Slovenia — National Math Olympiad

If the pair $(p, q)$ satisfies the conditions of the problem, then so does the pair $(q, p)$. It is therefore sufficient to only consider the case where $p \le q$. Obviously, $p = q = 2$ is not a solution. If $p$ and $q$ are both odd primes, then $2^2 + p^2 + q^2$ is an even integer greater than $2$, so it is not prime. Hence, $p = 2$.

Let us figure out when the number $8+q^2$ is prime. If $q = 3$, then $8+q^2 = 17$ is prime. Else, $3$ divides $q-1$ or $q+1$, so $3$ divides $9 + (q-1)(q+1) = 8+q^2$ which is then not prime.

We conclude that $2^2 + p^2 + q^2$ is prime only if $p = 2$ and $q = 3$ or $p = 3$ and $q = 2$.