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1 \chapter{Every primitive square survives}\label{ch:even}
2 \section{The local numerator introduced by content}
3 The transform in \eqref{eq:mobius} is a Dirichlet convolution of multiplicative functions, so $R$ is multiplicative. At a prime $p$, write
4 \[
5 a=\tau(p),\qquad b=p^{11},\qquad c=\chifive(p)p^{10}.
6 \]
7 The local generating factor is
8 \begin{equation}\label{eq:rlocal}
9 \boxed{\sum_{r\ge0}R(p^r)X^r
10 =\frac{1-cX^2}{1-aX+bX^2}.}
11 \end{equation}
12 It follows either by multiplying the Hecke factor by the square-supported Möbius factor, or by reading the two possible square divisors with nonzero Möbius value at a prime power. In particular,
13 \[
14 R(1)=1,\qquad R(p)=a,\qquad
15 R(p^r)=T_r-cT_{r-2}\quad(r\ge2).
16 \]
17 The subtraction begins at $r=2$. No undefined $T_{-1}$ or $T_{-2}$ is needed.
18 
19 The even primitive blocks are not the same as the even tau blocks. For example,
20 \[
21 R(p^2)=a^2-p^{10}(p+\chifive(p)).
22 \]
23 An argument that $\tau(p^{2m})\ne0$ does not alone imply the corresponding difference is nonzero. We now prove the primitive statement directly.
24 
25 \begin{theorem}\label{thm:primitiveeven}
26 For every prime $p$ and every integer $m\ge1$,
27 \[
28 R(p^{2m})\ne0.
29 \]
30 Consequently $R(s^2)\ne0$ for every positive integer $s$.
31 \end{theorem}
32 
33 The proof is stronger than the actual-tau specialization in a useful way. For $p\ne5$, it works for every integer recurrence parameter $a$ with $a\equiv p+\chifive(p)\pmod5$. At $p=5$, it works for every integer $a$. It therefore includes a hypothetical parameter $a=0$ whenever that parameter is compatible with the congruence. The square theorem has not secretly assumed prime nonvanishing.
34 
35 \section{A residue eliminates three prime classes}
36 Suppose $p\ne5$ and put $\chi=\chifive(p)$. Modulo $5$, $\chi\equiv p^2$, so the two recurrence roots may be taken as $p$ and $\chi$:
37 \[
38 p+\chi\equiv a,\qquad p\chi\equiv p^{11}\pmod5.
39 \]
40 The complete homogeneous recurrence coefficient is
41 \[
42 T_r\equiv C_r=\sum_{i=0}^r p^{r-i}\chi^i\pmod5.
43 \]
44 Also $\chi p^{10}\equiv1\pmod5$. Since $\chi^2=1$, for $r\ge2$ the sum telescopes:
45 \begin{equation}\label{eq:rmod5}
46 R(p^r)\equiv C_r-C_{r-2}
47 =p^{r-1}(p+\chi)\pmod5.
48 \end{equation}
49 The same expression holds at $r=1$. It is nonzero for $p\equiv1,2,3\pmod5$. It proves nonvanishing for all positive exponents in those prime classes. Only $p=5$ and $p\equiv4\pmod5$ remain.
50 
51 \section{The recurrence polynomial and untied valuations}
52 For $p\equiv4\pmod5$, $\chi=1$ and $a\equiv0\pmod5$. The exact recurrence expansion is
53 \begin{equation}\label{eq:trexpansion}
54 T_r=\sum_{j=0}^{\lfloor r/2\rfloor}
55 (-1)^j\binom{r-j}{j}b^ja^{r-2j}.
56 \end{equation}
57 It follows by induction: multiplying the previous row by $a$ and subtracting $b$ times the row before it combines neighboring binomial coefficients. Equivalently, expand the rational generating factor formally. The extreme coefficients are $1$, so a strictly least valuation at either end cannot be raised by a coefficient divisible by $p$.
58 
59 First suppose $a\ne0$, and write $t=v_p(a)$. In $T_{2m}$, the term indexed by $j$ has valuation at least
60 \[
61 2mt+j(11-2t).
62 \]
63 If $t\le4$, the $j=0$ term $a^{2m}$ has strictly least valuation $2mt$. Likewise $T_{2m-2}$ has valuation $(2m-2)t$. The correction $p^{10}T_{2m-2}$ has valuation
64 \[
65 2mt+(10-2t)>2mt.
66 \]
67 Thus
68 \begin{equation}\label{eq:eventlow}
69 v_p(R(p^{2m}))=2mt,\qquad t\le4.
70 \end{equation}
71 
72 If $t\ge6$, the constant-in-$a$ term $(-b)^m$ is uniquely least-valued in $T_{2m}$, with valuation $11m$. The analogous term gives $v_p(T_{2m-2})=11(m-1)$. The correction has valuation $11m-1$, now smaller than the first block. Hence
73 \begin{equation}\label{eq:eventhigh}
74 v_p(R(p^{2m}))=11m-1,\qquad t\ge6.
