\documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb,amsthm} \usepackage[margin=3cm]{geometry} \usepackage{hyperref} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newcommand{\Zt}{\mathbb{Z}_2} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\vv}{v} \title{The $2$-adic valuation of the constant\\ for symmetric ideal Prouhet--Tarry--Escott solutions} \author{David Niedbala Giraudin\\ \small ORCID 0009-0009-1526-1178} \date{3 November 2026} \begin{document} \maketitle \begin{abstract} For an ideal solution of the Prouhet--Tarry--Escott problem of size $n$, let $C$ denote the associated constant, and let $\overline{C_n}$ be the greatest common divisor of these constants. Filaseta and Markovich obtained $\vv_2(\overline{C_8})\ge6$ and determined $\vv_2(\overline{C_9})=9$, and asked for improved lower bounds in the range $10\le n\le12$; Bera and Nair gave bounds for $n=10$ and $n=12$. We treat the symmetric case at every size, even and odd, by $2$-adic arguments throughout. A single identity, the row law, governs both parities. For $n$ even, over $\Z$, it splits into two regular systems and yields $\vv_2(C)\ge6,8,11,12,12,16,18,18$ for $n=6,8,10,\dots,20$. The values at $n=6,8,10,12$ are attained and hence exact for the symmetric constant: in the symmetric case this closes both brackets of Bera--Nair, and gives $\vv_2(C'_8)=8$, for which the published information was $\vv_2(\overline{C_8})\ge6$. We also record that the two lower bounds of Bera--Nair, $9$ at $n=10$ and $11$ at $n=12$, are attained by $2$-adic solutions symmetric about $-1/2$, so that no argument using $2$-adic information alone can raise them. For $n$ odd, over $\Zt$, it combines with a finite-difference lemma refining a congruence of Filaseta--Markovich to give $\vv_2(C)\ge11,12,12$ for $n=11,13,15$, against $8$, $10$ and $11$ from Kleiman's bound $(n-1)!\mid C$; the chain also reproduces the known values $4,6,9$ at $n=5,7,9$. At $n=11$ the bound is sharp: the two quadratic near-solutions of Coppersmith, Mossinghoff, Scheinerman and VanderKam are symmetric ideal solutions over $\Zt$ with $\vv_2(C)=11$, since in both of the real quadratic fields involved the prime $2$ splits. This answers, for $n=11$ and $p=2$, two questions of Filaseta--Markovich, and exhibits for those sizes the case $p\le n$ of the question closing Rees--Smyth, who construct $p$-adic ideal solutions only for $p>n$: at the prime $2$ there is no local obstruction at these sizes. At $n=13$ we exhibit a symmetric ideal solution over $\Zt$, obtained from a second object of the same authors, although its constant has valuation $17$. \end{abstract} \section{Introduction} For a positive integer $n$, the Prouhet--Tarry--Escott problem asks for two distinct multisets $A$, $B$ of $n$ integers with $\sum_{a\in A}a^k=\sum_{b\in B}b^k$ for $1\le k\le n-1$. Such a pair is called an \emph{ideal} solution, and \[ \prod_{a\in A}(X-a)\;-\;\prod_{b\in B}(X-b)\;=\;C \] is then a nonzero constant. Writing $\overline{C_n}$ for the greatest common divisor of the constants $C$ over all ideal solutions of size $n$, Kleiman \cite{Kl} showed $(n-1)!\mid\overline{C_n}$, and Rees and Smyth \cite{RS}, Caley \cite{Ca,Ca2}, and others added further prime divisors. Filaseta and Markovich \cite{FM}, developing the master's thesis \cite{Ma}, used Newton polygons to obtain $\vv_2(\overline{C_8})\ge6$ and $\vv_2(\overline{C_9})=9$, and remarked that improved lower bounds for $10\le n\le12$ would be of interest. Bera and Nair \cite{BN} treated $n=10$ and $n=12$. A solution is \emph{symmetric} when $A=-A$ and $B=-B$ for $n$ even, and when $B=-A$ after translation for $n$ odd. We write $C'_n$ for the greatest common divisor of the constants over the symmetric ideal solutions of size $n$, so that $\overline{C_n}\mid C'_n$. All the arguments below are $2$-adic, but the two parities are not stated over the same ring, and it is worth saying at once why. For $n$ odd everything is carried out over $\Zt$ itself, so the results there bound the valuation for $2$-adic solutions and a fortiori for integer ones. For $n$ even the solutions are taken with entries in $\Z$. The arguments are again purely $2$-adic, but one step --- the equality of the numbers of odd entries on the two sides, in Proposition~\ref{prop:twosystems} --- needs to know that $C$ is even, which over $\Z$ is given by Kleiman's bound and which a $2$-adic solution of even size need not satisfy. The even-size results therefore extend verbatim to any symmetric ideal solution over $\Zt$ of even size whose constant is even, but not beyond. Throughout, $\vv=\vv_2$ denotes the $2$-adic valuation. The prime $2$ is the one the classical local theory does not reach. Rees and Smyth lift a solution modulo $p^k$ to one modulo $p^{k+1}$, and so obtain a $p$-adic ideal solution with $C\ne0$ for every $p>n$ \cite[Prop.~5.3 and Corollary]{RS}; their lifting requires the solution multiset to contain $n$ distinct residues modulo $p$, of which there are only two when $p=2$. Nothing in that direction constrains $\vv_2(C)$, which is why the $2$-adic case has to be treated by hand. Their paper closes by asking whether the corollary also holds for $p\le n$, so that there would be no local obstruction at any prime; the $2$-adic solutions of Sections~\ref{sec:sharp} and~\ref{sec:n13} answer that for $p=2$ at the sizes $11$ and $13$, where no integer solution is known. The two parities are governed by one identity, the row law of Section~\ref{sec:row}: if $f-g$ is a nonzero constant and $\beta$ is a root of $g$, then $f(\beta)$ is that constant, so the valuations of the differences between $\beta$ and the roots of $f$ sum to $\vv(C)$ --- and this holds for each root separately, not merely on average. What the two parities do with it differs, and the paper is organised accordingly: Sections~\ref{sec:red}--\ref{sec:evenmain} treat $n$ even, Sections~\ref{sec:fd}--\ref{sec:n13} treat $n$ odd. \begin{theorem}\label{thm:main} Let $A$ be a symmetric ideal solution of size $n$ with constant $C$, with entries in $\Z$ if $n$ is even and in $\Zt$ if $n$ is odd. Then \[ \begin{array}{c|cccccccc} n & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20\\\hline \vv(C)\ \ge & 6 & 8 & 11 & 12 & 12 & 16 & 18 & 18 \end{array} \qquad \begin{array}{c|cccccc} n & 5 & 7 & 9 & 11 & 13 & 15\\\hline \vv(C)\ \ge & 4 & 6 & 9 & 11 & 12 & 12 \end{array} \] The bounds at $n=5,6,7,8,9,10,11,12$ are attained, and those at $n=5,7,9$ are the known values of $\vv_2(\overline{C_n})$ \cite{RS,Ca,FM}. \end{theorem} At $n=10$ and $n=12$ the bounds are attained by integer solutions, so $\vv_2(C'_{10})=11$ and $\vv_2(C'_{12})=12$ exactly; these close the two brackets of Bera--Nair in the symmetric case. At $n=11$ the bound is attained only over $\Zt$, by objects already in the literature: \begin{theorem}\label{thm:sharp} For $n=11$ the bound is attained. Consequently \[ \min\{\vv(C) : A \text{ a symmetric ideal solution over } \Zt \text{ of size } 11\}=11 . \] \end{theorem} The witnesses are not new objects: they are the two near-solutions of size $11$ exhibited by Coppersmith, Mossinghoff, Scheinerman and VanderKam, each consisting of nine rational integers together with a conjugate pair of quadratic irrationalities. What is observed here is that both lie in $\Zt$ (Section~\ref{sec:sharp}). \section{The row law}\label{sec:row} \begin{proposition}[row law]\label{prop:row} Let $f=\prod_{i=1}^{N}(X-\alpha_i)$ and $g=\prod_{j=1}^{N}(X-\beta_j)$ be monic of the same degree over $\Zt$, with $f-g=C$ a nonzero constant. Then for every $j$, \[ \prod_{i=1}^{N}(\beta_j-\alpha_i)=C,\qquad\text{hence}\qquad \sum_{i=1}^{N}\vv(\beta_j-\alpha_i)=\vv(C), \] and symmetrically $\prod_{j}(\alpha_i-\beta_j)=-C$ for every $i$. \end{proposition} \begin{proof} Evaluating $f-g=C$ at $X=\beta_j$ gives $f(\beta_j)-0=C$, and $f(\beta_j)=\prod_i(\beta_j-\alpha_i)$. Evaluating at $X=\alpha_i$ gives $0-g(\alpha_i)=C$, and $g(\alpha_i)=\prod_j(\alpha_i-\beta_j)$. \end{proof} This evaluation is the one used by Rees--Smyth \cite{RS} and by Coppersmith--Mossinghoff--Scheinerman--VanderKam \cite{CMSV}; what is used below is its refinement at the level of valuations, namely that the $N$ sums $\sum_i\vv(\beta_j-\alpha_i)$ are all equal, and likewise the $N$ sums taken the other way. \section{Even sizes: reduction to squares and the two regular systems}\label{sec:red} Let $n=2r$ be even and let $A=\pm\{a_1,\dots,a_r\}$, $B=\pm\{b_1,\dots,b_r\}$ be a symmetric ideal solution with entries in $\Z$. \begin{proposition}\label{prop:squares} Put $\alpha_i=a_i^2$ and $\beta_j=b_j^2$. Then $\{\alpha_i\}$ and $\{\beta_j\}$ form an ideal solution of size $r$, all of whose entries are squares, and \[ C=(-1)^{\,r}\Bigl(\prod_{i=1}^{r}\alpha_i-\prod_{j=1}^{r}\beta_j\Bigr) , \qquad\text{so}\qquad \vv(C)=\vv\Bigl(\prod_i\alpha_i-\prod_j\beta_j\Bigr). \] \end{proposition} \begin{proof} The odd power sums of $A$ and of $B$ vanish by symmetry. For $1\le k\le r-1$ the power sums of order $2k$ agree, and $\sum_i a_i^{2k}=\tfrac12\sum_{a\in A}a^{2k}$, so the families of squares have equal power sums up to order $r-1$. Finally $\prod_{a\in A}(X-a)=\prod_i(X^2-\alpha_i)$ and likewise for $B$, and the constant is read at $X=0$. \end{proof} We write $\sigma$ and $\sigma'$ for the multisets $\{\vv(a_i)\}$ and $\{\vv(b_j)\}$, so that $\vv(\alpha_i)=2\vv(a_i)$ and $\vv(\beta_j)=2\vv(b_j)$. In the ancillary verifier these two profiles are printed $\sigma$ and $\tau$; the letter $\tau$ is reserved here for a scalar attached to the odd case (Section~\ref{sec:odd}). \begin{proposition}\label{prop:twosystems} Let $k_0$ be the number of odd entries among $a_1,\dots,a_r$ and $k_0'$ the number among $b_1,\dots,b_r$. Then $k_0=k_0'$, and the row law splits into two independent systems: a $k_0\times k_0$ matrix on the odd parts and an $(r-k_0)\times(r-k_0)$ matrix on the even parts, each with all row sums and all column sums equal to $\vv(C)$. \end{proposition} \begin{proof} If $\alpha_i$ is odd and $\beta_j$ even, or the reverse, then $\beta_j-\alpha_i$ is odd and contributes $0$. So for $\beta_j$ odd the row law reads $\sum_{\alpha_i\ \mathrm{odd}}\vv(\beta_j-\alpha_i)=\vv(C)$, and symmetrically. Let $M$ be the matrix of these valuations, with $k_0'$ rows and $k_0$ columns. Summing its entries in the two possible orders gives $k_0'\,\vv(C)=k_0\,\vv(C)$, and $\vv(C)\ne0$ because $(n-1)!$ divides $C$ by Kleiman's theorem and $n\ge4$; hence $k_0=k_0'$. The same argument applies to the even parts. \end{proof} \begin{corollary}\label{cor:3k0} $\vv(C)=3k_0+R$ with $R\ge0$. In particular $\vv(C)\ge3k_0$, and if every entry is odd then $\vv(C)\ge3r$. \end{corollary} \begin{proof} Odd squares are congruent to $1$ modulo $8$, so each of the $k_0$ entries of a row of the odd system is at least $3$; write $R$ for the total excess of that row over $3k_0$. \end{proof} Not every value of $R$ occurs. For a given $k_0$ the realisable values form a set with gaps, determined by the possible valuations of differences of odd squares; it is computed by a finite check in the ancillary verifier, which also confirms directly, modulo $8$, that $R=1$ cannot occur for $k_0=3$, and that $R=2$ can, consistently with the second solution of Borwein, Lison\v{e}k and Percival. \begin{lemma}\label{lem:equal} Suppose $\sigma=\sigma'$. Then $\vv(C)\ge2\sum\sigma+3$; and if moreover $k_0$ is odd, then $R\ge1$. \end{lemma} \begin{proof} Write $\alpha_i=2^{2\sigma_i}u_i$ and $\beta_j=2^{2\sigma'_j}u'_j$ with $u_i,u'_j$ odd squares (an entry $0$ cannot occur on both sides, since then $C$ would vanish by Proposition~\ref{prop:row}, and with $\sigma=\sigma'$ it occurs on neither). Then $\prod_i\alpha_i-\prod_j\beta_j=2^{2\sum\sigma}\bigl(\prod_iu_i-\prod_ju'_j\bigr)$, and both products are congruent to $1$ modulo $8$; this gives the first claim. For the second, suppose $R=0$. Then every entry of the odd system equals $3$, that is $\beta_j-\alpha_i\equiv8\pmod{16}$ for all odd $\alpha_i,\beta_j$. Hence all the odd $\alpha_i$ are congruent to one residue $X$ modulo $16$ and all the odd $\beta_j$ to $X+8$. Since $r\ge2$, the first power sums of the two families agree (Proposition~\ref{prop:squares}). Modulo $16$, an even $\alpha_i$ is $\equiv4$ if $\sigma_i=1$ and $\equiv0$ if $\sigma_i\ge2$, and likewise for the $\beta_j$; as $\sigma=\sigma'$, the even entries contribute the same amount on both sides. The odd entries contribute $k_0X$ and $k_0(X+8)$, which differ by $8k_0\equiv8\pmod{16}$ for $k_0$ odd, a contradiction. \end{proof} Finally, if $k_0=0$ then every entry is even, and $A/2$, $B/2$ is again a symmetric ideal solution with entries in $\Z$, whose constant is $C/2^n$. Iterating, every lower bound below that holds when $k_0\ge1$ holds a fortiori when $k_0=0$, since it then holds for $C/2^{n}$ and $\vv(C)=n+\vv(C/2^n)$. We may and do assume $k_0\ge1$, that is $\min\sigma=\min\sigma'=0$. \section{Even sizes: three instruments}\label{sec:tools} \subsection{The index bound} For integers $s,t\ge0$ put $\psi(s,t)=2\min(s,t)$ if $s\ne t$, and $\psi(s,s)=2s+3$. \begin{lemma}\label{lem:psi} If $\vv(a)=s$ and $\vv(b)=t$ then $\vv(a^2-b^2)\ge\psi(s,t)$, with equality when $s\ne t$. \end{lemma} \begin{proof} If $s\ne t$ then $\vv(a^2)=2s\ne2t=\vv(b^2)$ and the valuation of the difference is the smaller of the two. If $s=t$, write $a=2^sx$ and $b=2^sy$ with $x,y$ odd; then $a^2-b^2=2^{2s}(x-y)(x+y)$, and of the two factors $x\mp y$ one is congruent to $2$ modulo $4$ and the other divisible by $4$, so the valuation is at least $2s+3$. \end{proof} \begin{theorem}\label{thm:index} $\displaystyle \vv(C)\ \ge\ \max_{j}\ \sum_{i=1}^{r}\psi(\sigma_i,\sigma'_j)$, and symmetrically with the roles of the two families exchanged. \end{theorem} \begin{proof} Immediate from Proposition~\ref{prop:row} applied at $\beta_j$ and Lemma~\ref{lem:psi}. \end{proof} On the five symmetric ideal solutions of sizes $10$ and $12$ considered here this bound is tight in exactly three cases: it returns $14$ for the first solution of Borwein, Lison\v{e}k and Percival \cite{BLP}, $12$ for that of Kuosa, Meyrignac and Shuwen \cite{KMS}, and $12$ for that of Choudhry and Wr\'oblewski \cite{CW}, which are their exact values; on the second solution of Borwein, Lison\v{e}k and Percival it returns $9$ against the true value $11$, and on that of Broadhurst \cite{Br} $13$ against $14$. \subsection{Rigidity of the Newton polygon} \begin{theorem}\label{thm:rigid} In a symmetric ideal solution of even size, either $\sigma=\sigma'$ as multisets, or $\vv(C)=2\min\bigl(\textstyle\sum\sigma,\sum\sigma'\bigr)$. \end{theorem} \begin{proof} The elementary symmetric functions of order at most $r-1$ of $\{\alpha_i\}$ and of $\{\beta_j\}$ coincide, by Proposition~\ref{prop:squares}; only the constant terms may differ. Hence the Newton polygons of the two monic polynomials