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Thread: Any number theorists out there?

  1. #1

    Default Any number theorists out there?

    Hey everybody, I'm having trouble with this one math problem. It seems so simple, but I'm just not getting it -_-

    It goes like this:

    Show that there are infinitely many primes that are not twin primes.

    I feel like I can use Dirichlet's theorem here, but how, I'm not really sure... anyone have any idea? Any help at all would be appreciated :3

  2. #2


    The result follows from Wilson's Theorem and Dirichlet's Theorem. Consider p! and (p - 1)! - 1 for any prime p greater than five. By Wilson's Theorem, p divides (p - 1)! + 1, so p does not divide (p - 1)! - 1. It follows that p! and (p - 1)! - 1 are coprime. Thus, it follows from Dirichlet's Theorem that for infinitely many natural numbers n, n(p!) + (p - 1)! - 1 is prime.

    So, for infinitely many n, n*p*(p - 1)! + (p - 1)! - 1 is prime. For this to be a twin prime, then either n*p*(p - 1)! + (p - 1)! - 3 is prime or n*p*(p - 1)! + (p - 1)! + 1 is prime. The former is impossible because it is clearly divisible by 3. The second is impossible because, by Wilson's Theorem, it is divisible by p.

    So, there are infinitely many isolated primes.

    You could alternatively omit Wilson's theorem entirely and just examine 120n + 23 (equivalent to fixing p=5 in the proof above) under Dirichlet's theorem from the outset, as this can clearly never be a twin prime, but doing that provides no clue as to the motivation for examining those specific numbers (other than that they work).

    edit: Come to think of it, if you're gonna go down the "here are some numbers which work" route, just do 21n + 5.
    Last edited by Axiom; 25-Sep-2013 at 19:43.

  3. #3


    I had completely forgotten that I made this thread! xD

    Anyway, I ended up figuring it out on my own after another hour of staring at it. I ended up doing something close to the second method, I think the number I used was 15n + 23. Thanks for the help though!

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