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Math Help - Applying Fermat and Wilson's Theorem

  1. #1
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    Applying Fermat and Wilson's Theorem

    Hi everyone! I just had a couple questions about a few things I read in a textbook.

    There were a couple points made that did not show any proof. I was able to prove some of them but the two listed below gave me some trouble. I like to see why things are so and need the proof in order to accept its validity. I was wondering if anyone here could help me prove these?

    --- If b > 1 is not prime, then (b-1)! is not congruent to -1 (mod b)
    --- If p is an odd prime, then (p-3)! ≡ (p-1)/2 (mod p)

    Thank You!
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  2. #2
    o_O
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    #1: If you know Wilson's theorem, then what is the contrapositive of its statement?

    #2: By Wilson's theorem, (p-1)! \equiv - 1 \ (\text{mod } p)

    But: (p-1)! = (p-1)(p-2)(p-3)!

    Can you finish off?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by o_O View Post
    #1: If you know Wilson's theorem, then what is the contrapositive of its statement?
    #1 is not the contrapositive of Wilson's theorem!

    Suppose p|b, p \neq b. Then p|(b-1)!. If (b-1)!+1\equiv 0 \mod b then (b-1)!+1 = kb for some k \in \mathbb{Z}. Now p|b, p|(b-1)! but p \nmid 1, which is impossible.
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  4. #4
    o_O
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    Hmm .. you sure? Wilson's theorem is a biconditional statement. So if we take the direction that: (b-1)! \equiv -1 \ (\text{mod b}) \ \Rightarrow \ b \text{ is prime.}

    Then this is logically equivalent to saying if b is not prime, then etc. etc.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Oh, in that case you are correct! In my mind Wilson's theorem was "If p is prime, then (p-1)! \equiv -1 \mod p", but it makes sense to include the other direction as well.
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