are lower convex hulls of the same points, except the point of abscissa $0$. If the two constant terms have the same valuation, the polygons coincide, and so do the multisets of valuations of the roots, that is $2\sigma=2\sigma'$. Otherwise the valuation of the difference is the smaller of the two constant-term valuations, namely $2\min(\sum\sigma,\sum\sigma')$. \end{proof} \subsection{The predictor} \begin{proposition}\label{prop:pred} For a half-family of profile $\sigma$, let $T_j(\sigma)$ be the sum of the $j$ smallest entries of $\sigma$ and $N_j(\sigma)$ the number of $j$-element sub-multisets realising that minimum. Then \[ \vv(e_j)=2\,T_j(\sigma)+\vv\bigl(N_j(\sigma)\bigr), \] where $e_j$ is the $j$-th elementary symmetric function of the squares, provided $N_j(\sigma)$ is not divisible by $8$ and $\vv(N_j(\sigma))$ is smaller than twice the gap between $T_j$ and the next attainable weight. \end{proposition} \begin{proof} Write $\alpha_i=2^{2\sigma_i}u_i$ with $u_i$ an odd square, so that each $j$-element sub-multiset contributes $2^{2W}\prod u_i$ with $W$ its weight, and $\prod u_i\equiv1\pmod 8$ as a product of odd squares. The $N_j$ sub-multisets of minimal weight therefore contribute $2^{2T_j}(N_j+8k)$ for some $k$, of valuation $2T_j+\vv(N_j)$ as soon as $8\nmid N_j$. Every other sub-multiset has weight at least $T_j+g$, where $g$ is the gap to the next attainable weight, hence valuation at least $2T_j+2g$, which exceeds $2T_j+\vv(N_j)$ under the second proviso. \end{proof} This predictor is exact, at every order $j$, on the twelve half-solutions of the six even solutions of sizes $8$, $10$ and $12$ checked in the first ancillary file. \section{Even sizes: the main theorem and the extremal configurations}\label{sec:evenmain} \begin{theorem}\label{thm:even} For every symmetric ideal solution of even size $n$ with entries in $\Z$, the valuation of the constant satisfies the bounds below. At $n=6,8,10,12$ the bounds are attained, hence exact for $C'_n$. \[ \begin{array}{c|c|c|c|c|c} n & \text{Kleiman} & \text{previously} & \text{here} & \text{gain} & \text{status}\\\hline 6 & 3 & \text{---} & 6 & +3 & \text{exact}\\ 8 & 4 & \ge6\text{ (Filaseta--Markovich)} & 8 & +4 & \text{exact}\\ 10 & 7 & 9\text{ to }11\text{ (Bera--Nair)} & 11 & +4 & \text{exact}\\ 12 & 8 & 11\text{ to }12\text{ (Bera--Nair)} & 12 & +4 & \text{exact}\\ 14 & 10 & \text{---} & 12 & +2 & \text{lower bound}\\ 16 & 11 & \text{---} & 16 & +5 & \text{lower bound}\\ 18 & 15 & \text{---} & 18 & +3 & \text{lower bound}\\ 20 & 16 & \text{---} & 18 & +2 & \text{lower bound} \end{array} \] \end{theorem} \begin{proof} By the reduction following Lemma~\ref{lem:equal} we may assume $\min\sigma=\min\sigma'=0$, and by Theorem~\ref{thm:rigid} there are two cases. \emph{Case $\sigma=\sigma'$.} Then $\vv(C)$ is at least the bound of Theorem~\ref{thm:index}, at least $2\sum\sigma+3$, and of the form $3k_0+R$ with $R$ in the set of values realisable for $k_0$ (Corollary~\ref{cor:3k0}), with $R\ge1$ when $k_0$ is odd (Lemma~\ref{lem:equal}). An entry equal to $9$ or more gives $\vv(C)\ge\psi(9,9)=21$ by Theorem~\ref{thm:index}, beyond every bound of the table, so it suffices to enumerate the profiles with entries at most $8$ and to take, for each, the least value of $\vv(C)$ compatible with these conditions. \emph{Case $\sigma\ne\sigma'$.} Let $\sigma'$ denote the profile of smaller sum and put $s'=\sum\sigma'$, so that $\vv(C)=2s'$. Since $\vv(C)=3k_0+R$ is then even, $R=0$ is impossible for $k_0$ odd. The pair must satisfy the following necessary conditions: \begin{enumerate} \item[(a)] Theorem~\ref{thm:index}, with the roles of the two families in both orders; \item[(b)] $\vv(C)=3k_0+R$ with $R$ realisable for $k_0$; \item[(c)] compatibility of the predictor: since $e_j$ is the same for the two families for $1\le j\le r-1$, whenever Proposition~\ref{prop:pred} determines $\vv(e_j)$ for both profiles the two values agree, and whenever it determines it for one profile but not for the other, that value is at least $2T_j+\min(3,2g)$ computed for the other profile, which by the proof of Proposition~\ref{prop:pred} is a lower bound for $\vv(e_j)$ whenever the predictor does not determine it. \end{enumerate} For a value $B$ of the table, $2s'\le B$ bounds the entries of $\sigma'$. The entries of $\sigma$ are not bounded a priori (an entry $a_i=0$ counts as an infinite entry), but by Lemma~\ref{lem:sat} below it suffices to enumerate the profiles $\sigma$ whose entries are at most $s'+r+3$. Both enumerations are finite; they are carried out in the ancillary files for the eight sizes $6\le n\le20$, and the minimum over the two cases is the bound of the table. \end{proof} \begin{lemma}\label{lem:sat} Let $\sigma\ne\sigma'$ with $s'=\sum\sigma'<\sum\sigma$. There is an integer $c$ with $s'+1\le c\le s'+r+1$ such that no entry of $\sigma$ equals $c+1$. Let $\sigma^*$ be obtained from $\sigma$ by replacing every entry larger than $c$ by $c+2$. Then $\sigma^*$ has entries at most $s'+r+3$, and the pair $(\sigma^*,\sigma')$ satisfies each of the conditions \textup{(a)}, \textup{(b)}, \textup{(c)} if and only if $(\sigma,\sigma')$ does. \end{lemma} \begin{proof} The $r+1$ values $s'+2,\dots,s'+r+2$ cannot all be entries of $\sigma$, which has $r$ entries; this gives $c$. Call the entries larger than $c$ \emph{large}; they are at least $c+2$, and the other entries are at most $c$. Every entry of $\sigma'$ is at most $s'\max\sigma'$ one has $\psi(x,t)=2t$ for every entry $t$ of $\sigma'$, so the row and column sums of Theorem~\ref{thm:index} do not depend on the value of a large entry. (b) The number $k_0$ of zero entries and the value $\vv(C)=2s'$ are unchanged. (c) Let $p$ be the number of entries at most $c$. For $j\le p$, the $j$ smallest entries, $T_j$ and $N_j$ do not involve the large entries. The next attainable weight is either reached without large entries, which does not depend on them, or by exchanging a chosen entry, at most $c$, for a large one, which raises the weight by at least $2$. As the predictor depends on the gap $g$ only through $\min(3,2g)$ and through the comparison of $\vv(N_j)\le2$ with $2g$ (when $8\nmid N_j$), its output is the same for $\sigma$ and $\sigma^*$. For $j>p$, every $j$-element sub-multiset of $\sigma$ or of $\sigma^*$ contains a large entry, so $T_j\ge c+2\ge s'+3$ and the value or lower bound given by the predictor is at least $2s'+6$; for $\sigma'$ it is at most $2T_j(\sigma')+3\le2s'+3$. Hence, for both $\sigma$ and $\sigma^*$, the order $j$ is incompatible exactly when the predictor determines $\vv(e_j)$ for $\sigma'$, a condition that does not involve $\sigma$. \end{proof} The two smallest even sizes are not new, and serve as a calibration. At $n=6$ the bound returned is $6$, and it is attained by $\pm\{0,7,7\}$ against $\pm\{3,5,8\}$, whose constant is $14400=2^6\cdot225$. At $n=8$ the bound is $8$, and it is attained by equation~(4.7) of \cite{RS}, so that $\vv_2(C'_8)=8$; we have not found this exact value stated, the published information being $\vv_2(\overline{C_8})\ge6$ from \cite{FM}. The value $11$ at $n=10$ is attained already by the solution recorded as equation~(4.9) of \cite{RS}, which goes back to Letac and is recorded by Gloden \cite{Gl}, and again by the second solution of Borwein, Lison\v{e}k and Percival; the value $12$ at $n=12$ by that of Kuosa, Meyrignac and Shuwen \cite{KMS} as well as by