This is from the category "problems I cannot believe that are still open". But then again, I don't know whether it is still open; it seems to have escaped the attention of most number theorists and algebraists except for those in olympiad circles. This is the reason I am posting it here.

Let $p$ be a prime. Define a linear operator $F_p:\mathbb R^{\mathbb Z}\to\mathbb R^{\mathbb Z}$ by

$\left(F_p f\right)\left(n\right) = \dfrac{f\left(n\right)+f\left(n+1\right)+...+f\left(n+p-1\right)}{p}$ for every $n\in\mathbb Z$ and every $f\in\mathbb R^{\mathbb Z}$.

(Of course, elements of $\mathbb R^{\mathbb Z}$ are just two-sided infinite sequences of reals, written as functions from $\mathbb Z$ to $\mathbb R$. The operator $F_p$ replaces a sequence by the sequence of the arithmetic means of its $p$-windows.)

An element $f\in \mathbb Z^{\mathbb Z}$ is said to be *average-integral* if it satisfies $F_p^kf\in\mathbb Z^{\mathbb Z}$ for every nonnegative integer $k$.

For any $f\in\mathbb R^{\mathbb Z}$, define $f^p\in\mathbb R^{\mathbb Z}$ by

$f^p\left(n\right)=\left(f\left(n\right)\right)^p$ for every $n\in\mathbb Z$.

**Conjecture:** If $f\in \mathbb Z^{\mathbb Z}$ is average-integral, then so is $f^p$.

**Remarks:** For $p=2$, this was problem 8 for grade 10 in the Allrussian Mathematical Olympiad 1993, proposed by D. Tamarkin (the one of operad theory fame?). There is a discussion with several proofs of the $p=2$ case on MathLinks, and it shows that the $p=2$ case is actually a tip of an iceberg (namely, for $p=2$, the average-integral elements of $\mathbb Z^{\mathbb Z}$ form a ring, so not only squares but also pointwise products of average-integral elements are average-integral). For $p=3$, the conjecture is still true, but the iceberg apparently is not anmyore; it took me a lengthy computation with combinatorial divisibilities to verify the conjecture. For higher $p$, I don't know of any results at all. Has anything been done since 1993 at all?