi_p}\\Z)^{S_{ab}}\n\\]\nsatisfying\n\\[\n g_p\\cdot\\nu_0\n \\equiv0\n \\pmod{p^{\\chi_p}},\n\\]\nwhere $\\nu_0$ is the Kummer-normalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","er normalization, every output prime with nontrivial residual depth produces a primitive row\n\\[\n g_p\n \\in\n (\\Z/p^{\\chi_p}\\Z)^{S_{ab}}\n\\]\nsatisfying\n\\[\n g_p\\cdot\\nu_0\n \\equiv0\n \\pmod{p^{\\chi_p}},\n\\]\nwhere $\\nu_0$ is the Kummer-normalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","}\n\\]\nsatisfying\n\\[\n g_p\\cdot\\nu_0\n \\equiv0\n \\pmod{p^{\\chi_p}},\n\\]\nwhere $\\nu_0$ is the Kummer-normalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","ormalized input exponent vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","vector. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","r. The Chinese remainder theorem combines the local rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","l rows into one primitive row\n\\[\n g\\in(\\Z/C\\Z)^{S_{ab}}.\n\\]\nDefine the coherence character\n\\[\n \\boxed{\n \\chi_{\\mathrm{coh}}(x)\n =\n \\exp\\left(\\frac{2\\pi i}{C}g\\cdot x\\right).\n }\n\\]\nIts kernel is the adelic coherence lattice\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{x\\in\\Z^{S_{ab}}:g\\cdot x\\equiv0\\pmod C\\}.\n }\n\\]\nSince $g$ is primitive modulo $C$,\n\\[\n [\\Z^{S_{ab}}:\\Lambda_{\\mathrm{coh}}]=C\n\\]\nand\n\\[\n \\Z^{S_{ab}}/\\Lambda_{\\mathrm{coh}}\\cong\\Z/C\\Z.\n\\]\nThe normalized exponent vector lies in the lattice.","\\subsection{Worked lattice for \\texorpdfstring{$3+125=128$}{3+125=128}}\nHere\n\\[\n M=64\n\\]\nand\n\\[\n (\\epsilon_2,K,W,C)=(2,1,1,32).\n\\]\nThe input vector is\n\\[\n \\nu_0=(-1,3)\n\\]\non the prime coordinates $(3,5)$. The residual row is\n\\[\n (29,31)\\pmod{32}.\n\\]\nThus\n\\[\n \\boxed{\n \\Lambda_{\\mathrm{coh}}\n =\n \\{(x_3,x_5)\\in\\Z^2:29x_3+31x_5\\equiv0\\pmod{32}\\}.\n }\n\\]\nIndeed,\n\\[\n 29(-1)+31(3)=64\\equiv0\\pmod{32}.\n\\]\nThe five residual dyadic contact layers are exactly the index of this lattice.","\\begin{enumerate}[leftmargin=1.5em]\n\\item An entrywise defect can make every coefficient in one row divisible by a large power of $p$. This is the factor $W$.\n\\item Ordinary coefficients can cancel coherently against the specific exponent vector. This is the factor $C$.\n\\end{enumerate}","In the recovered 361-row coherence corpus, the recorded $W$ factor is usually one while $C$ can be large. The corpus is selected, rather than a statistical sample of all primitive triples. Its frequencies do not establish a generic population law. For its $W=1$ rows, the residual depth is represented by collective cancellation rather than an entrywise Wieferich defect.","\\subsection{Why abstract lattice geometry is insufficient}\nLet $v\\in\\Z^s$ be primitive, with $s\\ge2$. For every $C\\ge2$, there exists a primitive row $g\\pmod C$ satisfying\n\\[\n g\\cdot v\\equiv0\\pmod C.\n\\]\nThus a fixed short vector can lie in primitive congruence lattices of arbitrarily large index.","The proof completes $v$ to a unimodular basis and chooses a primitive row annihilating its first basis coordinate modulo $C$.","The missing theorem must use the arithmetic origin of $g$: its entries are simultaneous $p$-adic logarithms of the actual input primes, and the output primes are exactly the divisors of $a+b$.","\\subsection{Why positive divisibility is circular}\nIf positive integers $A$ and $B$ satisfy $c\\mid A+B$, then\n\\[\n A+B\\ge c,\n\\]\nso\n\\[\n \\max(A,B)\\ge c/2.