that of Choudhry and Wr\'oblewski. Both brackets of Bera--Nair therefore close in the symmetric case. Since every ideal solution known at these two sizes is symmetric, a smaller value for $\overline{C_n}$ would require a \emph{non-symmetric} ideal solution of size $10$ or $12$, and none is known. In the case $\sigma\ne\sigma'$, the enumeration shows that every pair compatible with the bound has its two profiles equal in every entry but the largest, the largest entry of $\sigma$ taking every value from some threshold on. This is an observation from the computation, not a consequence of Theorem~\ref{thm:rigid}. We call such a one-parameter family a \emph{form}. \begin{theorem}\label{thm:class} The enumeration yields more than a lower bound: the exhaustive list of the profiles that the necessary conditions above do not exclude at the bound. Any symmetric ideal solution attaining the bound has one of these profiles. \[ \begin{array}{c|l|c} n & \text{profiles }\sigma=\sigma'\text{ compatible with the bound} & \text{forms with }\sigma\ne\sigma'\\\hline 10 & (0,0,0,1,1),\ (0,0,0,1,2),\ (0,0,0,1,3) & 0\\ 12 & (0,0,0,0,1,1),\ (0,0,0,0,1,2),\ (0,0,0,0,1,3),\ (0,0,0,1,1,2) & 3\\ 14 & (0,0,0,0,1,1,2) & 1\\ 16 & \text{none} & 1\\ 18 & \text{fourteen profiles} & 6\\ 20 & \text{five profiles} & 2 \end{array} \] \end{theorem} Three consequences deserve to be recorded. At size $10$ the case of distinct profiles produces nothing at or below the bound, so any symmetric ideal solution of size $10$ attaining $11$ has two half-families of the \emph{same} profile, which the solution of Borwein, Lison\v{e}k and Percival confirms. At size $16$ no profile with $\sigma=\sigma'$ descends below $17$: if the two half-families have the same profile then $\vv(C)\ge17$, and the value $16$ can only be attained in the single form \[ \sigma'=(0,0,0,0,1,1,2,4),\qquad \sigma=(0,0,0,0,1,1,2,\Lambda),\qquad \Lambda\ge5 . \] At size $14$ the profile is \emph{unique} in the case $\sigma=\sigma'$, namely $(0,0,0,0,1,1,2)$, and the one form with distinct profiles is $\sigma'=(0,0,0,0,1,1,4)$ against $\sigma=(0,0,0,0,1,1,\Lambda)$ with $\Lambda\ge5$; at a size where no solution is known, this is a completely explicit search criterion. \section{Odd sizes: a finite-difference lemma}\label{sec:fd} \begin{definition}\label{def:sol} Let $n\ge5$ be odd. A \emph{symmetric ideal solution over $\Zt$ of size $n$} is a tuple $A=(a_1,\dots,a_n)\in\Zt^{\,n}$ such that \[ p_k(A):=\sum_{i=1}^{n}a_i^{\,k}=0\qquad\text{for every odd }k\text{ with }1\le k\le n-2, \] and such that $C(A):=-2\prod_{i=1}^n a_i$ is nonzero. \end{definition} Equivalently, with $f(X)=\prod_i(X-a_i)$ one has $f(X)+f(-X)=C(A)$. Indeed $\prod_i(X+a_i)=-f(-X)$ because $n$ is odd, so $\prod_i(X-a_i)-\prod_i(X+a_i)=f(X)+f(-X)$, which is twice the sum of the terms of $f$ of even degree. As $n$ is odd those are the terms $(-1)^ke_k(A)X^{n-k}$ with $k$ odd, and they reduce to the single term $-e_n(A)$ precisely when $e_1=e_3=\dots=e_{n-2}=0$. If $(A,-A)$ is a symmetric ideal PTE solution over $\Z$, then $A$ is a symmetric ideal solution over $\Zt$ in the above sense, with the same constant. We write \[ t(A):=\sum_{i=1}^{n}\vv(a_i), \] let $O$ and $E$ be the sets of indices with $a_i$ odd, respectively even, and put \[ w:=|O|,\qquad u:=|E|=n-w,\qquad c_i:=\vv(a_i)\ (i\in E),\qquad m_1:=\#\{i\in E:\ c_i=1\}. \] \begin{proposition}\label{prop:vC} $\vv(C)=1+t(A)$. \end{proposition} \begin{proof} Immediate from $C=-2\prod a_i$. \end{proof} \begin{corollary}\label{cor:rowodd} For every $j$, $\ \sum_{i=1}^{n}\vv(a_j+a_i)=\vv(C)$. \end{corollary} \begin{proof} Apply Proposition~\ref{prop:row} with $f=\prod_i(X-a_i)$ and $g=\prod_i(X+a_i)$, whose roots are the $-a_i$. \end{proof} \begin{lemma}\label{lem:fd} Let $A$ be as in Definition~\ref{def:sol} and let $0\le m\le (n-3)/2$. Then \[ \sum_{i=1}^{n} a_i\,(a_i^2-1)^m=0, \] and consequently \[ 8^{\,m}\ \Big|\ \sum_{i\in E} a_i\,(a_i^2-1)^m \qquad\text{in }\Zt . \] \end{lemma} \begin{proof} Expanding, $a(a^2-1)^m=\sum_{j=0}^{m}(-1)^{m-j}\binom{m}{j}a^{2j+1}$, so \[ \sum_i a_i(a_i^2-1)^m=\sum_{j=0}^{m}(-1)^{m-j}\binom{m}{j}\,p_{2j+1}(A)=0, \] every exponent $2j+1$ being odd and at most $2m+1\le n-2$. For $a$ odd one has $8\mid a^2-1$, so each term with $i\in O$ is divisible by $8^m$; the displayed divisibility follows. \end{proof} \begin{remark} The corresponding statement of Filaseta--Markovich \cite{FM} is recorded there in the form of a congruence modulo $16$ between the power sums of the odd parts, respectively of the even parts, of the two sides of a general ideal solution. Lemma~\ref{lem:fd} is a strengthening: instead of one fixed modulus it produces a tower of conditions modulo $8,64,512,4096,\dots$, indexed by $m$, and these are conditions on the even part alone. \end{remark} \begin{corollary}\label{cor:parity} $w$ is even and $u$ is odd. \end{corollary} \begin{proof} $p_1(A)=0$; reducing modulo $2$ gives $w\equiv 0$. \end{proof} \begin{corollary}\label{cor:cmin} Let $c_{\min}=\min\{c_i:i\in E\}$ and let $\mu$ denote its multiplicity. If $\mu$ is odd then $c_{\min}\ge 3\cdot\frac{n-3}{2}$. \end{corollary} \begin{proof} In $\sum_{i\in E}a_i(a_i^2-1)^m$ each term has valuation exactly $c_i$, since $a_i$ is even and $a_i^2-1$ is odd. The $\mu$ terms with $c_i=c_{\min}$ are each $2^{c_{\min}}$ times an odd number, so, $\mu$ being odd, their sum is again $2^{c_{\min}}$ times an odd number, while every other term has valuation greater than $c_{\min}$. Hence the whole sum has valuation exactly $c_{\min}$, and Lemma~\ref{lem:fd} with $m=(n-3)/2$ gives $3m\le c_{\min}$. \end{proof} \begin{corollary}\label{cor:m1}\label{cor:u1} The number $m_1$ is even; and if $u=1$ then $t\ge 3\cdot\frac{n-3}{2}$. \end{corollary} \begin{proof} If $m_1\ge1$ then $c_{\min}=1$, and $1<3(n-3)/2$ for $n\ge5$, so $\mu=m_1$ is even by Corollary~\ref{cor:cmin}; if $m_1=0$ there is nothing to prove. If $u=1$ then $\mu=1$ is odd and $t=c_{\min}$. \end{proof} \begin{lemma}\label{lem:oddsum} Put $s:=\max\{\min(2+\vv(k-1),\,k)\ :\ k \text{ odd},\ 3\le k\le n-2\}$. Then \[ \sum_{i\in O}a_i\equiv 0 \pmod{2^{\,s}} . \] One has $s\ge 3$ for $n\ge5$, $s\ge4$ for $n\ge7$ and $s\ge5$ for $n\ge 11$. \end{lemma} \begin{proof} Fix such a $k$. For $a$ odd and $M=k-1$ even, the lifting-the-exponent formula gives $\vv(a^{M}-1)=\vv(a-1)+\vv(a+1)+\vv(M)-1\ge 2+\vv(M)$, since one of $a\mp1$ is divisible by $4$. Hence $a^{k}\equiv a \pmod{2^{\,2+\vv(k-1)}}$ for $a$ odd, while $\vv(a^k)\ge k$ for $a$ even. Reducing $p_k(A)=0$ modulo $2^{\min(2+\vv(k-1),k)}$ leaves $\sum_{i\in O}a_i$. Taking the largest such modulus over the admissible $k$ gives the claim; the stated values come from $k=3,5,9$. \end{proof} \section{Odd sizes: the odd part}\label{sec:odd} \begin{proposition}\label{prop:oddrow} For every $j\in O$, \[ \sum_{i\in O,\ i\ne j}\vv(a_j+a_i)=t . \] In particular $t\ge w-1$, and $\tau:=t-(w-1)\ge 0$ does not depend on $j$. \end{proposition} \begin{proof} Apply Corollary~\ref{cor:rowodd} at $j\in O$. For $i\in E$ the sum $a_j+a_i$ is odd, so those terms vanish; the term $i=j$ equals $\vv(2a_j)=1$. By Proposition~\ref{prop:vC} the total is $1+t$. \end{proof} If $a_i\equiv a_j \pmod 4$ with both odd then $a_i+a_j\equiv 2\pmod 4$, so $\vv(a_i+a_j)=1$ exactly; if $a_i\not\equiv a_j\pmod 4$ then $\vv(a_i+a_j)\ge2$. Let $A_{(1)}$ and $A_{(3)}$ denote the elements of $O$ congruent to $1$, respectively $3$, modulo $4$. Then for $j\in A_{(1)}$, \[ \tau=\sum_{i\in A_{(3)}}\bigl(\vv(a_i+a_j)-1\bigr), \] and symmetrically for $j\in A_{(3)}$. \begin{proposition}\label{prop:tau} Assume $w\ge2$, let $h:=w/2$ and let $s$ be as in Lemma~\ref{lem:oddsum}, so $s\ge3$. Exactly one of the following holds. \begin{enumerate} \item[\textup{(i)}] $\tau=0$; then $O$ lies in a single class modulo $4$ and $w\equiv 0 \pmod 4$. \item[\textup{(ii)}] $\tau=h$; then $h$ is even, hence again $w\equiv 0 \pmod 4$. \item[\textup{(iii)}] $\tau\ \ge\ h+\lceil h/2\rceil$. \end{enumerate} Moreover if $w=2$ then \textup{(i)} and \textup{(ii)} are impossible and $\tau\ge s-1$. \end{proposition} \begin{proof} Suppose first $\tau=0$. Then no pair of elements of $O$ lies in distinct classes modulo $4$, so one of $A_{(1)},A_{(3)}$ is empty and $O$ is contained in a single class, say $\varepsilon\in\{1,3\}$. Then $\sum_{i\in O}a_i\equiv w\varepsilon \pmod 4$, and Lemma~\ref{lem:oddsum} forces $4\mid w\varepsilon$, i.e. $4\mid w$. Now suppose $\tau>0$, so both classes are nonempty; write $p=|A_{(1)}|$, $q=|A_{(3)}|$ and let $M=(M_{ab})$ be the $p\times q$ matrix with entries $M_{ab}=\vv(a+b)-1\ge 1$ for $a\in A_{(1)}$, $b\in A_{(3)}$. By the displayed formula all row sums and all column sums of $M$ equal $\tau$. Counting the total two ways gives $p\tau=q\tau$, hence $p=q=h$. Since all entries are at least $1$, $\tau\ge h$. Assume $\tau=h$. Then every entry of $M$ equals $1$, i.e. $\vv(a+b)=2$ for every cross pair, i.e. $a+b\not\equiv0\pmod 8$. Elements of $A_{(1)}$ are $\equiv1$ or $5$ modulo $8$ and elements of $A_{(3)}$ are $\equiv3$ or $7$; the forbidden sums modulo $8$ are exactly $(1,7)$ and $(5,3)$. If $A_{(1)}$ met both classes modulo $8$, then $A_{(3)}$ could contain neither a $3$ nor a $7$, which is absurd. So $A_{(1)}$ is homogeneous modulo $8$ and then so is $A_{(3)}$, the two possibilities being $(1,3)$ and $(5,7)$. In both cases $\sum_{i\in O}a_i\equiv 4h \pmod 8$, and Lemma~\ref{lem:oddsum} with $s\ge3$ forces $h$ even. Finally assume $\tau>h$ and put $N=M-J$, where $J$ is the all-ones matrix, so $N\ge0$ with all line sums $\rho=\tau-h\ge1$. One has $N_{ab}\ge1$ exactly when $a+b\equiv 0 \pmod 8$, that is on the set $(A_{(1)}^{1}\times A_{(3)}^{7})\cup(A_{(1)}^{5}\times A_{(3)}^{3})$, where the superscript records the class modulo $8$. Write $\kappa=|A_{(1)}^{1}|$ and $\lambda=|A_{(3)}^{3}|$. A row of $N$ indexed by an element of $A_{(1)}^{1}$ has at least $h-\lambda$ entries $\ge1$, whence $\rho\ge h-\lambda$ whenever $A_{(1)}^{1}\ne\emptyset$; similarly $\rho\ge\lambda$ if $A_{(1)}^{5}\ne\emptyset$, $\rho\ge h-\kappa$ if $A_{(3)}^{3}\ne\emptyset$, and $\rho\ge\kappa$ if $A_{(3)}^{7}\ne\emptyset$. If $0<\kappa\vv(a_k)$, a contradiction. So the maximum is attained exactly once. For the second assertion, write $f(X)=XH(X^2)+c_0$ with $H$ monic of degree $(n-1)/2$ and $C=2c_0$, as in Definition~\ref{def:sol}. From $f(a_i)=0$ one gets $a_iH(a_i^2)=-c_0$ and hence $a_i^2H(a_i^2)^2=c_0^2$, so each $a_i^2$ is a root of the monic polynomial $R(Y)=YH(Y)^2-c_0^2$ of degree $n$. In fact $R(X^2)=-f(X)f(-X)=\prod_i(X^2-a_i^2)$, since $f(-X)=-XH(X^2)+c_0$ and $n$ is odd, so the roots of $R$ are exactly the $a_i^2$ with multiplicity. Now $R$ and $YH(Y)^2$ have the same coefficients in every degree from $1$ to $n$ and differ only in degree $0$, so their Newton polygons share every vertex of abscissa at least $1$. By the first assertion the largest valuation among the $a_i^2$, namely $2M$, is attained once, so the initial segment of the Newton polygon of $R$ has length $1$; the remaining $n-1$ slopes are those of $YH(Y)^2$, that is the valuations of the roots of $H$, each occurring twice. Halving, every value other than $M$ occurs an even number of times among the $\vv(a_i)$. \end{proof} \begin{remark} Theorem~\ref{thm:profile} holds on the odd witnesses used in this paper: Letac's two solutions of size $9$ \cite{Le} have full valuation profiles $(4,1,1,1,1,0,0,0,0)$ and $(3,2,2,1,1,0,0,0,0)$, the two quadratic objects of Section~\ref{sec:sharp} have $(4,2,2,1,1,0^6)$, and the solution of Section~\ref{sec:n13} has $(6,3,3,1,1,1,1,0^6)$. In each case the maximum occurs once and every other value an even number of times. Taking the value $0$, whose multiplicity is $w$, recovers Corollary~\ref{cor:parity}. \end{remark} \section{Odd sizes: lower bounds}\label{sec:lower} \begin{lemma}[reduction]\label{lem:red} If $A$ is a symmetric ideal solution over $\Zt$ of size $n$ with $w=0$, then $A/2$ is one as well, and $t(A/2)=t(A)-n$. Consequently, in order to bound $t$ from below it suffices to treat the case $w\ge 2$. \end{lemma} \begin{proof} If every $a_i$ is even then $p_k(A/2)=2^{-k}p_k(A)=0$ and $C(A/2)=2^{-n}C(A)\ne0$. Given an arbitrary $A$, halving it repeatedly produces $A'$ with $w(A')\ge1$ and $t(A')\le t(A)$; by Corollary~\ref{cor:parity}, $w(A')$ is even, hence $\ge2$. \end{proof} \subsection{The sizes $n=5$ and $n=7$} \begin{proposition}\label{prop:57} For $n=5$ one has $t\ge3$, and for $n=7$ one has $t\ge5$; hence $\vv(C)\ge4$ and $\vv(C)\ge6$ respectively. Both bounds are attained. \end{proposition} \begin{proof} Let $n=5$, so $s=3$ and $w\in\{2,4\}$. If $w=2$ then $u=3$ and Corollary~\ref{cor:ubound} with $m_1$ even gives $t\ge4$. If $w=4$ then $u=1$ and Corollary~\ref{cor:u1} gives $t\ge3\cdot1=3$. Let $n=7$, so $s=4$ and $w\in\{2,4,6\}$. If $w=2$ then $u=5$ and Corollary~\ref{cor:ubound} gives $t\ge8$. If $w=6$ then $u=1$ and Corollary~\ref{cor:u1} gives $t\ge6$. If $w=4$ then $u=3$, $h=2$, and Corollary~\ref{cor:ubound} gives $t\ge4$; by Proposition~\ref{prop:tau} the possible values of $\tau$ are $0$, $2$ and $\tau\ge3$, giving $t=3+\tau\in\{3,5\}\cup[6,\infty)$, and $t=3$ is excluded. Hence $t\ge5$. The tuples $(9,5,1,-7,-8)$ and $(-50,-38,-13,-7,24,33,51)$ are symmetric ideal solutions over $\Z$ with $t=3$ and $t=5$, so $\vv(C)=4$ and $\vv(C)=6$. These are the known values of $\vv_2(\overline{C_5})$ \cite[Prop.~2.4]{RS} and of $\vv_2(\overline{C_7})$, for which Caley proved $2^6\mid\overline{C_7}$ \cite{Ca}; see \cite[\S2.1]{CMSV}. \end{proof} \subsection{The size $n=9$} \begin{theorem}\label{thm:n9} Let $A$ be a symmetric ideal solution over $\Zt$ of size $9$. Then $t\ge8$, hence $\vv(C)\ge9$. The bound is attained. \end{theorem} \begin{proof} Here $s=4$ and $(n-3)/2=3$; by Lemma~\ref{lem:red}, $w\in\{2,4,6,8\}$. \begin{itemize} \item $w=2$, $u=7$: Corollary~\ref{cor:ubound} gives $t\ge10$. \item $w=4$, $u=5$: Corollary~\ref{cor:ubound} gives $t\ge8$. \item $w=6$, $u=3$, $h=3$: case (i) of Proposition~\ref{prop:tau} requires $4\mid w$ and case (ii) requires $h$ even, so (iii) holds and $\tau\ge3+2=5$, whence $t=5+\tau\ge10$. \item $w=8$, $u=1$: Corollary~\ref{cor:u1} gives $t\ge3\cdot3=9$. \end{itemize} Thus $t\ge8$. The first solution of Letac of size $9$, namely \[ (-98,-82,-58,-34,13,16,69,75,99), \] has $w=4$, $u=5$ and even valuations $(4,1,1,1,1)$, so $t=8$ and $\vv(C)=9$. \end{proof} Theorem~\ref{thm:n9} recovers, for symmetric solutions, the value $\vv_2(\overline{C_9})=9$ obtained by Filaseta and Markovich by a different route, and shows in addition that the value cannot be improved by any $2$-adic argument, since it is attained already over $\Z$. \subsection{The size $n=11$} \begin{proposition}\label{prop:n11cases} Let $A$ be a symmetric ideal solution over $\Zt$ of size $11$ with $t\le9$. Then $w=8$, $u=3$, $\tau=0$, $t=7$, and