Subsets $E$ of $\mathbb{F}_{p^k}$ with vanishing polynomial subset sums 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).
 A: Let $E$ be a subset of any field. Then $\sum_{x \in E} x^m=0$ for all $m \leq t$ if and only if  $$\sum_m \sum_{x \in E}  u^{-m-1} x^m = \sum_{x \in E} \frac{1} {u-x} =  \sum_{x \in E} \frac{d}{du} \log (u-x)  = \frac{d}{du}  \log \left(\prod_{x \in E} (u-x)\right) $$ has degree $\leq -t -2$ in $u$.
Now, $$\frac{d}{du}  \log \left(\prod_{x \in E} (u-x)\right)= \frac{ \frac{d}{du}  \prod_{x \in E} (u-x)}{ \prod_{x \in E} (u-x)}$$
so this happens if and only if $$\deg   \frac{d}{du}  \prod_{x \in E} (u-x) \leq |E|-t-2 .$$
In other words, this happens if and only if the only terms of $ \prod_{x \in E} (u-x)$ occurring with degree greater than $|E|-t-1$ are those whose degree is a multiple of $p$ ($p$ the characteristic of the base field), as these are the only terms with vanishing derivative.
For instance, in the case when $|E|=p$, $t=p-2$, and $E$ is guaranteed to contain $0$, we see that this polynomial can only have nonvanishing coefficients of $u^p, u, $ and $1$, and the $1$ must vanish because $|E|$ contains $0$, so it must be of the form $u^p- au = 0$, which is easily seen to define a line through the origin.
In general, a polynomial of this special form will correspond to a set $E$ if and only if it has all distinct roots and these roots lie in the base field.
