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I'm self-learning number theory.
I want to prove the following statement:

$$p \text{ is prime } \land \text{ }p^2 + 2 \text{ is prime } \implies p^3 + 2 \text{ is prime }$$

I failed to do so, and I failed to find any proofs online.
My initial attempts involved using Fermat's Little Theorem:

$$\begin{align*} a^{p} &\equiv a \mod p \\ a^{p^2 + 2} &\equiv a \mod {p^2 + 2} \\ \end{align*}$$

But that form didn't really help me that much. Any hints?

user123
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1 Answers1

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Hint: if $n$ is not divisible by $3$, then $n^2\equiv1\pmod3$.

Hence if $p\neq3$, $3\mid p^2+2>3$ so $p^2+2$ cannot be prime. This means the implication says $\perp\implies p^3+2$ is prime. Indeed, $\perp$ implies everyting. If $p=3$, the implication appears true.

Bart Michels
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  • Your first statement allowed my proof to bloom on paper, thank you! In fact, this shows that $p^2 + 2$ is composite for all primes except for $3$. The statement to prove works for $3$ because it generates $11$ and $29$ which are prime. – user123 Nov 16 '14 at 09:30