Since $ p|(m^2 + n^2)(m + n) = m^3 + n^3 + mn(m + n)$, we have $ p|mn(m + n) + 4 \Rightarrow p|2mn(m + n) + 8$. But $ 2mn \equiv (m + n)^2 mod \text{ } p$, therefore $ p|(m + n)^3 + 2^3 = (m + n + 2)((m + n)^2 - 2(m + n) + 4)$. $ p|m + n + 2$ or $ p|2mn - 2(m + n) + 4$. In the first case with Cauchy ($ |m + n|^2 \le (1 + 1)(m^2 + n^2)$) we get $ \sqrt{2p} + 2 \ge p$. This limits $ p$, and some calculations suffice. In the second case $ p = 2$ or $ p|mn - m - n + 2$. Using AM-GM ($ mn \le \frac {m^2 + n^2}{2}$) we also limit $ p$.