first time doing such a proof. I somehow can't make it work, where's the problem?
Claim: If $p$ and $p^2 + 2$ is prime, then $p^3+2$ is prime. Proof:
Basic idea: Since we are always adding $2$, we are cheking the remainder of modulo $3$.
Case: $p=2$
this doesn't hold the claim, omit.
Case: $p=3$
since 3 is prim and $p^2+2=11$ is prime, we calculate $p^3+2=29$ which is also prime.
We now check for the remainder of mod $3$. The idea is, that if $p$ is a prime and we take it mod $3$, we get $R_3(p)\in\{1,2\}$.
We can now make a case distinction.
Case 1: $R_3(p) = 1$
If that is the case, $p$ is prime. We check, if $p^2+2$ is prime:
$R_3(p^2+2)=R_3(R_3(p^2)+R_3(2))=R_3(R_3(R_3(p)\cdot R_3(p))+2)=R_3(R_3(1+2))=R_3(0)=0$. So $p^2+2$ isn't a prime. Thus, we can omit this case.
Case 2: $R_3(p) = 2$
$R_3(p^2+2)=R_3(R_3(p^2)+2)=R_3(R_3(R_3(p)R_3(p))+2)=R_3(R_3(6))=0$
This case doesn't hold either.
So somewhere in my though process I made a terrible mistake but I can't see where. :/
This seems like a perfectly good solution to me. You have correctly proved that the only prime $p$ for which $p^2+2$ is also prime is $p=3$. And for $p=3$ you've checked that $p^3+2$ is also prime. So the result is true.