\n\\]\nA direct search for a small positive $S$-unit relation divisible by $c$ cannot produce an $abc$ saving. Signed cancellation, a determinant, a regulator, or a functional equation is necessary.","\\subsection{Exact reformulation of strong \\texorpdfstring{$abc$}{abc}}\nSince\n\\[\n c=R_c\\epsilon_2KWC\n\\]\nand $\\epsilon_2\\le2$ while $K\\le\\log_2c$, the all-$\\eps$ strong $abc$ statement is equivalent, after the usual absorption of bounded and logarithmic factors into an arbitrarily small power, to the target\n\\[\n \\boxed{\n WC\n \\le\n C_\\eps R_aR_bR^\\eps.\n }\n\\]\nThe logarithmic absorption is proved explicitly in Section~\\ref{sec:bridge}, without assuming an initial relation $c\\ll R$. Conversely, $WC\\le M_c$ makes the reverse implication immediate. In the recorded examples with $W=1$, this leaves the coherence conductor bound\n\\[\n C\\le C_\\eps R_aR_bR^\\eps.\n\\]","Since\n\\[\n c=R_c\\epsilon_2KWC\n\\]\nand $\\epsilon_2\\le2$ while $K\\le\\log_2c$, the all-$\\eps$ strong $abc$ statement is equivalent, after the usual absorption of bounded and logarithmic factors into an arbitrarily small power, to the target\n\\[\n \\boxed{\n WC\n \\le\n C_\\eps R_aR_bR^\\eps.\n }\n\\]\nThe logarithmic absorption is proved explicitly in Section~\\ref{sec:bridge}, without assuming an initial relation $c\\ll R$. Conversely, $WC\\le M_c$ makes the reverse implication immediate. In the recorded examples with $W=1$, this leaves the coherence conductor bound\n\\[\n C\\le C_\\eps R_aR_bR^\\eps.\n\\]","n_2KWC\n\\]\nand $\\epsilon_2\\le2$ while $K\\le\\log_2c$, the all-$\\eps$ strong $abc$ statement is equivalent, after the usual absorption of bounded and logarithmic factors into an arbitrarily small power, to the target\n\\[\n \\boxed{\n WC\n \\le\n C_\\eps R_aR_bR^\\eps.\n }\n\\]\nThe logarithmic absorption is proved explicitly in Section~\\ref{sec:bridge}, without assuming an initial relation $c\\ll R$. Conversely, $WC\\le M_c$ makes the reverse implication immediate. In the recorded examples with $W=1$, this leaves the coherence conductor bound\n\\[\n C\\le C_\\eps R_aR_bR^\\eps.\n\\]","tement is equivalent, after the usual absorption of bounded and logarithmic factors into an arbitrarily small power, to the target\n\\[\n \\boxed{\n WC\n \\le\n C_\\eps R_aR_bR^\\eps.\n }\n\\]\nThe logarithmic absorption is proved explicitly in Section~\\ref{sec:bridge}, without assuming an initial relation $c\\ll R$. Conversely, $WC\\le M_c$ makes the reverse implication immediate. In the recorded examples with $W=1$, this leaves the coherence conductor bound\n\\[\n C\\le C_\\eps R_aR_bR^\\eps.\n\\]","f bounded and logarithmic factors into an arbitrarily small power, to the target\n\\[\n \\boxed{\n WC\n \\le\n C_\\eps R_aR_bR^\\eps.\n }\n\\]\nThe logarithmic absorption is proved explicitly in Section~\\ref{sec:bridge}, without assuming an initial relation $c\\ll R$. Conversely, $WC\\le M_c$ makes the reverse implication immediate. In the recorded examples with $W=1$, this leaves the coherence conductor bound\n\\[\n C\\le C_\\eps R_aR_bR^\\eps.\n\\]","\\subsection{Adelic coherence regulator object}\nPackage the surviving data as\n\\[\n \\boxed{\n \\mathfrak R_{a,b,c}\n =\n \\left(\n U_{ab},\n \\beta^{1/K},\n \\{\\chi_p\\}_{p\\mid c},\n \\chi_{\\mathrm{coh}},\n \\Lambda_{\\mathrm{coh}},\n \\norm{\\cdot}_\\infty\n \\right).\n }\n\\]\nHere $U_{ab}$ is the input $S$-unit group, the local characters are the normalized logarithmic rows, and the archimedean norm records input height.","The open theorem is an adelic slope or regulator inequality bounding the finite conductor $\\log C$ by the input radical and an arbitrarily small global leakage term. General simultaneous linear-form theory contains the right kinds of local and archimedean quantities, but known general estimates are much too coarse for the $R^\\eps$ scale."]}