thanks because p=m^2+n^2 then p is 1(mod 4) or 2 when p is 2 m=n=1 so it is an answer when p is 1(mod 4) there is a number x which satisfies x^2+1 is 0(mod p) then m is nx(mod p) if we put it into the second equation x^3+1+8x is a multiple of p so 7x+1 is a multiple of p then p is a divisor of fifty then p=5 which has m=2 and n=1 so the answer is p=2,5

$m^2+n^2|m^3+n^2 \cdot m$ $m^2+n^2|n^3+8m \cdot n -n^2 \cdot m$ $m^2+n^2|n^3+mn(8-n)$ $m^2+n^2|m^2n+n^3$ $m^2+^2|mn(8-n)-m^2n$ $m^2+n^2|mn(8-n-m)$ $m^2+n^2$ is prime => $(m^2,n^2)=1$ => $(m^2+n^2,mn)=1$ We find the solutions for $m=n=1$ and $m=2,n=1$ and $n=2,m=1$