Problem 613

For the number $p+q$ to be prime the numbers $p$ and $q$ must be of different parity, which automatically means that $q=2$, since $p > q$. By the problem statement we then have that the numbers $p-2, p$, and $p+2$ should be prime. Since they obviously have different remainders in division by $3$, one of them must be equal to $3$. So we have three possibilities:
* If $p-2=3$, then $p=5$, $p+2=7$, and the pair $(5; 2)$ satisfies the problem statement;

* if $p=3$, then $p-2=1$, and is not prime;
* if $p+2=3$, then $p=1$, and is not prime.
So after considering all possible cases we end up with a single solution $p=5, q=2$.