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

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February 27th 2011, 05:30 PM
uberbandgeek6

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?
February 27th 2011, 05:59 PM
Bruno J.

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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