The following question arose in some discussions recently as a misunderstanding of another problem.

**Question:** Which subsets $E\subset \mathbb{F}_{p^k}$ satisfy the property that $ \sum\limits_{x\in E}Q(x)=0$ for all $Q(x)$ polynomials in $\mathbb{F_{p}}[x]$ of degree less or equal to $t$?

Zero here is the additive identity in the field $\mathbb{F}_{p^k}$.

A sub problem of the above which should be manageable is the following question:

**Question:** Does the collection of sets $E$, each of size $p$ and containing the origin which satisfy the above condition for $t \leq p-2$ consist precisely of the $\frac{p^{k}−1}{p−1}$ lines (containing $0$) in $\mathbb{F}_{p^k}$?

(Note the condition $t\leq p-2$ is needed by Fermat's little theorem, $0^{p-1}+1^{p-1}+...+(p-1)^{p-1}=1+1+...+1=p-1\neq 0$.)

This has been computationally verified for $\mathbb{F}_{5^3}$ below.

A simple example to clarify the first problem follows: the condition in the question can be reformulated as a collection of statements, namely $$\sum\limits_{x\in E} x^{m}=0\ \text{for all } m\leq t, $$ due to the fact that the first statement by restriction implies all of the second statements. Further, given $Q(x)=a_{0}+a_{1}x+\dots+a_{t}x^t$, the equalities in the second statement can be added to give that $\sum\limits_{x\in E} Q(x)=0$.

The first condition $\sum\limits_{x\in E} 1=0$ shows exactly that $N=|E|$ is a multiple of $p$.

The case $t=1$ also is not difficult. Here one gets a linear condition on the elements of $E$. The interesting question is what these sets look like, especially when $k$ and $t$ are not too small.

**Edit:** To make the problem more concrete, here is more information and an additional example.

There are some natural invariants of such sets $E$ like multiplication of elements in $E$ by a constant multiple (where constant here means constant in $\mathbb{F}_{p}$). For $y\in\mathbb{F}_{p}$ a constant, $E+y=\{x+y\ |\ x\in E\}$ is also a vanishing set if $E$ is. Both of these follow from the fact that the set of polynomials of degree bounded by $t$ are invariant under the corresponding operations.

Consider the vanishing condition for quadratics with $p=5$ and $k=3$. For example, let $\mathbb{F}_{5^3}$ be represented by the quotient of $\mathbb{F}_{5}[x]$ by $g(x)=x^3+x+1$ (which is irreducible in $\mathbb{F}_{5}[x]$). Letting the coefficients of each element in $\tilde{E}$ be given by the formula $p_{j}(x)=a_{j}x^2+b_{j}x+c_{j}$ ($\tilde{E}$ here is just the representatives of $E$ in reduced form in the quotient of $\mathbb{F}_{5}[x]$ by $g(x)=x^3+x+1$), then the condition $\sum\limits_{p(x)\in \tilde{E}} p(x)^2=0$ can be checked to be equivalent to three equalities : \begin{eqnarray*} \sum\limits_{j=1}^{N} 4a_{j}^2+2a_{j}c_{j}+b_{j}^2 \equiv_{5} 0 \\ \sum\limits_{j=1}^{N} 3a_{j}b_{j}+4a_{j}^2+2b_{j}c_{j} \equiv_{5} 0 \\ \sum\limits_{j=1}^{N} c_{j}^2+3a_{j}b_{j}\equiv_{5} 0 \end{eqnarray*}

which follows from reducing the quartic polynomial resulting from squaring a general polynomial of degree two in $\mathbb{F}_{5}[x]$ by $g$.

**Update:** Following the same procedure for cubics ($t=3$) and $p=5$, $k=3$ gives exactly $31$ $5$-element subsets containing the origin which are all lines, that is they are all of the form $\langle (a,b,c) \rangle=\{s(a,b,c) |\ s\in\mathbb{F}_{p}\}$. One natural generalization of this observation, is the possibility that such lines are exactly the *vanishing* sets of size $p$ containing the origin for polynomials of degree $t\geq k$ for $\mathbb{F}_{p^k}$ (see second question above).