75 \end{equation}
76 Actual nonzero tau parameters at $p\ge5$ cannot have $t\ge6$ by Deligne's bound. The wider argument is nonetheless a useful exact control for the recurrence class.
77 
78 If $a=0$, do not assign it a finite $t$. Direct substitution gives
79 \begin{equation}\label{eq:evenzeroa}
80 R(p^{2m})=(-1)^mp^{11m-1}(p+1),
81 \end{equation}
82 which is nonzero. This branch is indispensable: a theorem purporting to survive a hypothetical prime zero must actually evaluate that case.
83 
84 \section{The tied valuation \texorpdfstring{$t=5$}{t=5}}
85 The preceding endpoint arguments leave $t=5$. Write $a=p^5u$ with $p\nmid u$, and set
86 \[
87 H_r(u)=p^{-5r}T_r.
88 \]
89 Equation \eqref{eq:trexpansion} shows that $H_r$ is an integral polynomial. Its recurrence is
90 \[
91 H_0=1,\qquad H_1=u,\qquad H_r=uH_{r-1}-pH_{r-2}.
92 \]
93 The primitive block becomes
94 \begin{equation}\label{eq:tied}
95 R(p^{2m})=p^{10m}\bigl(H_{2m}(u)-H_{2m-2}(u)\bigr).
96 \end{equation}
97 The expression in parentheses is a monic polynomial $F_m(x)$ in $x=u^2$. Its constant term and reduction modulo $p$ are
98 \begin{equation}\label{eq:fm}
99 F_m(0)=(-1)^mp^{m-1}(p+1),\qquad
100 F_m(x)\equiv x^{m-1}(x-1)\pmod p.
101 \end{equation}
102 The leading term comes from $u^{2m}$. The constant term comes from $H_{2m}(0)=(-p)^m$ and $H_{2m-2}(0)=(-p)^{m-1}$.
103 
104 Suppose $R(p^{2m})=0$. Then $x=u^2$ is a positive integer root of the monic integer polynomial $F_m$. The integer-root theorem gives $x\mid p^{m-1}(p+1)$. Because $p\nmid u$, one obtains $x\mid p+1$. Reduction in \eqref{eq:fm} gives $x\equiv1\pmod p$. Since $1\le x\le p+1$, the only possibilities are
105 \[
106 u^2=1\quad\text{or}\quad u^2=p+1.
107 \]
108 The first is impossible: $a$ is divisible by $5$ and $p$ is not, so $u$ is divisible by $5$. The second gives
109 \[
110 p=(|u|-1)(|u|+1).
111 \]
112 For a prime this forces $|u|=2$ and $p=3$, outside the family $p\equiv4\pmod5$. Both possibilities are excluded. This closes the tied case for every exponent $m$, rather than for a tested list of exponents.
113 
114 For $m=1$, the polynomial is simply $F_1(x)=x-(p+1)$. For larger $m$, its degree grows, but the same constant-term and residue argument controls any integral square root. That is why no bound on $m$ is needed.
115 
116 \section{The prime 5 and multiplication over squares}
117 At $p=5$, $\chifive(5)=0$, so $R(5^r)=T_r$. If $a=0$, its even blocks are $(-5^{11})^m$. If $a\ne0$, then $2v_5(a)$ is even and cannot equal $11$. In the expansion \eqref{eq:trexpansion}, either the first term or the constant-in-$a$ term has a unique least valuation at an even exponent. Thus no even block vanishes.
118 
119 All prime cases have now been covered. If $s=\prod p^{m_p}$, multiplicativity gives
120 \[
121 R(s^2)=\prod_pR(p^{2m_p}).
122 \]
123 Every factor is nonzero, and $R(1)=1$. This proves Theorem~\ref{thm:primitiveeven}.
124 
125 The theorem gives survival, not a common sign. For example,
126 \[
127 R(4)=-448,\qquad R(9)=-54594,\qquad R(25)=-25499225,
128 \]
129 while other square values can have different signs. Nor does it furnish a fixed auxiliary-prime unit: a nonzero integer can be divisible by arbitrarily high preassigned moduli at suitable varying primes. Chapter~\ref{ch:congruences} makes that last observation into an actual-prime theorem.
130 
131 \figplaceholder{square_residuals}{Exact primitive residuals at square indices $s^2$ for $1\le s\le22$, displayed on a signed logarithmic scale. Their survival is a consequence of the universal theorem, whose proof also allows a compatible zero prime parameter. The graph illustrates the theorem rather than proves it.}
132 
133 \chapter{Odd primitive blocks retain exactly the prime obstruction}\label{ch:odd}
134 \section{Polynomial division before evaluation}
135 The odd Hecke polynomial $T_{2m+1}$ is divisible by $a$. For $m\ge1$, so is
136 \[
137 T_{2m+1}-cT_{2m-1}.
138 \]
139 Define its quotient as an integer polynomial in $a^2$ before substituting the actual value $a=\tau(p)$. This matters at a hypothetical zero: the quotient exists algebraically even when numerical division by $a$ would be undefined.
