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I searched for primes of the form $p^p+2$, where $p$ is prime for a range of $p \le 10^5$ on PARI/GP and found that 29 is the only prime of this form in this range.

Questions:

$(1)$ Is $29$ the only prime of the form $p^p+2$, where $p$ is prime?

$(2)$ If not, then are there a finite number of primes of the form $p^p+2$? Can you prove/disprove this?

Edit: Since $p^p$ grows really fast and primes get rarer and are spread farther out for large numbers,

I conjecture that $29$ is the only prime of the form $p^p+2$ where $p$ is a prime.

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    $\begingroup$ @user0410 $3^3+2=29$ $\endgroup$ – kccu May 8 at 21:06
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    $\begingroup$ I think (2) is likely. maybe (1) as well, and hard to decide. Primes are rare far out, and $p^p$ grows rapidly. Did the question occur to you for any reason other than random curiousity ? $\endgroup$ – Ethan Bolker May 8 at 21:09
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    $\begingroup$ @EthanBolker Well I like trying to find rare primes, and since $p^p$ grows rapidly I figured primes like that were rare. But still it would be really nice to prove that 29 is the only prime of this form $\endgroup$ – Mathphile May 8 at 21:14
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    $\begingroup$ @Mathphile For $p=239$ , there is no small factor. So, there is no easy way to show that $p=3$ is the only soluton, if this is actually the case. Did you really arrive at $p=10^6$ ? The number $p^p+2$ is huge for primes near $10^6$. $\endgroup$ – Peter May 8 at 21:29
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    $\begingroup$ I would guess that there are only a finite number of primes. There are about a $(p \log^2 p)^{-1}$-fraction of numbers of size roughly $p^p$ that are prime, by the Prime Number Thm. So--making some leaps of logic admittedly--it is tempting to say that the number of primes of this form is something like $\sum_{p} (p \log^2 p)^{-1}$ which is bounded $\endgroup$ – Mike May 8 at 22:09
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With pfgw, I checked $$p^p+2$$ for the primes from $\ 3\ $ to $\ 24\ 001\ $ The only prime occured for $\ p=3\ $ Hence if another prime of this form exist, it must have more than $\ 100\ 000\ $ digits

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    $\begingroup$ What is hardware configuration on which you are able to check primality of 100000 digit numbers? $\endgroup$ – Nilotpal Kanti Sinha May 9 at 8:09
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    $\begingroup$ The program pfgw is specially designed for testing huge numbers for primality. I used the default trial division. I also doublechecked with factordb. Upto $p=20\ 000$ , no other prime. $\endgroup$ – Peter May 9 at 8:22
  • $\begingroup$ So you ran the program on normal laptop? I mean what was the configuration and runtime? $\endgroup$ – Nilotpal Kanti Sinha May 9 at 8:24
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    $\begingroup$ On a normal PC. No special hardware. It took several hours however on my slow computer. $\endgroup$ – Peter May 9 at 8:25
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    $\begingroup$ now with my information linked you might be able to push that into millions. $\endgroup$ – Roddy MacPhee May 9 at 19:24
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Some working, but no solution:

Note that $p=2$ doesn't work, and $p=3$ does work. Now suppose $p > 3$.

Suppose $p \equiv 1 \pmod{3}$. Then $p^p + 2 \equiv 1^p + 2 \equiv 0 \pmod{3}$, so it isn't prime.

Therefore, $p \equiv -1 \mod{3}$ and $p$ odd so $p \equiv -1 \mod{6}$. This is as far as I could get. Using WolframAlpha for this case yields numbers that do not have many prime factors, and these factors being distinct and large, so I don't see any way to progress from here.

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    $\begingroup$ Euler's totient theorem and a larger modulus might help. $\endgroup$ – Roddy MacPhee May 9 at 17:51

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