# Integral zeros of a multivariate polynomial

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?

• Doesn't the sufficient magnitude of $m$ depend on the odd prime? – Max Alekseyev Nov 15 '19 at 20:52

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 $$u+v+w+t. So, we constructed a zero of $$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.

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