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Math Help - Proof involving Wilson's theorem.

  1. #1
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    Proof involving Wilson's theorem.

    Hey, I was just wondering gif someone could help me prove the following.

    ((p-1)/2)!)^ 2=(+ or -) 1 mod (p), where p is prime.

    THanks
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  2. #2
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    Quote Originally Posted by HelloWorld2 View Post
    Hey, I was just wondering gif someone could help me prove the following.

    ((p-1)/2)!)^ 2=(+ or -) 1 mod (p), where p is prime.

    THanks

    For p an odd prime, we have:

    \displaystyle{(p-1)! = 1\cdot 2\cdot\ldots\cdot(p-1)=1\cdot\ldots\cdot \frac{p-1}{2}\cdot \frac{p+1}{2}\cdot\ldots\cdot (p-1)!=(-1)^{\frac{p-1}{2}}\left(1\cdot\ldots\cdot \frac{p-1}{2}\right)^2} (why?),

    so...

    Tonio

    Pd. If you in fact understand the above, then you must be able even to tell when exactly we'll

    get 1 and when -1 modulo p.
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  3. #3
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    well, we'll get 1 when p=1mod(4), and -1 when p=-1mod(4). I realy don't get how you got that equaility.
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  4. #4
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    Quote Originally Posted by HelloWorld2 View Post
    well, we'll get 1 when p=1mod(4), and -1 when p=-1mod(4). I realy don't get how you got that equaility.


    What're the inverses modulo p of 1\,,\,2\,\ldots,\frac{p-1}{2} ? Well, there you go.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    What're the inverses modulo p of 1\,,\,2\,\ldots,\frac{p-1}{2} ? Well, there you go.

    Tonio
    But, for example, 3\cdot4\equiv1\pmod{11}, and \displaystyle3,4\leq\frac{11-1}{2} (If I understand what you meant.).

    The observation here is that for \displaystyle1\leq x\leq\frac{p-1}{2}, we have \displaystyle\frac{p+1}{2}\leq p-x\leq p-1.
    Then noting that p-x\equiv-x\pmod p shows that \displaystyle\frac{p+1}{2}\cdots(p-2)(p-1)\equiv(-\frac{p-1}{2})\cdots(-2)(-1)\pmod p.
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  6. #6
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    Quote Originally Posted by melese View Post
    But, for example, 3\cdot4\equiv1\pmod{11}, and \displaystyle3,4\leq\frac{11-1}{2} (If I understand what you meant.).


    I don't think you do, but for sure I don't understand you: "inverse" means "additive inverse", not "multiplicative inverse", as

    I think it was clear from the context of my answer, so your example is irrelevant, I believe, and what is relevant is what you wrote

    after that, which is what I was trying the OP to realize by himself.

    Tonio



    The observation here is that for \displaystyle1\leq x\leq\frac{p-1}{2}, we have \displaystyle\frac{p+1}{2}\leq p-x\leq p-1.
    Then noting that p-x\equiv-x\pmod p shows that \displaystyle\frac{p+1}{2}\cdots(p-2)(p-1)\equiv(-\frac{p-1}{2})\cdots(-2)(-1)\pmod p.
    .
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