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3 Problems

  1. Oct 31, 2005 #1
    1.) Show that (n+1)! = 2(n-1)! mod n+2
    I finished this one. Actually very easy.

    2.) Let n > 2 be odd. Prove that if 4[(n-1)! + 1] + n = 0 mod n(n+2) holds, then n, n+2 are twin primes. Hint says to use the previous problem.

    I don't even know what to do for this problem.

    3.) Prove the converse of the theorem in the preceeding problem is also true. Thus, the two problems together constitute a necessary and sufficient condition for (n, n+2) to be a pair of twin primes.

    Obviously, if I can figure out #2, this one will be a walk in the park.

    Thanks for any help given, I appreciate it.
  2. jcsd
  3. Oct 31, 2005 #2


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    For 2), consider 4[(n-1)! + 1] + n = 0 modulo n and modulo n+2 separately. Remember Wilson's theorem.
  4. Oct 31, 2005 #3
    Alright, I think I got it. Let me just make sure before moving on, Here is what I did:

    (1) 4(n-1)!+4+n = 0 mod n
    (2) 4(n-1)!+4+n = 0 mod n+2

    (1) tells us that (n-1)! = -1 mod n since n = 0 mod n. So, by Wilson's theorem, n is prime. There was a little more work with the second equation, but came to the result of (n+1)! = -1 mod n+2 and made a similiar conclusion. Is this correct?
  5. Oct 31, 2005 #4


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    Looks good. Part 3) should be no problem now?
  6. Oct 31, 2005 #5
    I haven't looked at it, but our Professor said the first and 3rd questions are really easy. If I have any questions i'll post on here. Thanks for your help man, I really do appreciate it.
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