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\title{Two improved lower bounds for the $2$-adic valuation\\ of the constant of the
Prouhet--Tarry--Escott problem}
\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 $n$ the associated constant
$C$ is divisible by a large power of $2$. Restricting to symmetric solutions, we prove two new
lower bounds for $v_2(C)$: $v_2(C)\ge14$ at $n=15$, improving $12$, and $v_2(C)\ge19$ at $n=20$,
improving $18$; both earlier bounds are those of the companion article hexagon:2611.00057. The first rests on two identities satisfied by every symmetric solution of odd
size, one relating $v_2(C)$ to a difference of two products and one turning the index relation
into an equality, together with an arithmetic criterion for a block of entries of equal
valuation. The second rests on an identity internal to each of the two sides, from which we
deduce that the vanishing of the cross invariant $R$ forces $k_0\equiv0\pmod4$ when $k_0\ge4$.
Two self-contained verifiers accompany the note.
\end{abstract}

\section{Setting and statements}

Let $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_n\}$ be multisets forming an ideal solution of the
Prouhet--Tarry--Escott problem, that is $\sum_ia_i^k=\sum_ib_i^k$ for $1\le k\le n-1$ while the two
multisets differ. The associated constant is
\[
  C=\prod_i a_i-\prod_i b_i .
\]
A solution is \emph{symmetric} when $B=-A$ (after a translation) for $n$ odd, and when $A=\{\pm a_1,\dots,\pm a_m\}$,
$B=\{\pm b_1,\dots,\pm b_m\}$ with $n=2m$ for $n$ even. As in \cite{V7}, solutions of odd size are
taken with entries in $\Zt$ and solutions of even size with entries in $\Z$. Throughout, $v_2$
denotes the $2$-adic valuation and all statements are about symmetric solutions.

Kleiman's bound gives $v_2(C)\ge v_2((n-1)!)$ \cite{Kl}. In \cite{V7} the symmetric case is
treated at every size up to $20$: the values $4,6,9,11$ at $n=5,7,9,11$ are attained (at $n=11$
over $\Zt$), the bound $v_2(C)\ge12$ is proved at $n=13$ and at $n=15$, and the bounds
$8,11,12,12,16,18,18$ are proved at $n=8,10,12,14,16,18,20$. The present note improves two of
these.

\begin{theorem}\label{thm:15}
Every symmetric ideal solution of size $15$ over $\Zt$ satisfies $v_2(C)\ge14$.
\end{theorem}

\begin{theorem}\label{thm:20}
Every symmetric ideal solution of size $20$ with entries in $\Z$ satisfies $v_2(C)\ge19$.
\end{theorem}

Sections~\ref{sec:odd} and~\ref{sec:even} develop the tools and prove these. Section~\ref{sec:status}
records what is proved, what is verified by computation, and what remains open.

\section{Odd sizes}\label{sec:odd}

Let $n$ be odd and $B=-A$. The ideal conditions read $p_k(A)=0$ for every odd $k\le n-2$, where
$p_k$ is the $k$-th power sum. Equivalently, by Newton's identities, $e_j(A)=0$ for every odd
$j\le n-2$; the only surviving odd index is $j=n$. Writing $m=(n-1)/2$ we therefore obtain the
following normal form.

\begin{lemma}\label{lem:normal}
Let $F(x)=\prod_{i=1}^n(x-a_i)$. Then there are a monic $H\in\Zt[y]$ of degree $m$ and $c\in\Zt$
with
\[
  F(x)=x\,H(x^2)-c,\qquad c=\prod_ia_i,
\]
and $C=2c$, so that $v_2(C)=1+\sum_iv_2(a_i)$.
\end{lemma}

We write $t=\sum_iv_2(a_i)$, so $v_2(C)=1+t$, and call the multiset of the nonzero $v_2(a_i)$
the \emph{profile}; $w$ denotes the number of odd entries, so that $n=w+u$ with $u$ the number of
even entries. Reducing $p_1(A)=0$ modulo $2$ shows that $w$ is even.