the multiset of even valuations is $(5,1,1)$ or $(3,2,2)$. Writing $a_2,a_3$ for the two entries of equal valuation, one has $\vv(a_2+a_3)=5$ in the first case and $\vv(a_2+a_3)=3$ in the second. \end{proposition} \begin{proof} Here $s=5$ and, by Lemma~\ref{lem:red}, $w\in\{2,4,6,8,10\}$. \begin{itemize} \item $w=2$, $u=9$: Corollary~\ref{cor:ubound} gives $t\ge14$. \item $w=4$, $u=7$: Corollary~\ref{cor:ubound} gives $t\ge10$. \item $w=6$, $u=5$, $h=3$: cases (i) and (ii) of Proposition~\ref{prop:tau} are excluded ($4\nmid6$; $h$ odd), so $\tau\ge3+2=5$ and $t=5+\tau\ge10$. \item $w=10$, $u=1$: Corollary~\ref{cor:u1} gives $t\ge3\cdot4=12$. \item $w=8$, $u=3$, $h=4$: Proposition~\ref{prop:tau} leaves $\tau=0$, $\tau=4$ or $\tau\ge6$, that is $t=7+\tau\in\{7\}\cup\{11\}\cup[13,\infty)$. \end{itemize} So $t\le9$ forces $w=8$, $u=3$, $\tau=0$ and $t=7$. The multiset $(c_1,c_2,c_3)$ is a partition of $7$ into three positive parts: $(5,1,1)$, $(4,2,1)$, $(3,3,1)$ or $(3,2,2)$. In $(4,2,1)$ and $(3,3,1)$, $c_{\min}=1$ has multiplicity $1$, so Corollary~\ref{cor:cmin} would force $c_{\min}\ge12$, which fails. For $(5,1,1)$, Proposition~\ref{prop:evenrow} at an index of valuation $1$ reads $2+\min(5,1)+\vv(a_2+a_3)=1+t=8$, so $\vv(a_2+a_3)=5$. For $(3,2,2)$, at an index of valuation $2$ it reads $3+\min(3,2)+\vv(a_2+a_3)=8$, so $\vv(a_2+a_3)=3$. \end{proof} \begin{proposition}\label{prop:kill} Neither configuration of Proposition~\ref{prop:n11cases} occurs. \end{proposition} \begin{proof} \emph{The profile $(5,1,1)$.} Write $E_k=\sum_{i\in E}a_i^{\,k}$ and $\Delta^m E=\sum_{j=0}^m(-1)^{m-j}\binom{m}{j}E_{2j+1}=\sum_{i\in E}a_i(a_i^2-1)^m$, so that $8^m\mid\Delta^m E$ for $m\le4$ by Lemma~\ref{lem:fd}. Write $a_1=32b_1$, $a_2=2b_2$, $a_3=2b_3$ with $b_i$ odd. For $k$ odd, $b_2^{\,k}+b_3^{\,k}=(b_2+b_3)q_k$ with $q_k$ odd, being a sum of $k$ odd terms. The condition $\vv(a_2+a_3)=5$ gives $\vv(b_2+b_3)=4$, say $b_2+b_3=16c$ with $c$ odd; put $z:=b_1+c$, an even integer. Then \[ E_1=32z,\qquad E_3=2^{15}b_1^3+2^7cq_3,\qquad E_5=2^{25}b_1^5+2^9cq_5, \] \[ E_7=2^{35}b_1^7+2^{11}cq_7,\qquad E_9=2^{45}b_1^9+2^{13}cq_9 . \] Modulo $2^9$ the only surviving terms of $\Delta^3E=E_7-3E_5+3E_3-E_1$ are $3E_3$ and $-E_1$, so $384\,cq_3-32z\equiv0\pmod{2^9}$, that is $z\equiv12\,cq_3\pmod{16}$. As $cq_3$ is odd, $12\,cq_3\equiv4$ or $12\pmod{16}$, whence $z\equiv4\pmod 8$. Modulo $2^{12}$ the only surviving terms of $\Delta^4E=E_9-4E_7+6E_5-4E_3+E_1$ are $6E_5$, $-4E_3$ and $E_1$, so $3072\,cq_5-512\,cq_3+32z\equiv0\pmod{2^{12}}$, that is $96\,cq_5-16\,cq_3+z\equiv0\pmod{128}$; reducing modulo $16$ gives $z\equiv0\pmod{16}$, hence $z\equiv0\pmod 8$. This contradicts the previous congruence. \emph{The profile $(3,2,2)$.} Write $a_1=8b_1$, $a_2=4b_2$, $a_3=4b_3$ with $b_i$ odd. Now $\vv(a_2+a_3)=3$ gives $b_2+b_3=2c$ with $c$ odd, and as before $b_2^{\,k}+b_3^{\,k}=2cq_k$ with $q_k$ odd for $k$ odd. Put $z:=b_1+c$. Then \[ E_1=8z,\qquad E_3=2^{9}b_1^3+2^7cq_3,\qquad E_5=2^{15}b_1^5+2^{11}cq_5, \] \[ E_7=2^{21}b_1^7+2^{15}cq_7,\qquad E_9=2^{27}b_1^9+2^{19}cq_9 . \] Modulo $2^9$, $\Delta^3E\equiv3E_3-E_1\equiv384\,cq_3-8z$, so $z\equiv48\,cq_3\pmod{64}$ and $\vv(z)=4$. Modulo $2^{12}$, the terms $E_9$, $4E_7$ and $6E_5$ vanish (note $6\cdot2^{11}=3\cdot2^{12}$), so $\Delta^4E\equiv-4E_3+E_1\equiv-2^{11}b_1^3-2^9cq_3+8z$, whence $z\equiv64\,cq_3+256\,b_1^3\pmod{512}$ and $\vv(z)=6$. The two values of $\vv(z)$ are incompatible. \end{proof} \begin{remark} Proposition~\ref{prop:kill} has been confirmed independently by exhaustive computation, for both profiles, in the second ancillary file: no even triple of either profile satisfies the two conditions at $m=3$ and $m=4$ simultaneously. \end{remark} \begin{theorem}\label{thm:n11} Let $A$ be a symmetric ideal solution over $\Zt$ of size $11$. Then $t\ge10$, hence \[ \vv(C)\ \ge\ 11 . \] \end{theorem} \begin{proof} Immediate from Propositions~\ref{prop:n11cases} and \ref{prop:kill}. \end{proof} For comparison, Kleiman's bound $(n-1)!\mid\overline{C_n}$ gives $\vv_2\ge\vv_2(10!)=8$. \subsection{The sizes $n=13$ and $n=15$} \begin{theorem}\label{thm:n13} Let $A$ be a symmetric ideal solution over $\Zt$ of size $13$. Then $t\ge11$, hence $\vv(C)\ge12$. \end{theorem} \begin{proof} Here $s=5$ and $(n-3)/2=5$; by Lemma~\ref{lem:red}, $w\in\{2,4,6,8,10,12\}$. \begin{itemize} \item $w=2$, $u=11$: Corollary~\ref{cor:ubound} gives $t\ge16$. \item $w=4$, $u=9$: Corollary~\ref{cor:ubound} gives $t\ge14$. \item $w=10$, $u=3$, $h=5$: cases (i) and (ii) of Proposition~\ref{prop:tau} are excluded, so $\tau\ge5+3=8$ and $t=9+\tau\ge17$. \item $w=12$, $u=1$: Corollary~\ref{cor:u1} gives $t\ge3\cdot5=15$. \item $w=8$, $u=5$, $h=4$: $\tau\in\{0,4\}\cup[6,\infty)$, so $t=7+\tau\in\{7,11\}\cup[13,\infty)$; and $t\ge8$ by Corollary~\ref{cor:ubound}, so $t\ge11$. \item $w=6$, $u=7$, $h=3$: as at $n=11$, $\tau\ge5$ and $t=5+\tau\ge10$. Suppose $t=10$. Corollary~\ref{cor:ubound} forces $\max(m_1+6,\,14-m_1)\le10$, so $m_1=4$ and the profile is $(2,2,2,1,1,1,1)$. Applying Proposition~\ref{prop:evenrow} at an index of valuation $2$ gives a row sum at least $3+3+3+1+1+1+1=13>1+t=11$, a contradiction. Hence $t\ge11$. \end{itemize} \end{proof} \begin{theorem}\label{thm:n15} Let $A$ be a symmetric ideal solution over $\Zt$ of size $15$. Then $t\ge11$, hence $\vv(C)\ge12$. \end{theorem} \begin{proof} Here $s=5$ and $(n-3)/2=6$; by Lemma~\ref{lem:red}, $w$ is even with $2\le w\le14$. Corollary~\ref{cor:ubound} gives $t\ge20$, $16$ and $14$ for $u=13$, $11$ and $9$, that is for $w=2$, $4$ and $6$. For $w=10$ one has $u=5$ and $h=5$, and cases (i) and (ii) of Proposition~\ref{prop:tau} are excluded, so $\tau\ge5+3=8$ and $t=9+\tau\ge17$. For $w=14$ one has $u=1$, and Corollary~\ref{cor:m1} gives $t\ge3\cdot6=18$. For $w=8$ one has $u=7$ and $h=4$, so $\tau\in\{0,4\}\cup[6,\infty)$ and $t=7+\tau\in\{7,11\}\cup[13,\infty)$, while Corollary~\ref{cor:ubound} gives $t\ge10$; hence $t\ge11$. Finally for $w=12$ one has $u=3$ and $h=6$, so $\tau\in\{0,6\}\cup[9,\infty)$ and $t=11+\tau\ge11$. \end{proof} \begin{corollary}\label{cor:integers} Let $(A,-A)$ be a symmetric ideal Prouhet--Tarry--Escott solution with entries in $\Z$ and constant $C$. Then $2^{9}\mid C$ if $A$ has size $9$, $2^{11}\mid C$ if it has size $11$, and $2^{12}\mid C$ if it has size $13$ or $15$. \end{corollary} \begin{proof} Such a solution is a symmetric ideal solution over $\Zt$ in the sense of Definition~\ref{def:sol}; apply Theorems~\ref{thm:n9}, \ref{thm:n11}, \ref{thm:n13} and \ref{thm:n15}. \end{proof} \section{Sharpness at $n=11$}\label{sec:sharp} Coppersmith, Mossinghoff, Scheinerman and VanderKam exhibit two multisets of size $11$, each consisting of nine rational integers together with a conjugate pair of quadratic irrationalities, whose odd power sums $p_1,p_3,p_5,p_7,p_9$ all vanish: \[ A_1=\{-95,\ -68-\alpha,\ -52,\ -48,\ -13,\ -9,\ 30,\ 34,\ 61,\ 65,\ 95+\alpha\}, \] \[ A_2=\{-589,\ -546+\beta,\ -363,\ -185,\ -170,\ 41,\ 87-\beta,\ 234,\ 355,\ 548,\ 588\}, \] where $\alpha^2+163\alpha-8=0$ and $\beta^2-633\beta+30=0$. \begin{theorem}\label{thm:witness} Each root $\alpha$ of $X^2+163X-8$ and each root $\beta$ of $X^2-633X+30$ lies in $\Zt$, and the corresponding tuples $A_1$, $A_2$ are symmetric ideal solutions over $\Zt$ of size $11$ with $\vv(C)=11$. Consequently the bound of Theorem~\ref{thm:n11} is attained and \[ \min\{\vv(C)\}=11 . \] \end{theorem} \begin{proof} The discriminants are $163^2+32=26601$ and $633^2-120=400569$, both congruent to $1$ modulo $8$, hence squares in $\Zt$; their square roots are odd, so $-163\pm\sqrt{26601}$ and $633\pm\sqrt{400569}$ are even and the four roots lie in $\Zt$. The identities $p_1=p_3=p_5=p_7=p_9=0$ hold identically in $\Z[\alpha]$ and in $\Z[\beta]$, as a direct expansion confirms, hence in $\Zt$. Finally $C=-2\prod a_i$ is a rational integer in both cases, of $2$-adic valuation $11$. \end{proof} \begin{remark} The structure of these two witnesses matches the case analysis exactly: in all four embeddings one finds $w=6$, $u=5$, $\tau=5$ and even valuations $(4,2,2,1,1)$, the six odd row sums of Proposition~\ref{prop:oddrow} all equal $10$ and the five even row sums of Proposition~\ref{prop:evenrow} all equal $11$. \end{remark} \begin{remark} Since $2$ splits in both fields, the same argument applies to every prime $p$ that splits in $\Q(\sqrt{26601})$ or in $\Q(\sqrt{400569})$: for such $p$ there exists a $p$-adic ideal solution of size $11$, necessarily not rational. Among the $30$ primes below $120$, exactly $21$ are of this kind, namely \[ 2,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,31,\,37,\,41,\,43,\,59,\,61,\,71,\,73,\,83,\,89,\, 103,\,109,\,113 , \] the remaining nine being $3,29,47,53,67,79,97,101,107$. Filaseta and Markovich ask whether a $2$-adic ideal solution exists for every $n\ge3$, and whether a $p$-adic ideal solution necessarily exists at $n=11$; both questions are repeated by Foster \cite{Fo}. The first is vacuous for $n\le10$ and $n=12$, where ideal solutions exist over $\Z$, so its content lies at $n=11$ and at $n\ge13$; Theorem~\ref{thm:witness} settles $n=11$ and Proposition~\ref{prop:n13sol} settles $n=13$, the first two open cases, and in both the solution is irrational. For the second question the present section gives an affirmative answer for every $p$ that splits in one of the two fields, and says nothing for the nine primes below $120$ that are inert or ramified in both. \end{remark} \begin{remark} It may be worth recording that $C(A_1)$ satisfies every divisibility known to be required of the constant of a symmetric ideal solution of size $11$ over $\Z$ except one: its valuations at $2,3,5,7,11,13,17,19$ are $11,5,3,2,1,3,1,1$, all at or above the required values $8,4,3,2,1,1,1,1$, while $31\nmid C(A_1)$. The exception is accounted for by the fact that $31$ is inert in $\Q(\sqrt{26601})$, since $26601\equiv3\pmod{31}$ is not a square, so that the local argument producing the factor $31$ does not apply. In $\Q(\sqrt{400569})$ the prime $31$ splits, and indeed $31\mid C(A_2)$. \end{remark} \section{A $2$-adic solution at $n=13$}\label{sec:n13} No ideal solution of size $13$ is known over $\Z$. Over $\Zt$ one exists, and it too is obtained from an object already in the literature: among the configurations reported by Coppersmith, Mossinghoff, Scheinerman and VanderKam at size $13$, consisting of eight rational integers together with the roots of a totally real quintic, the second one splits $2$-adically. \begin{proposition}\label{prop:n13sol} Let $f(x)=x^5-4278125x^3+3664306448884x-975527373189600$. Then $f$ splits completely over $\Q_2$, and the multiset \[ \{-1827,\ -989,\ -715,\ -405,\ 64,\ 902,\ 1330,\ 1640\}\ \cup\ \{\alpha:f(\alpha)=0\} \] is a symmetric ideal solution over $\Zt$ of size $13$, with $\vv(C)=17$. \end{proposition} \begin{proof} The six identities $p_1=p_3=\dots=p_{11}=0$ hold exactly: the power sums of the roots of $f$ are computed from its coefficients by Newton's identities, and added to those of the eight integers they vanish. The Newton polygon of $f$ at $2$ has vertices $(0,5)$, $(1,2)$, $(3,0)$, $(5,0)$, hence segments of slopes $-3$, $-1$, $0$ and lengths $1$, $2$, $2$; a $2$-adic lifting shows that each segment splits, so $f$ has five roots in $\Zt$, of valuations $3,1,1,0,0$. The eight rational integers have valuations $0,0,0,0,1,1,3,6$. Hence $t=16$ and $\vv(C)=1+t=17$ by Proposition~\ref{prop:vC}. \end{proof} \begin{remark}[certificate] The splitting of $f$ over $\Q_2$ is checked as follows. Each of the five residues $r$ below satisfies $\vv(f(r))>2\,\vv(f'(r))$, so by Hensel's lemma there is a unique $\alpha\in\Zt$ with $f(\alpha)=0$ and $\vv(\alpha-r)\ge \vv(f(r))-\vv(f'(r))\ge32$; the five residues being pairwise distinct modulo $2^{32}$, the five roots are distinct, and $\vv(\alpha)=\vv(r)$. \[ \begin{array}{r|ccc} r \bmod 2^{32} & \vv(f(r)) & \vv(f'(r)) & \vv(r)\\\hline 2723974027 & 33 & 1 & 0\\ 1263895509 & 34 & 1 & 0\\ 3276945658 & 35 & 3 & 1\\ 3386758734 & 36 & 3 & 1\\ 2233327960 & 34 & 2 & 3 \end{array} \] \end{remark} \begin{corollary} At $n=13$ one has $12\le\min\{\vv(C)\}\le17$. In particular a symmetric ideal solution over $\Zt$ of size $13$ exists, no ideal solution of that size being known over $\Z$. \end{corollary} \begin{remark} The structure of this witness is $w=6$, $u=7$, $\tau=11$ and even valuations $(6,3,3,1,1,1,1)$; the six odd row sums of Proposition~\ref{prop:oddrow} all equal $16$ and the seven even row sums of Proposition~\ref{prop:evenrow} all equal $17$. Whether the lower bound $12$ of Theorem~\ref{thm:n13} is attained remains open. \end{remark} \begin{remark} The first configuration reported at size $13$ in the same place consists of the eight integers $-1026,-532,-248,-245,207,286,533,1025$ together with the roots of a quintic printed, in the arXiv version of~\cite{CMSV}, as $x^5-1215245x^3+170781322924x+931022099280$. With that constant term the required identities fail: one finds $p_5=-10\cdot931022099280\neq0$. With the constant term $-931022099280$ all six identities $p_1=p_3=\dots=p_{11}=0$ hold exactly, so the sign appears to be a misprint. With that sign, however, its Newton polygon at $2$ has segments of slopes $-2,-1,0$ and lengths $1,2,2$, and only three of the five roots lie in $\Q_2$; the configuration therefore lives in a quadratic extension and is not a symmetric ideal solution over $\Zt$. For the second configuration, the arXiv version of \cite{CMSV} prints the coefficient of $x$ as $366430644884$; with that value $p_9$ and $p_{11}$ do not vanish, whereas the value $3664306448884$ used in Proposition~\ref{prop:n13sol} makes all six identities hold exactly. \end{remark} \section{Summary, scope and credits}\label{sec:summary} All entries in the second and third columns are bounds on $\vv_2(C)$; at $n=5,7,9$ the value in the third column is the known exact value of $\vv_2(\overline{C_n})$. The value at $n=5$ is Proposition~2.4 of \cite{RS}, the divisibility $2^6\mid\overline{C_7}$ is due to Caley \cite{Ca}, and the value at $n=9$ is due to Filaseta and Markovich \cite{FM}. \begin{center} \small \begin{tabular}{c|c|c|l} $n$ & here & previously & attained by\\\hline $5$ & $\ge4$ & $=4$ & $(9,5,1,-7,-8)$\\ $6$ & $\ge6$ & $\ge3$ (Kleiman) & $\pm\{0,7,7\}$ against $\pm\{3,5,8\}$\\ $7$ & $\ge6$ & $=6$ (Caley) & $(-50,-38,-13,-7,24,33,51)$\\ $8$ & $\ge8$ & $\ge6$ (Filaseta--Markovich) & $\pm\{5,14,23,24\}$ against $\pm\{2,16,21,25\}$\\ $9$ & $\ge9$ & $=9$ & Letac\\ $10$ & $\ge11$ & $9$ to $11$ (Bera--Nair) & Gloden; Borwein--Lison\v{e}k--Percival\\ $11$ & $\ge11$ & $\ge8$ (Kleiman) & $A_1$, $A_2$ over $\Zt$\\ $12$ & $\ge12$ & $11$ to $12$ (Bera--Nair) & Kuosa--Meyrignac--Shuwen\\ $13$ & $\ge12$ & $\ge10$ (Kleiman) & --- (but $17$ occurs)\\ $14$ & $\ge12$ & $\ge10$ (Kleiman) & ---\\ $15$ & $\ge12$ & $\ge11$ (Kleiman) & ---\\ $16$ & $\ge16$ & $\ge11$ (Kleiman) & ---\\ $18$ & $\ge18$ & $\ge15$ (Kleiman) & ---\\ $20$ & $\ge18$ & $\ge16$ (Kleiman) & ---\\ \end{tabular} \end{center} \textbf{Scope.