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Math Help - Prime number

  1. #1
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    Prime number

    Hi,

    How to show that (n-1)!\equiv-1[n] \Longrightarrow n \in \mathbb{P}????
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  2. #2
    Super Member Deadstar's Avatar
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    I take it \mathbb{P} are the prime numbers? This is Wilsons theorem.

    For any a \in \{1, 2, \dots, p-1\} there's an a' \in \{1, 2, \dots, p-1\} with aa' \equiv 1 (mod p). Further, a' = a i ff a = 1 or p-1. Hence, in (p-1)!, the numbers forming this product can be cancelled in pairs, apart from 1 and p-1. Hence result.
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  3. #3
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    Quote Originally Posted by Deadstar View Post
    I take it \mathbb{P} are the prime numbers? This is Wilsons theorem.

    For any a \in \{1, 2, \dots, p-1\} there's an a' \in \{1, 2, \dots, p-1\} with aa' \equiv 1 (mod p).


    I'm afraid this doesn't work: this presumes you already know p is prime, which is precisly what you're tring to prove!


    Further, a' = a i ff a = 1 or p-1. Hence, in (p-1)!, the numbers forming this product can be cancelled in pairs, apart from 1 and p-1. Hence result.
    .
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  4. #4
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    Quote Originally Posted by bhitroofen01 View Post
    Hi,

    How to show that (n-1)!\equiv-1[n] \Longrightarrow n \in \mathbb{P}????

    Suppose n=dk\,,\,\,d>1\Longrightarrow d\mid (n-1)!\,\,\,and\,\,\,also\,\,\,d\mid n\Longrightarrow d\mid 1 , contradiction!

    Tonio
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by tonio View Post
    Suppose n=dk\,,\,\,d>1\Longrightarrow d\mid (n-1)!\,\,\,and\,\,\,also\,\,\,d\mid n\Longrightarrow d\mid 1 , contradiction!

    Tonio
    Oh yeah! I never noticed the =>. I misread it as prove Wilsons formula with n \in prime numbers...
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