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[((n-1)/2)!]^2 = -(-1)^[(n-1)/2] (mod n)

  1. Apr 5, 2008 #1

    xax

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    p is prime, not 2. Thanks in advance
     
  2. jcsd
  3. Apr 5, 2008 #2

    mathman

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    There is no p in your expression!
     
  4. Apr 5, 2008 #3
    he had to have meant (mod p)
     
  5. Apr 6, 2008 #4

    xax

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    p = n
     
  6. Apr 6, 2008 #5
    This is not a difficult problem, but you must understand Wilson's Theorem. Wilson, by the way, was never a mathematician, and as a student went on to become a lawyer.

    He is credited with noticing the theorem, but was unable to prove it. So much for the immortal glory of a name theorem!
     
  7. Apr 6, 2008 #6

    :rofl:

    So what is your question and what have you tried to solve it?
     
  8. Apr 7, 2008 #7

    xax

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    I thought it was clear what the question was(since the others understood): prove that this is true for every n prime, n!= 2. Using the Wilson theorem, I didn't go far:
    ((n-1)/2)!*((n+1)/2)*((n+3)/2)*...*(n-1) = -1 (mod n).
     
  9. Apr 7, 2008 #8

    xax

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    Nevermind guys, I figured it out. Thanks for your input.
     
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