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Math Help - Primitive roots

  1. #1
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    Primitive roots

    Suppose g is a primitive root modulo p (a prime) and suppose m|p-1 (1<m<p-1). How many solutions are there to the congruence:

    x^m-g \equiv 0 (mod p)

    My attempt:

    m*ind_gx \equiv ind_gg (mod \phi(p))
    \therefore m*ind_gx \equiv p-1 \equiv 0 (mod p-1)
    \therefore ind_gx \equiv 0 (mod p-1)
    \therefore x \equiv 1 (mod p)

    Hence g=1 which is a contradiction so there are no solutions.

    I don't believe my own answer! I have not used the statement that
    m|p-1 which tells me that g.c.d(m, \phi(p)) = m
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  2. #2
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    Quote Originally Posted by Kiwi_Dave View Post
    Suppose g is a primitive root modulo p (a prime) and suppose m|p-1 (1<m<p-1). How many solutions are there to the congruence:

    x^m-g \equiv 0 (mod p)
    If there is a solution x then x\equiv g^y~ (p).
    Thus, g^{my} \equiv g^1 ~ (p).
    Therefore, my\equiv 1 ~ (p-1).
    Now \gcd(m,p-1) =m and m\not | 1.
    Thus, there are no solutions to this congruence.
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