Prime, Square, Cube!!!

Lemma :- Every prime number except 2 and 3 are of the form 6n+1 or 6n+5. Proof :- When we take p=6n+1, neither p^8 nor p^3+4 are of the form 6k+1. Same happens when we put p=6n+5. So taking the exceptions to the lemma, we get the only possible answer is 3, because 3^2 + 8 = 17 and 3^3+4=31. So both are primes. Q.E.D.

Clearly $2$ doesn't work, for $p>3$ , we have... $p\equiv -1,1 \pmod{3} ~~~ \Longrightarrow ~~~ p^2\equiv 1 \pmod{3} ~~~ \Longrightarrow ~~~ p^2+8\equiv 1+8=9\equiv 0 \pmod{3}$

Hence, for $p>3$ , $p^2+8$ is not prime. We have one try left which is $p=3$ and so $p^2+8=17$ and $p^3+4=31$ . Therefore $\fbox{3}$ is the only solution.

The sum of the digits of any prime number squared + 8 is a multiple of 3, therefore 3 is the only one solution.

If $p=3$ , $p^2+8= 17$ and $p^3+4= 31$ , which are both primes. We claim that there are no other solutions. First, note that the only prime divisible by $3$ is $3$ itself. All primes $p \neq 3$ are $\equiv -1, 1 \pmod{3}$ , and thus $p^2 \equiv 1 \pmod{3}$ . Note that $p^2+8 \equiv 1 + 8 \equiv 0 \pmod{3}$ and $p^2+8 \geq 2^2+8 > 3$

It immediately follows that $p^2+8$ is never a prime for $p \neq 3$ . The sum of all possible values of $p$ is, thus, $\boxed{3}$ .