\subsection{Two identities}

\begin{lemma}[two products]\label{lem:twoprod}
With the notation of Lemma~\ref{lem:normal},
\[
  \prod_i(1+a_i)=H(1)+c,\qquad \prod_i(1-a_i)=H(1)-c,
\]
hence $2c=\prod_i(1+a_i)-\prod_i(1-a_i)$ and
$v_2(C)=v_2\bigl(\prod_i(1+a_i)-\prod_i(1-a_i)\bigr)$.
\end{lemma}

\begin{proof}
$F(1)=\prod_i(1-a_i)$ and $F(1)=H(1)-c$ give the second identity. Since $n$ is odd,
$F(-1)=\prod_i(-1-a_i)=-\prod_i(1+a_i)$, while $F(-1)=-H(1)-c$; this gives the first.
Subtracting yields $2c$, and $C=2c$.
\end{proof}

\begin{lemma}[index identity]\label{lem:index}
For every index $j$,
\[
  \prod_i(a_j+a_i)=2c,\qquad\text{hence}\qquad\sum_iv_2(a_j+a_i)=v_2(C).
\]
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:normal}, $F(x)+F(-x)=-2c$ identically. Taking $x=a_j$ and using $F(a_j)=0$
gives $F(-a_j)=-2c$; and $F(-a_j)=\prod_i(-a_j-a_i)=-\prod_i(a_j+a_i)$.
\end{proof}

Lemma~\ref{lem:index} is an equality for every single index, not merely an inequality for the
best index. Written out, it determines the internal structure of each block of entries of equal
valuation. Let $f$ be a valuation occurring with multiplicity $m_f$ in the profile and write
those entries as $2^fb_i$ with $b_i$ odd. Separating the contributions in Lemma~\ref{lem:index}
according to the valuation of the partner gives, for each $i$ in the block,
\begin{equation}\label{eq:block}
  \sum_{\substack{k\ne i\\ v_2(a_k)=f}}v_2(b_i+b_k)=R_f,\qquad
  R_f=v_2(C)-(1+f)-\sum_{v\ne f}m_v\min(f,v)-(m_f-1)f,
\end{equation}
a quantity depending only on $v_2(C)$ and the profile (here $m_0=w$). The same computation
applied to an odd entry gives $\sum_{k\ne j}v_2(a_j+a_k)=t$, the sum running over the $w$ odd
entries.

\subsection{An arithmetic criterion for a block}

\begin{lemma}\label{lem:mod4}
For $b,b'$ odd, $\min\bigl(v_2(b-b'),v_2(b+b')\bigr)=1$, and $v_2(b+b')=1$ if and
only if $b\equiv b'\pmod4$.
\end{lemma}

\begin{proof}
$(b+b')+(b-b')=2b$ has valuation $1$, so the minimum of the two valuations is $1$. Both $b\pm b'$
are even; $b+b'\equiv0\pmod4$ holds exactly when $b'\equiv-b\pmod4$, that is exactly when $b$
and $b'$ lie in different classes modulo $4$.
\end{proof}

\begin{proposition}[block criterion]\label{prop:block}
Let $b_1,\dots,b_m$ be odd with $\sum_{k\ne i}v_2(b_i+b_k)=R$ for every $i$. Then
\[
  R=m-1,\qquad\text{or}\qquad m\text{ is even and }R\ge\tfrac{3m}{2}-1 .
\]
In particular $R=m$ is impossible as soon as $m\ge3$, and for $m$ odd the value $R=m-1$ is
forced.
\end{proposition}

\begin{proof}
By Lemma~\ref{lem:mod4} the indices split into $P=\{i:b_i\equiv1\ (4)\}$ and
$Q=\{i:b_i\equiv3\ (4)\}$, with $v_2(b_i+b_k)=1$ inside a part and $\ge2$ across. Put $p=|P|$,
$q=|Q|=m-p$. For $i\in P$ the row sum is $(p-1)+\sum_{k\in Q}v_2(b_i+b_k)=R$, so the total of
the cross entries, counted from the $P$ side, equals $p(R-p+1)$; counted from the $Q$ side it
equals $q(R-q+1)$. Equating and writing $s=R+1$ gives $s(p-q)=(p-q)(p+q)$, hence $p=q$ or $s=m$,
that is $R=m-1$. If $p=q=m/2$ then each row carries $m/2-1$ entries equal to $1$ and $m/2$
entries at least $2$, whence $R\ge3m/2-1$. If one part is empty all entries equal $1$ and
$R=m-1$.
\end{proof}

