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Find all primes $p$ such that $$p^p+2$$ is also prime

I have been struggling with this problem the past few days and I can't seem to figure a way around it. I conjecture that the only solution is $p=3$. Hence my first approach was $mod\ 3$ and contradiction, which didn't lead me far, as the cases $p\equiv 0,1\pmod{3}$ conclude that either $p=3$ or $p^p+2=3$, a contradiction. But the case: $$p \equiv -1\pmod{3}\Leftrightarrow\ p^p+2\equiv 1\pmod{3}$$ doesn't help me deduce something and I don't know how to continue on with this case. I have also proved that $$p^p+2\equiv 1\pmod{6},\forall\ p>3, p\equiv -1\pmod{3}$$ which I also can't seem to find useful. Any hints would be appreciated.

TOP STIN ELLADA
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    Note: contrary to the claim in the post, $p\equiv 1 \pmod 6\implies p^p+2\equiv 3\pmod 6$. – lulu Aug 16 '20 at 12:38
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    @Vepir It's not a great duplicate since there is no answer provided. The posted "solutions" provide some numerical evidence and some heuristic arguments, but there is no resolution there that I can see. – lulu Aug 16 '20 at 12:41
  • @lulu when $p\equiv -1\pmod{3}$ then you can easily deduct $p\equiv -1\pmod{6}$ and as such $p^p+2\equiv 1\pmod{6}$ – TOP STIN ELLADA Aug 16 '20 at 12:57
  • You made the claim that $p^p+2\equiv 1 \pmod 6,\forall, p>3$. This claim is false. The claim is false for $p=7$, for example...indeed it is false for every prime $p\equiv 1 \pmod 6$. – lulu Aug 16 '20 at 13:00
  • @lulu I'm sorry, perhaps I wasn't clear enough with my phrasing. I edited my post. – TOP STIN ELLADA Aug 16 '20 at 13:05

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