140 
141 \begin{theorem}\label{thm:oddmultiplier}
142 For every odd prime $p\ne5$ and every $m\ge0$, an odd integer $U_{p,m}$ satisfies
143 \[
144 R(p^{2m+1})=\tau(p)U_{p,m}.
145 \]
146 The multiplier is defined by the recurrence polynomial, including when $\tau(p)=0$. If $\tau(p)\ne0$, then
147 \[
148 v_2(R(p^{2m+1}))=v_2(\tau(p)).
149 \]
150 \end{theorem}
151 \begin{proof}
152 For $m=0$, take $U_{p,0}=1$. For $m\ge1$, use the quotient
153 \[
154 U(a^2)=\frac{T_{2m+1}-cT_{2m-1}}a
155 \]
156 as polynomial division. The constant term in $a^2$ comes from the linear-in-$a$ terms in the two blocks. By \eqref{eq:trexpansion}, it is
157 \begin{equation}\label{eq:oddconstant}
158 (-1)^mb^{m-1}\bigl((m+1)b+mc\bigr).
159 \end{equation}
160 For an odd prime $p\ne5$, both $b=p^{11}$ and $c=\chifive(p)p^{10}$ are odd. The bracket in \eqref{eq:oddconstant} is congruent to $(m+1)+m=2m+1$ modulo $2$. Thus the constant term is odd.
161 
162 The actual $a=\tau(p)$ is even by \eqref{eq:parity}. Every nonconstant term in the polynomial in $a^2$ is even after evaluation. Hence the evaluated quotient is odd. It cannot be zero. Multiplication by $a$ gives the theorem, and multiplying a nonzero integer by an odd integer preserves its $2$-adic valuation.
163 \end{proof}
164 
165 This proof uses parity rather than a delicate size estimate for the quotient. It also exhibits the zero parameter honestly: if $a=0$, every odd primitive block is zero and every even primitive block remains nonzero by Chapter~\ref{ch:even}.
166 
167 \section{The primes 2 and 5}
168 At $p=2$, the residue formula \eqref{eq:rmod5} gives
169 \[
170 R(2^r)\equiv2^{r-1}(2-1)\not\equiv0\pmod5
171 \]
172 for every $r\ge1$. Thus no local block at $2$ vanishes.
173 
174 At $p=5$, the character is zero and $R(5^r)=\tau(5^r)$. The actual parameter is $\tau(5)=4830$, with $v_5(a)=1$. In \eqref{eq:trexpansion}, the $j=0$ term has valuation $r$ and every other term has valuation at least $r+9j$. Hence
175 \[
176 v_5(R(5^r))=r
177 \]
178 for every positive exponent, including odd ones.
179 
180 \section{The exact zero-equivalence theorem}
181 \begin{theorem}\label{thm:zeroequivalence}
182 For every positive integer $n$,
183 \[
184 R(n)=0\quad\Longleftrightarrow\quad\tau(n)=0.
185 \]
186 More precisely, either side is zero exactly when some prime occurring to an odd exponent in $n$ has $\tau(p)=0$.
187 \end{theorem}
188 \begin{proof}
189 The primitive transform is multiplicative. Its even prime-power blocks are nonzero by Theorem~\ref{thm:primitiveeven}. Its odd blocks at odd primes other than $5$ are the prime parameter times an odd, hence nonzero, integer by Theorem~\ref{thm:oddmultiplier}. The exceptional local blocks at $2$ and $5$ are nonzero by the preceding calculations. Thus its product over the prime factorization of $n$ has precisely the claimed zero criterion. Theorem~\ref{thm:primereduction} gives the same criterion for $\tau(n)$.
190 \end{proof}
191 
192 Primitive extraction introduces no new zero. It also removes no zero of the original coefficient sequence. This is stronger and more informative than agreement at a finite list of shells. It describes exactly how the two arithmetic functions encode the same conjecture.
193 
194 \section{The theorem's gain and its limit}
195 The primitive square theorem and the odd multiplier theorem classify the new local factors. They show that the extraction does not add an accidental vanishing caused by the numerator $1-\chifive(p)p^{10}X^2$. They also give a precise dyadic valuation law at every odd nonzero block.
196 
197 The remaining obstruction is concentrated in the first prime parameter. For $p\equiv4\pmod5$, the hypotheses used in the even theorem permit $a=0$. The resulting blocks are
198 \[
199 \begin{aligned}
200 R(p^{2m})&=(-1)^mp^{11m-1}(p+1),&&m\ge1,\\
201 R(p^{2m+1})&=0,&&m\ge0.
202 \end{aligned}
203 \]
204 The constant block is $R(1)=1$. This is consistent with the congruence, parity and recurrence inputs. Therefore those inputs and the new square theorem cannot jointly establish $a\ne0$. They have isolated its role.
205 
206 At a prime shell there is no imprimitive part, so the universal target is still
207 \[
208 \sum_{\substack{v\in\mathcal V_p\\g(v)=1}}W(v)
209 =R(p)=\tau(p)\ne0.
210 \]
211 The content geometry has clarified the radial arithmetic. The prime total still needs a global or uniform arithmetic mechanism excluding cancellation to zero.

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