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For a prime $p$ and $a_1,\dotsc,a_n\in\mathbb F_p^\times$, consider the system of equations $$ \begin{cases} \begin{align} a_1 + \dotsb + a_n &= 0 \\ a_1x_1 + \dotsb + a_nx_n &= 0 \\ \qquad &\ \vdots \\ a_1x_1^K + \dotsb + a_nx_n^K &= 0 \end{align} \end{cases} $$ How large can $K$ be given that this system has a solution in the variables $x_1,\dotsc,x_n\in\mathbb F_p^\times$?

By the Chevalley–Warning theorem, if $n>K(K+1)/2$, then there are solutions with at least some of the variables distinct from $0$. Is it true, say, that if $K>C\sqrt n$ with a suitable absolute constant $C$, then the system does not have non-zero solutions? (For my purposes, it would be fine to know that there are no non-zero solutions with at least $2n/3$ among $x_1,\dotsc,x_n$ pairwise distinct.)

An essentially equivalent restatement of the problem is as follows:

What is the smallest possible number of non-zero coefficients of a polynomial $P\in\mathbb F_p[x]$ of degree $\deg P\le p-1$ given that $P$ has a root of multiplicity $K$ at $x=1$?

There are many similar results and problems around (the Tarry-Escott problem, Linnik lemma, Vinogradov system, Borwein-Erdelyi-Kos results for polynomials over the integers, etc), and I suspect that the problem stated above has also been studied. Any references will be appreciated.

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If $P(x)$ is divisible by $(x-1)^K$ and $\deg P<p$, then $P$ has at least $K+1$ non-zero coefficients. Proof: divide $P$ by maximal possible power of $x$, take the derivative and use induction in $K$. Or what I get wrong?

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  • $\begingroup$ This is perfectly correct and is attained on the polynomial $(x-1)^K$. I thought at some point of this polynomial, but somehow convinced myself that, translated back to the original system, it yields a solution with the $x_i$ equal to each other - which was totally wrong, as I now see. I will post a follow-up which is a closer approximation to the real problem that i need. $\endgroup$
    – Seva
    Commented Apr 10, 2019 at 16:47

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