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Math Help - Wilson's theorem

  1. #1
    Ant
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    Wilson's theorem

    Hi,

    Would the following constitute a proof of "if n is compositie then (n-1!)=0 mod n"

    We can use the chinese remainder theorem to find (n-1) mod n.

    Suppose that  n = p_{1}^{a1}, p_{2}^{a2}...p_{r}^{ar}

    Then because in each case  p_{i}^{ai} is less than (n-1) and so it divides (n-1!). We find that  n-1 =0 mod p_{i}^{ai} for all i. Then the CRT implies that n-1 =0 mod n


    Does the above make sense??

    Thanks for any help!
    Last edited by Ant; May 16th 2012 at 04:57 PM.
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  2. #2
    Member Sylvia104's Avatar
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    Re: Wilsonís theorem

    Quote Originally Posted by Ant View Post
    in each case  p_{i}^{ai} is less than (n-1)
    This is only true if i>1; it is not true if i=1 (i.e. if n is a single-prime power).

    By the way, (4-1)!\not\equiv0\mod4. The converse of Wilsonís theorem only states that if n is composite, then (n-1)!\not\equiv-1\mod n.
    Last edited by Sylvia104; May 17th 2012 at 04:15 AM.
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  3. #3
    Ant
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    Re: Wilsonís theorem

    ah yes, I didn't deal with that case. Thank you
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    Member Sylvia104's Avatar
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    Re: Wilsonís theorem

    You were almost correct though. If n is composite, we in fact have

    (n-1)!\ \equiv\ \left\{\begin{array}{rl} 2\mod n & \text{if}\ n=4 \\\\ 0\mod n & \text{if}\ n>4 \end{array}\right.

    If n>4 and n=ab where a,b\in\{2,3,\ldots,n-1\}, then (i) if a\ne b we have straightaway n=ab\mid(n-1)! (ii) if a=b then:

    a^2=n>4

    \Rightarrow\ a>2

    \Rightarrow\ n=a^2>2a

    \Rightarrow\ a,2a\in\{2,\ldots,n-1\}

    \Rightarrow\ n\mid2n=2a^2\mid(n-1)!
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  5. #5
    Ant
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    Re: Wilson's theorem

    I see, so you really needn't split it into coprime divisors of n like I did.

    I had the above proof as an exam questions - hopefully I'll still get some marks for getting close!
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