Showing that if $p$ and $p^2+8$ are primes then $p^3+4$ is also prime

Question

The question is as stated above. here is my attempt; $p^3+4=p(p^2+8-8)+4=p(p^2+8)-8p+4$, hence, $p^3+4=p(p^2+8)-4(2p-1)$, so i supposed that $p^3+4$ is composite so that there exist an odd number $n=2k+1$ which divides it. MY PROBLEM IS; how can i show that $k=p$, for if that can be done, the last equation gives a contradiction, showing that $p^3+4$ is prime. Any other approach is welcome please

Answer 1

hint If $3 \nmid p$ then $p^2 \equiv 1 \pmod 3$.

Answer 2

Hint: If $|p|>3$ is prime, then $p^2+8>3$ is divisible by $3$. Thus, the only possible answers are $p=\pm 3$.