math.NT — Number Theory

The 2-adic valuation of the constant for symmetric ideal Prouhet-Tarry-Escott solutions

Contributed by David Niedbala Giraudin

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 $v_2(\overline{C_8})\ge6$ and determined $v_2(\overline{C_9})=9$, 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 $\mathbb{Z}$, it splits into two regular systems and yields $v_2(C)\ge6,8,11,12,12,16,18,18$ 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 $v_2(C'_8)=8$, for which the published information was $v_2(\overline{C_8})\ge6$. 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 $\mathbb{Z}_2$, it combines with a finite-difference lemma refining a congruence of Filaseta--Markovich to give $v_2(C)\ge11,12,12$ 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 $\mathbb{Z}_2$ with $v_2(C)=11$, 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 $\mathbb{Z}_2$, obtained from a second object of the same authors, although its constant has valuation $17$.