Saudi Arabian Mathematical Competitions

If $m = 0$ then $p = n^{2}$ is a prime, which is impossible. Similar case happens when $n = 0$. So we may assume $m n \neq 0$. Then $p > |m|$ and $p > |n|$. Note that
$$m^{3} + n^{3} = (m + n)\left(m^{2} - m n + n^{2}\right) \equiv -m n(m + n) \pmod{p}$$
So the problem statement is equivalent to
$$m n(m + n - 8) \equiv 0 \pmod{p}.$$
Since $p > m, n$ and $p$ is prime, then $p \mid m + n - 8$, so $m^{2} + n^{2} \leq |m + n - 8|$. We have two cases:

1. If $m + n - 8 \geq 0$ then $m^{2} + n^{2} \leq m + n - 8$ which means $m(m - 1) + n(n - 1) \leq -8$. This is impossible, since $m(m - 1)$ and $n(n - 1)$ are integers and non-negative.

2. If $m + n - 8 < 0$ then we get $m(m + 1) + n(n + 1) \leq 7$. So we conclude $m, n = \pm 3, \pm 2, \pm 1$.

By case work we get 3 solutions as follows:
- $p = 2$ when $(m, n) = (1, 1)$,
- $p = 5$ when $(m, n) = (2, 1)$,
- $p = 13$ when $(m, n) = (-3, -2)$.