As the title implies.
I'm proving it by contradiction and I've already ruled out $p \equiv 0$ and $2$ (mod $4$).
I'm having trouble proving that it cannot be $\equiv 3$ (mod $4$). Namely, if $p \equiv 3$ (mod 4), then $m^2 + n^2 \equiv 1$ or $3$ (mod $4$). I know that $m^2 + n^2$ cannot be congruent to $3$ (mod $4$); but what about $1$?