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Let $n\geq 1$ an integer, in this post we denote the greatest prime dividing $n$ as $\operatorname{gpf}(n)$.

See it you want the article from MathWorld Greatest Prime Factor.

While I was writing equations and congruence relations involving different arithmetic functions and $\operatorname{gpf}(n)$ I wondered about if there exist infinitely many primes of the form $$2\cdot n^{\operatorname{gpf}(n)}+1.\tag{1}$$

I was doing experiments with different factors instead our first factor $2$. Now I have no intuition if should have infinitely many primes of the form $(1)$.

Question. What work can be done about the existence of infinitely many primes of the form $$2\,m^{\operatorname{gpf}(m)}+1$$ when $m\geq 1$ runs over positive integers? Or well, can you provide a heuristic or reasoning whereby we should think that there exist only a finite number of them? Many thanks.

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    $\begingroup$ usually the first thing one tries with problems like this is a search. How many primes have you found for $m≤1000$, say? $\endgroup$ – lulu Jun 27 '18 at 13:49
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    $\begingroup$ Worth noting: very little is known about prime values of algebraic expressions. Even for a comparatively simple form like $n^2+1$ almost nothing is known. Sometimes, of course, one can show that there are only finitely many such primes (if there is a congruence or an algebraic factoring). Here, though, taking $m=6,9,12,27$ (based on random search) certainly give primes which tends to suggest that there is no trivial division out there. $\endgroup$ – lulu Jun 27 '18 at 13:53
  • $\begingroup$ Up to $m=1000$ I found about $35$ primes (I believe that different primes, my output provide me those primes, but maybe there is some repetition). Many thanks for your comment and attention @lulu $\endgroup$ – user243301 Jun 27 '18 at 13:53
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    $\begingroup$ My guess is that for particular sorts of $m$ you'll be able to show that there aren't any primes, but in general it won't be possible to find affirmative results. That, unfortunately, is how these sort of questions tend to play out. $\endgroup$ – lulu Jun 27 '18 at 13:55
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    $\begingroup$ Chances are good that infinite many primes of the form $2^m\cdot 3^n+1$ with $m=3j+1$ and $n=3k$ exist, which would be enough to prove the conjecture. A proof that infinite many primes of the given form exist will probably be out of reach. $\endgroup$ – Peter Jun 27 '18 at 21:03
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Trivially for every prime $n$, $gpf(n)=n$. So your conjecture or question is the same as the Sophie Germain prime conjecture, which says infinitely many primes $n$ such that $2n+1$ is also prime. This has yet to be proven.

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    $\begingroup$ The OP would be asking about $2n^n+1$, not $2n+1$, for $n$ prime. $\endgroup$ – Barry Cipra Jul 7 '18 at 2:34
  • $\begingroup$ Many thanks as you see from previous comment was a typo in your expression. I believe that my question is similar than: are there infinitely/finitely many Fermat primes? $\endgroup$ – user243301 Jul 7 '18 at 7:01
  • $\begingroup$ @yesterday that also remains unproven and I personally argue infinitely many Fermat primes but I will not explain further because this is no place to discuss debate $\endgroup$ – J. Linne Jul 8 '18 at 16:41
  • $\begingroup$ @BarryCipra was right. The $n$ such that $2n^n+1$ is prime are rare: 12, 18, 251, 82992, and only the term $251$ is a prime. On the other hand, the $n$ such that $2 n^{\operatorname{gpf}(n)}+1$ is prime are plenty enough: 6, 9, 12, 14, 21, 27, 30, 72, 92, 96, 126, 150, 162, 165, 192, 225, 242, 251, 261, 341, 345, 384, 392, 420, 455, 462, 500, 540, 629, 684, 704, 779, 800, 825, 990, 1350, 1386, 1404, 1482, 1512, 1625, 1725, 1808, 2001, .... $\endgroup$ – Jeppe Stig Nielsen Mar 3 at 8:05

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