Slovenia — National Math Olympiad

We have $r \ge 2$, so $p$ and $q$ must be odd primes. Thus, $p-q$ is even and equal to $2$. We get $p-q=2$. The numbers $p-r=q-r+2$ and $q-r$ are prime and differ by $2$, so they have the same parity. We conclude that both must be odd. Since $q$ and $q-r$ are odd, $r$ must be even. Thus, $r=2$.

The numbers $q$, $p = q+2$ and $q-r = q-2$ are prime. Since $q-2$ is an odd prime it must be at least $3$. But exactly one of the prime numbers $q-2$, $q$, $q+2$ is divisible by $3$, so $q-2=3$. This implies $q=5$ and $p=7$.

We get $r=2$, $q=5$ and $p=7$. These numbers satisfy the conditions of the problem since $p-r=5$, $p-q=2$ and $q-r=3$ are also prime.