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Math Help - Proof of a congruence problem

  1. #1
    Super Member Bacterius's Avatar
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    Proof of a congruence problem

    Hello ! I'm having this little problem.
    Prove that if m is prime, then (m - 2)! \equiv 1 \pmod{m}.
    I thought it would be possible to state that (m - 2)! = \frac{(m - 1)!}{m - 1}, thus we have \frac{(m - 1)!}{m - 1} \equiv 1 \pmod{m}. From Wilson's Theorem, since m is prime, (m - 1)! \equiv m - 1 \pmod{m}, thus we are left with \frac{m - 1}{m - 1} \equiv 1 \pmod{m}, which concludes the proof. But it is actually right ? I am not really confident with division inside \mathbb{Z}/m \mathbb{Z}.
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    Quote Originally Posted by Bacterius View Post
    Hello ! I'm having this little problem.


    I thought it would be possible to state that (m - 2)! = \frac{(m - 1)!}{m - 1}, thus we have \frac{(m - 1)!}{m - 1} \equiv 1 \pmod{m}. From Wilson's Theorem, since m is prime, (m - 1)! \equiv m - 1 \pmod{m}, thus we are left with \frac{m - 1}{m - 1} \equiv 1 \pmod{m}, which concludes the proof. But it is actually right ? I am not really confident with division inside \mathbb{Z}/m \mathbb{Z}.


    Yes, it is right...you can freely divide in \mathbb{Z}\slash m\mathbb{Z} as long as you don't divide by zero, or what is the same: as long as the divisor is not a multiple of m.

    Tonio
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    Super Member Bacterius's Avatar
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    Thanks !
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