\subsection{Finite differences}

Let $E_j=\sum_ic_i^{2j+1}$, the sum running over the even entries $c_i$, and let $\Delta$ be the
forward difference in $j$. Since $\Delta^m$ applied to $j\mapsto x^j$ equals $(x-1)^m$ at $j=0$,
\begin{equation}\label{eq:delta}
  \Delta^mE=\sum_ic_i\,(c_i^2-1)^m .
\end{equation}
On the other hand, for an odd entry $a$ write $a^2=1+8k_a$ with $k_a\in\Zt$; then
$a^{2j+1}=a(1+8k_a)^j$, and $\Delta^m$ of that sequence equals $a(8k_a)^m$ at $j=0$. Since
$E_j=-\sum_{a\text{ odd}}a^{2j+1}$ for $0\le j\le M$ with $M=(n-3)/2$, we obtain
\begin{equation}\label{eq:div}
  \Delta^mE=-8^m\sum_{a\text{ odd}}a\,k_a^m,\qquad\text{in particular}\qquad
  2^{3m}\mid\Delta^mE\quad(0\le m\le M).
\end{equation}
This is the finite-difference lemma of \cite{V7}.

\begin{lemma}\label{lem:scale}
Let $c_1,\dots,c_u$ be even elements of $\Zt$ satisfying the conditions $2^{3m}\mid\Delta^mE$
for $0\le m\le M$, and let $\lambda$ be a unit of $\Zt$. Then $\lambda c_1,\dots,\lambda c_u$
satisfy the same conditions. Consequently, when these conditions are imposed, the odd part of
one even entry may be normalised to $1$.
\end{lemma}

\begin{proof}
Write $\lambda^2=1+8\kappa$. For each $c$,
\[
  (\lambda^2c^2-1)^m=\bigl((c^2-1)+8\kappa c^2\bigr)^m=\sum_{k=0}^m\binom mk(8\kappa)^kc^{2k}(c^2-1)^{m-k},
\]
and $c^{2k}=\bigl(1+(c^2-1)\bigr)^k=\sum_{l=0}^k\binom kl(c^2-1)^l$. Summing over the even
entries,
\[
  \sum_i\lambda c_i(\lambda^2c_i^2-1)^m=\lambda\sum_{k=0}^m\sum_{l=0}^k\binom mk\binom kl
  (8\kappa)^k\,\Delta^{m-k+l}E .
\]
Since $m-k+l\le m\le M$, each $\Delta^{m-k+l}E$ is divisible by $2^{3(m-k+l)}\ge2^{3(m-k)}$, so
each term is divisible by $2^{3k}\cdot2^{3(m-k)}=2^{3m}$. For the normalisation take
$\lambda=b^{-1}$, $b$ the odd part of the chosen entry; the block conditions \eqref{eq:block}
involve only valuations of sums and are unchanged by the scaling.
\end{proof}

\subsection{Proof of Theorem~\ref{thm:15}}

The admissible pairs $(\text{profile},w)$ at a given $n$ and $t$ are those passing the necessary
conditions of \cite[Sections 6--9]{V7}. At $n=15$ they leave none at $t\le10$, none at $t=12$,
and four candidates at $t=11$. We eliminate the four.

The candidate $w=8$, profile $(3,2,2,1,1,1,1)$. Here $v_2(C)=12$ and the block of valuation $1$
has multiplicity $m_1=4$; formula~\eqref{eq:block} gives $R_1=12-2-3-3=4$. By
Proposition~\ref{prop:block} the value $R=m$ is impossible for $m\ge3$, so this candidate does
not occur.

The candidates $w=12$, profiles $(9,1,1)$, $(7,2,2)$, $(5,3,3)$. Each has three even entries,
and \eqref{eq:block} pins the single value $v_2(b_1+b_2)$ of the repeated block to $8$, $5$ and
$2$ respectively. The odd parts of the three even entries are then lifted one $2$-adic digit at
a time; by Lemma~\ref{lem:scale} one of them is fixed to $1$. At level $L$ the odd parts are known modulo $2^L$, and a residue class is
discarded only when the constraint of \eqref{eq:block} or one of the divisibilities
\eqref{eq:div}, evaluated to the precision then available, already fails for every lift of it.
In each of the three cases the set of surviving residue classes becomes empty, at levels $11$,
$10$ and $9$ respectively, the largest intermediate set having $512$, $256$ and $128$ elements.
Hence none of the three occurs.

