Show that if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime, then $(m,n)=(2,3)$ or $(3,2)$ if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime
then $(m,n)=(2,3)$ or $(3,2)$
I found out that one of the $m$ and $n$ must be two since $n^2+m^3$ will be even if $m$ and $n$ have the same parity.
However, how to prove that if $4+n^3$ and $8+n^2$ are both prime, then $n=3$?

If $n \neq 3$ and is also prime, then $n \equiv 1 \pmod{3} \implies 8 + n^2 \equiv 0 \pmod{3}$, or $n \equiv 2 \pmod{3} \implies 4 + n^3 \equiv 0 \pmod{3}$. Since both $4 + n^3$ and $8 + n^2$ are greater than $3$, this means at least one of them can't be prime.
