<s>[INST] Show that $p^3+4$ is prime If $p$ and $p^2 +8$ are both prime number, prove that $p^3 +4$ is also prime. 
I was thinking using two cases for even and odd. So $p$ is even thus there is no prime, the statement does not hold. So, $p$ is odd, $p$ has to be of the form of $2k+1$. 
let $p=2k+1$ which is prime
$(2k+1)^2 + 8 = 4k^2+4k+1+8 = 2(2k^2+2k+4)+1$ which is prime 
so $(2k+1)^3 +4 = 8k^3+12k^2+6k+1 +4= 2(4k^3+6k^2+3k+2)+1$ which is also prime. 
is that all that i need to show 
 [/INST] Hint: if you don't know the theory of quadratic residue, you can note that every prime $p>3$ can be written in the form $6n\pm1$. It has remainder $\pm1$ when divided by $3$. Note therefore that for $p>3$, $p^2+8$ is divisible by $3$.
Indeed $$p^2+8=(6n\pm1)^2+8\equiv (\pm1)^2\!+8\equiv 0\pmod{\!3}$$
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