Consequently $t\ge13$, and $v_2(C)=1+t\ge14$. \qed

\begin{remark}
At $n=13$ the same analysis, applied to the four candidates at $t=11$ left by the conditions used
in the lower-bound proofs of \cite{V7}, removes two of them, namely $w=6$ with
profile $(3,2,2,1,1,1,1)$, by Proposition~\ref{prop:block}, and $w=8$ with profile $(4,3,2,1,1)$,
because \eqref{eq:block} returns $R_3=1$ for a block of multiplicity $1$, where the left-hand side
is an empty sum (it is also excluded by \cite[Theorem~8.3]{V7}). Two candidates remain, $w=8$ with profiles $(7,1,1,1,1)$ and $(5,2,2,1,1)$, and
the bound at $n=13$ stays at $12$.
\end{remark}

\section{Even sizes}\label{sec:even}

Let $n=2m$ and put $u_i=a_i^2$, $v_j=b_j^2$. The odd power sums vanish by symmetry, and the even
ones agreeing up to degree $2(m-1)$ say exactly that
\[
  P(y)=\prod_{i=1}^m(y-u_i)\qquad\text{and}\qquad Q(y)=\prod_{j=1}^m(y-v_j)
\]
share every coefficient but the constant one. Write $Q=P+d$; then $d=\prod_j(u_i-v_j)$ for every
$i$, $d=\pm C$ and $v_2(d)=v_2(C)$. Evaluating $Q$ at $u_i$ recovers the index identity in this
setting,
\begin{equation}\label{eq:index-even}
  \sum_jv_2(u_i-v_j)=v_2(C)\ \text{for each }i,\qquad
  \sum_iv_2(u_i-v_j)=v_2(C)\ \text{for each }j.
\end{equation}
Let $k_0$ be the number of odd entries on each side (it is the same on both sides
\cite[Prop.~3.2]{V7}); for odd entries write $a^2=1+8\xi$ and $b^2=1+8\eta$. Separating
\eqref{eq:index-even} at an odd index gives $v_2(C)=3k_0+R$ with
\begin{equation}\label{eq:R}
  R=\sum_{j\text{ odd}}v_2(\xi_i-\eta_j)\quad\text{independent of }i,
\end{equation}
and symmetrically along columns.

\subsection{An identity internal to each side}

\begin{lemma}\label{lem:deriv}
$P'=Q'$, and therefore for each $i$
\[
  \prod_{k\ne i}(u_i-u_k)=\sum_j\prod_{l\ne j}(u_i-v_l)=d\sum_j(u_i-v_j)^{-1},
\]
with the symmetric statement along the other side.
\end{lemma}

\begin{proof}
$P$ and $Q$ differ by a constant, so their derivatives agree. The left member is $P'(u_i)$; the
right one is $Q'(u_i)$, and $\prod_{l\ne j}(u_i-v_l)=d/(u_i-v_j)$ because
$\prod_l(u_i-v_l)=Q(u_i)=d$.
\end{proof}

\begin{corollary}\label{cor:internal}
For each $i$,
\[
  \sum_{k\ne i}v_2(a_i^2-a_k^2)\ \ge\ v_2(C)-\max_jv_2(a_i^2-b_j^2),
\]
with equality whenever that maximum is attained at a single $j$.
\end{corollary}

\begin{proof}
By Lemma~\ref{lem:deriv}, the left-hand side equals $v_2(C)+v_2\bigl(\sum_j(u_i-v_j)^{-1}\bigr)$.
Each term of the sum has valuation $-v_2(u_i-v_j)\ge-\max_jv_2(u_i-v_j)$, with equality for a
single term when the maximum is attained once; the ultrametric inequality gives the claim.
\end{proof}

