Denote $$S(p):=2^2+3^3+5^5+\cdots +p^p$$

$S(p)$ is prime for $p=3,7,89$.

Is there another prime $p$ such that $S(p)$ is prime ? Is the number of primes $p$ such that $S(p)$ is prime, finite ?

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    $\begingroup$ I'm astonished there is a third case. I'm sure the probabilities indicate that the expected number is finite; effectively you have a number with no prior known divisibility roughly the size of $p^p$, which has a descending chance of being prime, $<(1/(p\ln p))$, in a population becoming sparse. $\endgroup$ – Joffan Feb 27 '18 at 1:02
  • $\begingroup$ Up to what $p$ did you check? $\endgroup$ – Tito Piezas III Feb 27 '18 at 2:20
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    $\begingroup$ Nothing more up to p < 10000 from what I see. $\endgroup$ – DanaJ Feb 27 '18 at 4:05
  • $\begingroup$ @DanaJ: For $p=10000$, then $S(p)$ has more than 40000 decimal digits. What primality test did you use? ECM? $\endgroup$ – Tito Piezas III Feb 27 '18 at 7:31
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    $\begingroup$ @TitoPiezasIII ECM is usually used to find factors. But with elliptic curves, prime proving is also possible. But the usual approach to decide whether a number is prime is the strong-probable-prime test. Only if a number passes it, we cannot be sure whether it is prime, although it is very likely in general. If a number fails that test, it must be composite. $\endgroup$ – Peter Feb 27 '18 at 12:05

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