} All the statements above concern \emph{symmetric} solutions, hence $C'_n$ and not $\overline{C_n}$; for $n$ even they concern solutions with entries in $\Z$, for $n$ odd solutions over $\Zt$, for the reason given in the introduction. At $n=5,7,9$ the two agree, the minimum being already attained by a symmetric solution over $\Z$. At $n=10$ and $n=12$ the equalities hold for $C'_n$; a smaller value for $\overline{C_n}$ would require a non-symmetric ideal solution of that size, and whether one exists is open. The restriction to symmetry about $0$ is not vacuous, and at these two sizes one can say exactly how much is lost. Over $\Zt$ there exist ideal solutions of size $10$ and of size $12$, symmetric about $-1/2$ rather than about $0$, whose constants have valuation exactly $9$ and exactly $11$; they are exhibited with Hensel certificates in the third ancillary file. The lower bounds of Bera and Nair at those two sizes are therefore attained $2$-adically, so no argument using $2$-adic information alone can raise them, and deciding the exact values of $\vv_2(\overline{C_{10}})$ and $\vv_2(\overline{C_{12}})$ requires global input. The equalities proved here for $C'_{10}$ and $C'_{12}$ and this limitation on $\overline{C_{10}}$ and $\overline{C_{12}}$ are consistent: they bear on different classes of solutions. The lower bounds at $n=13,\dots,20$ are not asserted to be optimal. For $n$ odd the same case analysis applies at every size, but beyond $n=15$ it is weaker than Kleiman's bound: at $n=17$, $19$ and $21$ it gives $\vv(C)\ge12$, $12$ and $16$, against $\vv_2((n-1)!)=15$, $16$ and $18$. \textbf{Verification.} Three self-contained verifiers are attached, written in Python with no dependencies. The even-size verifier replays sixty-seven checks: ideality and valuations of the six even solutions of sizes $8$, $10$ and $12$, regularity of the two systems, validity and tightness of the index bound, exactness of the predictor on the twelve half-solutions, slopes of the Newton polygons, the minimum of the bound for each of the eight sizes $6\le n\le20$ in both cases of the dichotomy (in the case $\sigma\ne\sigma'$ with all entries of the larger profile up to $s'+r+3$, as allowed by Lemma~\ref{lem:sat}), the counts and the explicit lists of Theorem~\ref{thm:class} together with the one-parameter shape of every form, and structural statements on the odd witnesses; it runs in about ten minutes. The odd-size verifier replays twenty-five checks: the witnesses of Sections~\ref{sec:sharp} and~\ref{sec:n13}, the Hensel certificate, the congruences of Proposition~\ref{prop:kill} for both profiles, exhaustively, and the chain of Section~\ref{sec:lower} at every odd size up to $21$. The third file replays twenty-eight checks on the two $2$-adic witnesses of the scope paragraph above: Hensel certificates for every root of the four polynomials involved, a calibration at size $8$ against the bound of Filaseta and Markovich, and the position of the size-$10$ witness within the case analysis of Bera and Nair. \textbf{Credits.} The reference lower bound is due to Kleiman (1975). The use of Newton polygons on this constant is due to Filaseta and Markovich (2017), who also raised the question of the sizes $10$ to $12$; the germ of Lemma~\ref{lem:fd} is their congruence modulo $16$ between power sums of odd and even parts, and the parameters $w$ and $m_1$ used here are their $k_1$ and $k_2$. The published lower bounds at sizes $10$ and $12$, which the present work closes, are due to Bera and Nair (2024). The construction of $2$-adic ideal solutions by successive lifting, which is the counterpart of the obstruction developed above, is due to Foster, in his 2021 thesis, who carries it out for $3\le n\le8$. The explicit ideal solutions used as witnesses are catalogued by Coppersmith, Mossinghoff, Scheinerman and VanderKam, and are due to Letac (1942) at size $9$, to Smyth (1991) \cite{Sm} for the infinite family at size $10$ and to Borwein, Lison\v{e}k and Percival (2003) for the two smaller solutions of that size, and at size $12$ to Kuosa, Meyrignac and Shuwen (1999), to Broadhurst (2007) and to Choudhry and Wr\'oblewski (2008). The two quadratic objects of Section~\ref{sec:sharp} and the quintic configuration of Section~\ref{sec:n13} are those of Coppersmith, Mossinghoff, Scheinerman and VanderKam; what is added here is the observation that they are $2$-adic. Technical assistance: Claude (Anthropic). \begin{remark}[on Bera--Nair] The paper of Bera and Nair treats the sizes $n=10$ and $n=12$ only, as their abstract states. Their results at those two sizes are the brackets recorded in Theorem~\ref{thm:even}, which the present work closes in the symmetric case; they give nothing at $n=11$ or $n=13$, so the statements of Sections~\ref{sec:lower}--\ref{sec:n13} do not overlap with theirs. \end{remark} \begin{thebibliography}{99} \bibitem{BN} R. Bera and S. G. Nair, \emph{Newton polygons and the constant associated with the Prouhet--Tarry--Escott problem}, Acta Arith. \textbf{213} (2024), 75--95. \bibitem{BLP} P. Borwein, P. Lison\v{e}k and C. Percival, \emph{Computational investigations of the Prouhet--Tarry--Escott problem}, Math. Comp. \textbf{72} (2003), 2063--2070. \bibitem{Br} D. Broadhurst, \emph{A new Prouhet--Tarry--Escott solution}, NMBRTHRY mailing list, 26 September 2007. \bibitem{Ca} T. Caley, \emph{The Prouhet--Tarry--Escott problem}, Ph.D. thesis, University of Waterloo, 2012. \bibitem{Ca2} T. Caley, \emph{The Prouhet--Tarry--Escott problem for Gaussian integers}, Math. Comp. \textbf{82} (2013), 1121--1137. \bibitem{CW} A. Choudhry and J. Wr\'oblewski, \emph{Ideal solutions of the Tarry--Escott problem of degree eleven with applications to sums of thirteenth powers}, Hardy--Ramanujan J. \textbf{31} (2008), paper no.~1. \bibitem{CMSV} D. Coppersmith, M. J. Mossinghoff, D. Scheinerman and J. M. VanderKam, \emph{Ideal solutions in the Prouhet--Tarry--Escott problem}, preprint, arXiv:2304.11254. \bibitem{FM} M. Filaseta and M. Markovich, \emph{Newton polygons and the Prouhet--Tarry--Escott problem}, J. Number Theory \textbf{174} (2017), 384--400. \bibitem{Fo} J. C. Foster, \emph{Polynomials, primes and the PTE problem}, Ph.D. thesis, University of South Carolina, 2021. \bibitem{Gl} A. Gloden, \emph{Mehrgradige Gleichungen}, 2nd ed., Noordhoff, Groningen, 1944. \bibitem{Kl} H. Kleiman, \emph{A note on the Tarry--Escott problem}, J. Reine Angew. Math. \textbf{278/279} (1975), 48--51. \bibitem{KMS} P. Kuosa, J.-C. Meyrignac and C. Shuwen, ideal solution of size $12$ announced in 1999 on J.-C. Meyrignac's website on equal sums of like powers; see \cite{CMSV}. \bibitem{Le} A. Letac, Gazeta Matematic\u{a} \textbf{48} (1942), 68--69. \bibitem{Ma} M. E. Markovich, \emph{On a constant associated with the Prouhet--Tarry--Escott problem}, Master's thesis, University of South Carolina, 2016. \bibitem{RS} E. Rees and C. J. Smyth, \emph{On the constant in the Tarry--Escott problem}, Lecture Notes in Math. \textbf{1415}, Springer, 1990, 196--208. \bibitem{Sm} C. J. Smyth, \emph{Ideal 9th-order multigrades and Letac's elliptic curve}, Math. Comp. \textbf{57} (1991), no.~196, 817--823. \end{thebibliography} \end{document}