# A Vandermonde-type system

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.

If $$P(x)$$ is divisible by $$(x-1)^K$$ and $$\deg P, then $$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?
• 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. – Seva Apr 10 '19 at 16:47