8

Are there infinitely many primes $p$ such that $2^p-1$ is composite? Can someone help with another possible case other than the one listed below? Thanks for help in advance.

If the prime $p \equiv 3 \bmod 4$ and $p$ is a Sophie Germain prime ($2p+1$ is also prime), then $2p+1$ divides $2^p-1$, and therefore not prime. From this there should be infinitely many primes $p$ such that $2^p-1$ is composite, but the problem is the infinitude of Sophie Germain primes remains unproven. Second, if the Sophie Germain primes were proven to be infinite, we would still need to prove there are inifnitely many $3\bmod 4$, then this can be a proof that there are infinitely many primes $p$ such that $2^p-1$ is composite. Is there an alternate solution to proving my question?

Michael Hardy
  • 1
J. Linne
  • 3,238
  • 1
    To me, it is not clear what you are asking for. If there are infinite Sophie Germain primes, there are infinite composite numbers of the form $2^p-1$, that is a fact. However, both the infinitude of Sophie Germain primes and the infinitude of composite numbers of the form $2^p-1$ are open problems. Are you asking us a way to tackle an open problem among them? – Jack D'Aurizio Sep 15 '16 at 14:34
  • 1
    I think you are right. I am aware of the theorem that for any odd prime p, 2 is a square modulo p if and only if p is congruent to 1 or 7 modulo 8. – Timothy Mar 18 '20 at 02:17

1 Answers1

5

According to this page on Wikipedia, it is not known whether infinitely many Mersenne numbers with prime exponents are composite.

So the answer to the title of your question is currently no.

barak manos
  • 43,717