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Math Help - show the inverse?

  1. #1
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    show the inverse?

    show that if r is a primitive root modulo the positive integer m then the r^-1 is also a primitive root modulo m, where r^-1 is an inverse of r modulo m
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  2. #2
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    r^{-1} \equiv r^{\phi(m)-1}(\bmod m). Now use the fact that r^k is a primitive root if and only if \gcd(k,\phi(m)) = 1 where k is a positive integer.
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  3. #3
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    Hello,

    A precision (just in case) :

    \phi(m) stands for the Euler Totient function. It also designs the cardinal of the subgroup containing the invertible elements taken from Z/mZ.
    This is why a^{\phi(m)} is always \equiv 1 [m].
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