\subsection{The vanishing of $R$}

\begin{theorem}\label{thm:R}
Let a symmetric ideal solution of even size satisfy $R=0$, so that $v_2(C)=3k_0$, and assume
$k_0\ge4$. Then $k_0\equiv0\pmod4$.
\end{theorem}

\begin{proof}
By \eqref{eq:R} every $\xi_i-\eta_j$ is odd, so the family $(\xi_i)$ and the family $(\eta_j)$
have opposite parities; in particular $\xi_i-\xi_k$ is even for all $i\ne k$. Separating the
terms of $D_i=\sum_{k\ne i}v_2(u_i-u_k)$ at an odd index $i$, the even entries contribute $0$ and
each other odd entry contributes $3+v_2(\xi_i-\xi_k)\ge4$, so
\[
  D_i\ \ge\ 4(k_0-1)\ \ge\ 3k_0=v_2(C)\qquad(k_0\ge4).
\]
By Lemma~\ref{lem:deriv}, $D_i=v_2(C)+v_2\bigl(\sum_j(u_i-v_j)^{-1}\bigr)$, whence
$v_2\bigl(\sum_j(u_i-v_j)^{-1}\bigr)\ge0$. In that sum the $k_0$ odd indices $j$ contribute
$\bigl(8(\xi_i-\eta_j)\bigr)^{-1}$ and the remaining ones contribute units of $\Zt$; writing
$S_i=\sum_{j\text{ odd}}(\xi_i-\eta_j)^{-1}$, the sum equals $\frac18S_i$ plus an element of
$\Zt$, so $v_2(S_i)\ge3$. The same argument along columns gives $v_2(S'_j)\ge3$ with
$S'_j=\sum_{i\text{ odd}}(\eta_j-\xi_i)^{-1}$.

Every odd element of $\Zt$ is its own inverse modulo $8$, so
\[
  S_i\equiv k_0\xi_i-\sum_j\eta_j,\qquad S'_j\equiv k_0\eta_j-\sum_i\xi_i\pmod 8 .
\]
Suppose $k_0\equiv2\pmod4$ and, first, that the $\xi_i$ are even and the $\eta_j$ odd. Writing
$k_0=2\kappa$ with $\kappa$ odd and $\xi_i=2x_i$ we get $k_0\xi_i=4\kappa x_i\equiv4x_i\pmod8$, so
the row conditions force $\sum_j\eta_j\equiv0\pmod4$. The column conditions force all $k_0\eta_j$
equal modulo $8$; as $k_0=2\kappa$ with $\kappa$ odd, all $\eta_j$ are congruent modulo $4$, say
to $c\in\{1,3\}$, whence $\sum_j\eta_j\equiv k_0c\equiv2c\equiv2\pmod4$. This contradicts
$\sum_j\eta_j\equiv0\pmod4$. If instead the $\xi_i$ are odd and the $\eta_j$ even, the same
computation with the roles of the two families exchanged gives the same contradiction. Hence
$k_0\not\equiv2\pmod4$.

For $k_0$ odd the row conditions give $k_0\xi_i\equiv\sum_j\eta_j\pmod8$ for every $i$ and the
column conditions $k_0\eta_j\equiv\sum_i\xi_i\pmod8$ for every $j$; as $k_0$ is invertible modulo
$8$, all $\eta_j$ are congruent to one $\eta$ modulo $8$, so $\sum_j\eta_j\equiv k_0\eta$ and the
rows give $\xi_i\equiv\eta\pmod8$, contradicting the opposite parities. Hence
$k_0\equiv0\pmod4$.
\end{proof}

For $k_0$ odd the conclusion $R\ge1$ was already known on both branches of \cite{V7}, by
\cite[Lemma~3.4]{V7} when $\sigma=\sigma'$ and by parity when the sums differ; the new content of
Theorem~\ref{thm:R} is the case $k_0\equiv2\pmod4$.

\subsection{Proof of Theorem~\ref{thm:20}}

The lower bound of \cite[Theorem~5.1]{V7} at a given even $n$ is the smallest value compatible
with the necessary conditions listed there, minimised over the profiles with $\sigma=\sigma'$
and over the pairs with $\sigma\ne\sigma'$, the latter enumerated exhaustively thanks to the
saturation lemma \cite[Lemma~5.2]{V7}. Theorem~\ref{thm:R} adds one condition: $R\ge1$ whenever $k_0\ge4$
and $k_0\equiv2\pmod4$. Replaying the minimisation with this condition leaves $n=8,10,12,14,16,18$
unchanged at $8,11,12,12,16,18$ and raises $n=20$ from $18$ to $19$: at that size every
configuration compatible with the value $18$, on either branch, has $k_0=6$ and $R=0$, and is now
excluded (on the branch $\sigma=\sigma'$ these five profiles move to $21$, since $R=1$ and $R=2$ are not
realisable for $k_0=6$). The next compatible value on the branch $\sigma=\sigma'$ is $19$, reached
by the profile $(0,0,0,0,0,1,1,1,2,3)$ with $k_0=5$ and $R=4$, while the branch with distinct
profiles does not descend below $20$. \qed

