If p, q, r are distinct primes, does there exist infinitely many (p, q, r) such that p^q+q^r+r^p is prime

number theory

The question is as stated in the title. I want to know if $\exists$ infinitely many prime triplets $(p, q, r)$ such that \begin{equation} p^q+q^r+r^p \end{equation} is prime.

Wrote some code and found that for primes up to $200$, $(3, 5, 11), (3, 5, 107), (3, 11, 131), (3, 13, 61), (3, 17, 107), (3, 17, 113), (3, 23, 167), (5, 11, 43), (5, 29, 127), (5, 41, 67), (5, 53, 109), (5, 67, 71), (5, 79, 149), (11, 23, 127), (11, 53, 109), (11, 67, 79), (11, 103, 109), (11, 137, 163), (13, 41, 43), (13, 41, 59), (13, 107, 109), (13, 131, 179), (17, 19, 41), (17, 37, 199), (17, 53, 79), (19, 23, 83), (19, 47, 61), (19, 67, 113), (19, 103, 191), (23, 31, 37), (23, 43, 73), (23, 43, 109), (23, 97, 101), (23, 131, 181), (29, 31, 131), (29, 61, 137), (31, 47, 157), (31, 59, 113), (37, 97, 173), (41, 67, 113), (43, 47, 71), (43, 89, 193), (43, 179, 181), (47, 79, 163), (47, 167, 181), (61, 67, 113), (61, 71, 127), (61, 101, 131), (79, 83, 103), (83, 127, 151), (89, 137, 151), (97, 131, 197), (103, 107, 139), (107, 151, 167), (113, 151, 173), (113, 151, 179), (163, 167, 197), (181, 191, 197), (191, 193, 199)$ satisfy the given condition.

Bonus: Extend this question to prime sets of length $2n+1$.

Best Answer

The conjecture that there are infinite many primes of the desired form can be strengthened to this

Conjecture : For every prime pair $(p,q)$ ($2<p<q$) , there is at least one prime $r>q$ such that $p^q+q^r+r^p$ is prime.

The growth rate and that small prime factors can be avoided speak for the existence. Even if for infinite many pairs $(p,q)$ (instead of all) , we can find such a prime $r$, this would still give infinite many solutions.

How can we rule out small factors ?

$2$ is impossible anyway. So assume $s$ is an odd prime. If we choose a prime $r$ with $$r\equiv 0\mod s-1$$ and $$r\equiv 1\mod s$$ we get $$p^q+q^r+r^p\equiv p^q+2\mod s$$ unless $s=q$ , in which case we get $p^q+1$ and if we choose $r$ such that $$r\equiv 0\mod s-1$$ and $$r\equiv -1\mod s$$ we get $$p^q+q^r+r^p\equiv p^q\mod s$$ unless $s=q$ in which case we get $p^q-1$.

At least one of the numbers is not divisible by $s$. Of course, this construction requires large primes $r$ , so that it is unclear whether we will ever get a prime this way. But probably most of the numbers are not divisible by a given prime factor $s$ anyway.

Maybe it pays out to verify this stronger conjecture.

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