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Math Help - Wilsons Proof help please

  1. #1
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    Wilsons Proof help please

    The proof is:

    Prove that the value of (p-1)! mod p is always p-1.

    Not really sure how to approach this so any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by clockingly View Post
    The proof is:

    Prove that the value of (p-1)! mod p is always p-1.

    Not really sure how to approach this so any help would be greatly appreciated!
    This is Wilson's theorem.

    The idea is to pair elements say p=13 and a_1,...,a_{13} where a_i\in \{1,2,...,12\}. Given a_1 \not 1,12 we can find a_2 so that a_1a_2 = 1 because this is the solution to congruence. And a_3a_4 = 1 and so on ... until we only have a_{10}=1 and a_{11}=11. So have a_1a_2...a_{10}a_{11} = 11.
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  3. #3
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    Thanks! So basically the proof is this?:

    In order to compute (p-1)!, you have to find the product of the integers 1 through p-1. In this set of integers, there are groups of two elements (excluding the group 1 and p-1) that can be multiplied together to get a number (p-1)!/(p-1), which equals p-1, which is congruent to 1 mod p.
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  4. #4
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    Quote Originally Posted by clockingly View Post
    Thanks! So basically the proof is this?:

    In order to compute (p-1)!, you have to find the product of the integers 1 through p-1. In this set of integers, there are groups of two elements (excluding the group 1 and p-1) that can be multiplied together to get a number (p-1)!/(p-1), which equals p-1, which is congruent to 1 mod p.
    Exactly. It is like the trick Gauss did when he was a little boy when he was asked to compute 1+2+...+100 he paired them like (100+1)+(99+2)+... where 101 appears 50 times. Now here the idea is similar except that the pairing is not as straightforward as in Gauss' trick. But however such a pairing exists. And that is what we do.
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