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Consider the multivariate polynomial $$f(x_1,\ldots,x_m)=mk\sum_{i=1}^mx_i^2-mk(k-1)\sum_{i=1}^mx_i-\left(\sum_{i=1}^mx_i\right)^2,$$ for integers $m,k\ge2$. We are looking for integral zeros of $f$ with $0\le x_i\le k$. If $mk$ is square free, then it is easily seen that the only zeros of $f$ under the above conditions are $(0,\ldots,0)$ and $(k,\ldots,k)$. We believe that given $k$ if $m$ is large enough and $mk$ is divisible by square of an odd prime, then $f$ has more zeros. Could somebody provide a proof?

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    $\begingroup$ Doesn't the sufficient magnitude of $m$ depend on the odd prime? $\endgroup$ – Max Alekseyev Nov 15 '19 at 20:52
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First, I will consider the case of $k\geq 4$.

Let $mk=aq^2$, where $q$ is an odd prime. Since $f(x_1,\dots,x_m)=0$ implies that $mk$ divides $(\sum x_i)^2$, we will look for a zero with $\sum x_i = aq$. Then $f(x_1,\dots,x_m)=0$ will follow from the two equations: $$\begin{cases} \sum_{i=1}^m x_i = aq, \\ \sum_{i=1}^m x_i^2 = (k-1)aq + a. \end{cases} $$ We will show that there is a solution containing only $0$'s, $1$'s, $2$'s, $(k-1)$'s and $k$'s, i.e. $$\begin{cases} u + 2v + (k-1)w + kt = aq,\qquad(\star) \\ u + 4v + (k-1)^2w + k^2t = (k-1)aq + a, \end{cases} $$ where $u,v,w,t$ are the multiplicities of $1$'s, $2$'s, $(k-1)$'s, $k$'s, respectively.

This system has an integer solution whenever \begin{split} (k-1) &\mid k^2(aq-u-2v) - k((k-1)aq + a - u - 4v),\\ k&\mid (k-1)^2(aq-u-2v) - (k-1)((k-1)aq + a - u - 4v), \end{split} that is $$\begin{cases} 2v \equiv - a(q-1) \pmod{k-1},\\ 2u + 6v \equiv a\pmod{k}. \end{cases}$$

We fix integers: \begin{split} \text{if $k$ is even:} & \qquad\begin{cases} v := \frac{-a(q-1)}2\bmod (k-1), \\ u := \frac{a}{2} - 3v\bmod{\frac{k}2}, \end{cases} \\ \\ \text{if $k$ is odd:} & \qquad \begin{cases} v := \frac{-a(q-1)}2\bmod \frac{k-1}2, \\ u := \frac{a}{2} - 3v\bmod{k}, \end{cases} \end{split} to further obtain integers: \begin{split} w &:= \frac{a(q-1)-(k-1)u-2(k-2)v}{k-1},\\ t &:= \frac{a+(k-2)u + 2(k-3)v}k. \end{split}

From definition of $u,v$, we have that they are smaller than $k$. It further follows that for a fixed $k$ and large enough $m$ (and thus large enough $aq$), values $w$ and $t$ are positive. From equation $(\star)$ it also follows that $u+v+w+t<m$. So, we constructed a zero of $f$ different from the trivial ones.


For $k=3$, the above construction works but $2$'s and $(k-1)$'s collapse and have total multiplicity $v+w$. Similarly, for $k=2$, we get a solution with $1$'s and $2$'s having multiplicities $u+w$ and $v+t$, respectively.

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  • $\begingroup$ Thanks Max for your nice argument. I just saw your answer as the internet was down this week in our country! $\endgroup$ – Ebrahim Nov 22 '19 at 14:33

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