# Is it known whether there are ever infinitely many primes of the form $\prod_i p_i^{n_i} + 1$ where the $p_i$ are fixed primes but the $n_i$ can vary?

So if we fix finitely many primes $p_i$, where one $p_i$ is $2$, but let the powers $n_i$ on the $p_i$ vary, is it known whether it is ever possible to have infinitely many primes of the form $\prod_i p_i^{n_i} + 1$? Note this is sort of a soft question, if there are any choice of distinct primes $p_i$ that make it possible then I would be happy with that. The negation of the statement would be that for any choice of primes $p_i$, the set of primes so formed according to the question would be finite.

• – Michael Burr Jul 9 '15 at 23:19
• @MichaelBurr Yes thanks, that's exactly what motivated my question. But now I'm allowing any arbitrary products of powers of an a priori fixed set of primes. I was hoping my question would strike at the heart of the matter, or at least reveal our limitations in being able to answer the question. I.e., is it true that for ANY fixed finite set of primes, we cannot yet determine the answer? – user2566092 Jul 9 '15 at 23:22
• It's certainly not possible if the $p_i$ are all odd. – Robert Israel Jul 9 '15 at 23:25
• The next refinement would be what is the best $f$ we can use in the statement that there are an infinite number of primes $q$ of the form $q = 1 + \prod_i p_i ^ {n_i}$ with all $p_i \le f(q)$. This question asks if $f(q) = c$, and $f(q) = q + 1$ works. – Dan Brumleve Jul 9 '15 at 23:27
• @RobertIsrael Fair enough, I should have clarified because I implicitly knew that that was a trivial answer if all primes were odd. I'm modifying my question now. – user2566092 Jul 9 '15 at 23:28

What we can do (using Dirichlet's theorem on primes in arithmetic progressions) is say there are primes $q$ such that $q-1$ is divisible by each of a given finite set of primes, and not divisible by each of another given finite set of primes. You want $q-1$ to be divisible only by primes in a given finite set. Much, much harder.