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Math Help - primitive root

  1. #1
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    Post primitive root

    Show that if n has a primitive root and if the odd prime p divides n, then n is of the form p^j or 2p^j.

    Could you please show me how to do this question? Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Show that if n has a primitive root and if the odd prime p divides n, then n is of the form p^j or 2p^j.

    Could you please show me how to do this question? Thank you very much.
    Not necessarily.*)

    It really is a long proof.
    First done by Gauss (I think).
    It is long because it is divided into cases.

    *)The theorem says it is either 2,4 p^n, 2*p^n where p is odd prime.
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  3. #3
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    Can somebody help me out, I was thinking about a similar attempt.

    If n has a primitive root that means \mathbb{Z}_n under addition and multiplication will be a finite field.
    Thus,
    |\mathbb{Z}_n|=n
    That means,
    n=p^m
    Because finite fields are power of primes.

    Apparently, I missed something. I happen to know that I am wrong because it is a well known theorem that I heard of it before. Can someone spot an error?


    EDIT: I spotted my error. I was assuming \phi (n)=n-1 which is only true when it is a prime. I will now see if I can answer this question elegantly via finite fields.
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