
> 
> 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.



A Vandermonde-type system
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? 


