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Math Help - If a has order n - 1 modulo n, then n is prime.

  1. #1
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    If a has order n - 1 modulo n, then n is prime.

    I'm stuck on this question. Prove that if a has order n - 1 modulo n, then n is prime.

    So this means a^(n-1) is congruent to 1 (mod n). This looks exactly like Fermat's Theorem for primes but I'm not sure how to prove that n is prime. I tried contradiction but am getting nowhere. Hints please. Thank you.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: If a has order n - 1 modulo n, then n is prime.

    Quote Originally Posted by mathguy71421 View Post
    I'm stuck on this question. Prove that if a has order n - 1 modulo n, then n is prime.

    So this means a^(n-1) is congruent to 1 (mod n). This looks exactly like Fermat's Theorem for primes but I'm not sure how to prove that n is prime. I tried contradiction but am getting nowhere. Hints please. Thank you.
    Let n be composite.

    \phi(n)<n-1

    O_n(a) | \phi(n)

    Clearly, O_n(a)\neq n-1 - Contradiction!
    Last edited by alexmahone; November 10th 2011 at 12:40 PM.
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  3. #3
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    Re: If a has order n - 1 modulo n, then n is prime.

    Ah Thank you so much. I figured that phi(n) =/= n - 1 but couldn't quite connect that to a contradiction in my head. Thanks so much.
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