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
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.