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Math Help - Prove if a is a primitive root mod p^2, then it is a primitive root mod p

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    Prove if a is a primitive root mod p^2, then it is a primitive root mod p

    Prove that if a is a primitive root mod p^2, then it is a primitive root mod p.


    I have no idea where to go with this. I know that if a is a primitive root mod p^2 then: ord(mod p^2) a = phi(p^2) = p(p - 1), and a^(p(p-1)) = 1 mod p^2.

    Where do I go from here?
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    MHF Contributor Bruno J.'s Avatar
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    Suppose a is not a primitive root mod p, so that a^k=1 \mod p for some k<p-1.

    Note that a^{kp}-1 = (a^k-1)(a^{k(p-1)}+a^{k(p-2)}+\dots+1).

    Moreover, p \mid a^k-1 and p\mid a^{k(p-1)}+a^{k(p-2)}+\dots+1 since each of the p terms in the sum is =1\mod p. Hence p^2 \mid (a^k-1)(a^{k(p-1)}+a^{k(p-2)}+\dots+1), hence a^{kp}\equiv 1 \mod p^2, so the order of a is strictly less than p(p-1), contradicting the assumption.
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