Number Theory – Infinitely Many Primes of the Form 2^n+c

nt.number-theoryprime numbers

At the time of writing, question 5191 is closed with the accusation of homework. But I don't have a clue about what is going on in that question (other than part 3) [Edit: Anton's comments at 5191 clarify at least some of the things going on and are well worth reading] [Edit: FC's excellent answers shows that my lack of clueness is merely due to ignorance on my own part] so I'll ask a related one.

My impression is that it's generally believed that there are infinitely many Mersenne primes, that is, primes of the form $2^n-1$. My impression is also that it's suspected that there are only finitely many Fermat primes, that is, primes of the form $2^n+1$ (a heuristic argument is on the wikipedia page for Fermat primes). [EDIT: on the Wikipedia page there is also a heuristic argument that there are infinitely many Fermat primes!]

So I'm going to basically re-ask some parts of Q5191, because I don't know how to ask that a question be re-opened in any other way, plus some generalisations.

1) For which odd integers $c$ is it generally conjectured that there are infinitely many primes of the form $2^n+c$? For which $c$ is it generally conjectured that there are only finitely many? For which $c$ don't we have a clue what to conjecture? [Edit: FC has shown us that there will be loads of $c$'s for which $2^n+c$ is (provably) prime only finitely often. Do we still only have one $c$ (namely $c=-1$) for which it's generally believed that $2^n+c$ is prime infinitely often?]

2) Are there any odd $c$ for which it is a sensible conjecture that there are infinitely many $n$ such that $2^n+c$ and $2^{n+1}+c$ are simultaneously prime? Same question for "finitely many $n$".

3) Are there any pairs $c,d$ of odd integers for which it's a sensible conjecture that $2^n+c$ and $2^n+d$ are simultaneously prime infinitely often? Same for "finitely often".

Best Answer

Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.

[edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]

Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences). I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:

www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf

Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339.

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