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 $mn\ne 0$. Then $p>|m|$ and $p>|n|$. Note that $$m^3+n^3=(m+n)\left(m^2-mn+n^2\right)\equiv -mn(m+n)(\mathrm{mod}p)$$ So the problem statement is equivalent to $$mn(m+n-8)\equiv 0(\mathrm{mod}p).$$ Since $p>m,n$ and $p$ is prime, then $p\mid m+n-8$, so $m^2+n^2\le |m+n-8|$. We have two cases:

1. If $m+n-8\ge 0$ then $m^2+n^2\le m+n-8$ which means $m(m-1)+n(n-1)\le -8$. 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 $m(m+1)+n(n+1)\le 7$. 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 $(m,n)=(-3,-2)$.