-4

By asking this question, I hope that someone, somewhere, has an answer, not to the conjecture, but the question itself. Then, that person has proved an unsolved conjecture. I do have a proof, which is quite long, that all Germain primes are of the form 6n-1. For more details, go to the question: Are there an infinity of Sophie Germain primes? My second answer provides a full proof.

mngi
  • 19
  • 2
    What is the question? And why on earth is the answer urgently needed? – lulu Aug 29 '18 at 20:11
  • I believe the maximum we can use is the Green–Tao theorem – rtybase Aug 29 '18 at 20:12
  • I'm not sure what you are trying to ask, but one interpretation is equivalent to "are there infinitely many Sophie Germain primes?" and this is (so far as I know, at time of posting) unknown. It would help if you could state your question clearly in the body of the question - where it is easier to format, and also you don't have to summarise. Is it "are there infinitely many positive integers $n$ for which $6n-1$ and $12n-1$ are both prime?" ? – Mark Bennet Aug 29 '18 at 20:12
  • Note: I don't believe you have shown that there are infinitely many Sophie Germain primes. Perhaps you meant something else? – lulu Aug 29 '18 at 20:15
  • Note that every prime of the form $12n-1$ is also of the form $6n-1$ so if there are infinitely many primes of the form $12n-1$ (which there are) then these same primes will be an infinite set of primes of the form $6n-1$ (there are also infinitely many primes of the form $12n+5$). Simply because the two sets are infinite does not guarantee that there are infinitely many pairs. – Mark Bennet Aug 29 '18 at 20:40
  • Voting to close the question as it is not clear what you are asking. If you can, please edit your post for clarity. – lulu Aug 29 '18 at 20:40
  • Mark, sorry, but I think I implied it in the notice underneath the question. – mngi Aug 30 '18 at 16:05
  • Responding to your first comment, lulu, I do mean to prove the infinity of Germain primes. I asked this question to get someone else's perspective. Thank you for your comment, though. – mngi Aug 30 '18 at 16:34
  • rtybase, sorry, but I am not sure how to answer your comment. – mngi Aug 30 '18 at 16:56
  • The question in the title is probably not what you intent. There are infinite many pairs $(a/b)$ such that both $6a-1$ and $12b-1$ are prime if $a$ and $b$ are not related. If they have to be equal, it is an open question whether infinite many positive integers $n$ exist such that $6n-1$ and $12n-1$ are both prime. The Bunyakovsky conjecture implies that this is the case. – Peter Sep 01 '18 at 19:49

1 Answers1

2

$2(6n-1)+1=12n-1$. Thus, if $6n-1$ and $12n-1$ are both prime, then $6n-1$ is a Sophie Germain prime and $12n-1$ is the associated safe prime. Thus, for both to be prime infinitely often, there would have to be infinitely many Sophie Germain primes. As this is currently an unproved conjecture (at least according to Wikipedia), you probably won't get a proof here.

joriki
  • 238,052
  • joriki, I am actually trying to prove the infinity of Germain primes. Thanks for clarification, though, but I still edited it for reasons of the question being unclear. – mngi Aug 30 '18 at 16:32
  • @mngi: I don't get it. You asked for a quick answer that proves a well-known open conjecture? – joriki Aug 30 '18 at 16:38
  • I'm trying to phrase it in a way which maybe someone had the answer to, therefore proving the conjecture. Thank you for your question. – mngi Aug 30 '18 at 16:59