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Let $p$ be a prime. Let $F_p$ be the finite field of $p$ elements. Let $A$ be a subset of $F_p$ of size $s$. Assume that $s > 2$ is polylogarithmic in $p$.

Suppose that we want to count number of solutions of $$ x_1 + x_2 + ... + x_k = t (\mod\ p) $$ under the restriction that $x_i \in A $ for all $1\leq i\leq k$. The expectation seems to be $s^k/p$ if $k$ is large, independent of $t$.

My question is: Does there exist $A$ such that the expectation is known to be quite sharp (namely that the error term is poly(k p) in absolute value)?

Thanks a lot

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I am a little be confused by your terminology: what does "$s$ polylogarithmic in $p$" mean? – Wadim Zudilin Jun 21 '10 at 4:19
"$s$ is polylogarithmic in $p$" generally means $s = \Theta(\log^c p)$ for a fixed constant $c > 0$. – Ryan Williams Jun 21 '10 at 4:23
But then I can't follow the claim about the expectation to be independent of $t$: take $A$ to be $\lbrace 0,1,\dots,s-1\rbrace$, so that only $t<ks$ can be expressed in the required form... By the way for this form of $A$ one can easily count the number of solutions. – Wadim Zudilin Jun 21 '10 at 4:29
In my question k is large. In fact, the case most interesting to me is when k = p. So the expectation of number of solutions is independent of t, I guess. Error term will be dependent on t of course. Qi – user3208 Jun 21 '10 at 6:16

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