\section{Tables}

\[
\begin{array}{c|ccc}
n & \text{Kleiman} & \text{\cite{V7}} & \text{this note}\\\hline
5 & 3 & 4 & 4\\
7 & 4 & 6 & 6\\
9 & 7 & 9 & 9\\
11 & 8 & 11 & 11\\
13 & 10 & 12 & 12\\
15 & 11 & 12 & \mathbf{14}
\end{array}
\qquad\qquad
\begin{array}{c|ccc}
n & \text{Kleiman} & \text{\cite{V7}} & \text{this note}\\\hline
8 & 4 & 8 & 8\\
10 & 7 & 11 & 11\\
12 & 8 & 12 & 12\\
14 & 10 & 12 & 12\\
16 & 11 & 16 & 16\\
18 & 15 & 18 & 18\\
20 & 16 & 18 & \mathbf{19}
\end{array}
\]

\section{Status and verification}\label{sec:status}

\textbf{Proved.} Lemmas~\ref{lem:normal}, \ref{lem:twoprod}, \ref{lem:index}, \ref{lem:mod4},
\ref{lem:scale} and \ref{lem:deriv}, formulas \eqref{eq:block}--\eqref{eq:R},
Proposition~\ref{prop:block}, Corollary~\ref{cor:internal} and Theorem~\ref{thm:R}.

\textbf{Verified.} Theorems~\ref{thm:15} and~\ref{thm:20} rest, besides the statements above, on
finite computations: an exhaustive digit-by-digit lifting for the three candidates at $n=15$,
and a replay of the bound minimisation of \cite{V7} for the even sizes. Both are carried out in
the ancillary files, which are pure Python~3, with no dependency, and take no argument.
The file \texttt{VERIF\_PTE2ADIC\_N15.py} replays the odd-size assertions and reports $36$ PASS
and $0$ FAIL in about ten seconds; the file \texttt{VERIF\_PTE2ADIC\_N20.py} replays the
even-size ones and reports $51$ PASS and $0$ FAIL in about twenty minutes, the case of distinct
profiles being enumerated exhaustively, with all entries of the larger profile up to $s'+m+3$.
The identities of Sections~\ref{sec:odd} and~\ref{sec:even} are checked on the known symmetric
odd solutions of sizes $5,7,9,11,13$ and on the six symmetric even solutions of sizes $8,10,12$;
the block criterion of Proposition~\ref{prop:block} is checked against exhaustive enumeration of
odd residues modulo a power of $2$ for $m=2,3,4,5$; the congruences in the proof of Theorem~\ref{thm:R} are checked by exhaustive
enumeration modulo $8$ for $k_0=2,\dots,6$; and the bound engine for the even sizes is checked to
return the values of \cite{V7} before the new condition is added.

\textbf{Open.} At $n=13$ two candidate profiles remain at $t=11$, namely $w=8$ with profiles
$(7,1,1,1,1)$ and $(5,2,2,1,1)$, so the bound there stays $v_2(C)\ge12$; the smallest valuation
realised by a known $2$-adic witness at that size is $17$ \cite{V7}. Whether $R=0$ can occur at
all for $k_0\equiv0\pmod4$ is not settled here.

\section*{Credits}

The germ of the finite-difference lemma for the odd sizes is a congruence modulo $16$ of
Filaseta and Markovich \cite{FM}. The construction of $2$-adic ideal solutions by successive lifting, for $3\le n\le8$, is due to
Joseph Foster, in his 2021 dissertation \cite{Fo}. The near-solutions of size $11$ used as $2$-adic
witnesses in \cite{V7} are those of Coppersmith, Mossinghoff, Scheinerman and VanderKam
\cite{CMSV}. The sizes $10$ and $12$ were treated by Bera and Nair \cite{BN}. The bound
$v_2(C)\ge v_2((n-1)!)$ is Kleiman's \cite{Kl}. A first version of this note is dated
19 October 2026. Technical assistance: Claude (Anthropic).

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\emph{Ideal solutions in the Prouhet--Tarry--Escott problem}, preprint, arXiv:2304.11254.
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