66th Belarusian Mathematical Olympiad

Let $n-1$, $n$, $n+1$, and $n+2$ denote four consecutive positive integers with the sum of the squares equaled $2p$. Then $$2p=(n-1{)}^2+n^2+(n+1{)}^2+(n+2{)}^2=4n^2+4n+6,$$ whence $p=2n^2+2n+3=2n(n+1)+3$. Now if one of the numbers $n$ and $n+1$ is divisible by $3$, then $p$ is divisible by $3$ hence $p$ is not a prime. So, $n$ and $n+1$ are not divisible by $3$, whence $n=3k+1$, where $k\in \mathbb{N}$. Then $p=2(3k+1)(3k+2)+3=18k(k+1)+7$, so $p-7=18k(k+1)$ is divisible by $18$. Since at least one of the numbers $k$ and $k+1$ is even, we obtain that $p-7$ is divisible by $36$, as required.