# Is $29$ the only prime of the form $p^p+2$?

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.

• @user0410 $3^3+2=29$ – kccu May 8 at 21:06
• 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 ? – Ethan Bolker May 8 at 21:09
• @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 – Mathphile May 8 at 21:14
• @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 $10^6$. – Peter May 8 at 21:29
• 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 – Mike May 8 at 22:09

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

• What is hardware configuration on which you are able to check primality of 100000 digit numbers? – Nilotpal Kanti Sinha May 9 at 8:09
• 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. – Peter May 9 at 8:22
• So you ran the program on normal laptop? I mean what was the configuration and runtime? – Nilotpal Kanti Sinha May 9 at 8:24
• On a normal PC. No special hardware. It took several hours however on my slow computer. – Peter May 9 at 8:25
• now with my information linked you might be able to push that into millions. – Roddy MacPhee May 9 at 19:24

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.

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