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).



Subsets $E$ of $\mathbb{F}_{p^k}$ with vanishing polynomial subset sums
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.


