Problem 1075

Let $\nu_{p}(n)$ denote the largest exponent of $p$ dividing $n$. The problem follows from the following proposition.
Proposition
Let $\left(a_{n}\right)$ be a sequence of integers and let $p$ be a prime. Suppose that every consecutive group of $a_{i}$ 's with length at most $p$ averages to a perfect square. Then $\nu_{p}\left(a_{i}\right)$ is independent of $i$.
We proceed by induction on the smallest value of $\nu_{p}\left(a_{i}\right)$ as $i$ ranges (which must be even, as each of the $a_{i}$ are themselves a square). First we prove two claims.
Claim - If $j \equiv k(\bmod p)$ then $a_{j} \equiv a_{k}(\bmod p)$.

Proof. Taking groups of length $p$ in our given, we find that $p \mid a_{j}+\cdots+a_{j+p-1}$ and $p \mid a_{j+1}+\cdots+a_{j+p}$ for any $j$. So $a_{j} \equiv a_{j+p}(\bmod p)$ and the conclusion follows.
Claim - If some $a_{i}$ is divisible by $p$ then all of them are.

Proof. The case $p=2$ is trivial so assume $p \geq 3$. Without loss of generality (via shifting indices) assume that $a_{1} \equiv 0(\bmod p)$, and define
$$S_{n}=a_{1}+a_{2}+\cdots+a_{n} \equiv a_{2}+\cdots+a_{n} \quad(\bmod p) .$$

Call an integer $k$ with $2 \leq k<p$ a pivot if $1-k^{-1}$ is a quadratic nonresidue modulo $p$.
We claim that for any pivot $k, S_{k} \equiv 0(\bmod p)$. If not, then
$$\frac{a_{1}+a_{2}+\cdots+a_{k}}{k} \text { and } \frac{a_{2}+\cdots+a_{k}}{k-1}$$
are both qudaratic residues. Division implies that $\frac{k-1}{k}=1-k^{-1}$ is a quadratic residue, contradiction.

Next we claim that there is an integer $m$ with $S_{m} \equiv S_{m+1} \equiv 0(\bmod p)$, which implies $p \mid a_{m+1}$. If 2 is a pivot, then we simply take $m=1$. Otherwise, there are $\frac{1}{2}(p-1)$ pivots, one for each nonresidue (which includes neither 0 nor 1 ), and all pivots lie in $[3, p-1]$, so we can find an $m$ such that $m$ and $m+1$ are both pivots.

Repeating this procedure starting with $a_{m+1}$ shows that $a_{2 m+1}, a_{3 m+1}, \ldots$ must all be divisible by $p$. Combined with the first claim and the fact that $m<p$, we find that all the $a_{i}$ are divisible by $p$.

The second claim establishes the base case of our induction. Now assume all $a_{i}$ are divisible by $p$ and hence $p^{2}$. Then all the averages in our proposition (with length at most $p$ ) are divisible by $p$ and hence $p^{2}$. Thus the map $a_{i} \mapsto \frac{1}{p^{2}} a_{i}$ gives a new sequence satisfying the proposition, and our inductive hypothesis completes the proof.
Remark. There is a subtle bug that arises if one omits the condition that $k \leq p$ in the proposition. When $k=p^{2}$ the average $\frac{a_{1}+\cdots+a_{p^{2}}}{p^{2}}$ is not necessarily divisible by $p$ even if all the $a_{i}$ are. Hence it is not valid to divide through by $p$. This is why the condition $k \leq p$ was added.