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An Introduction to the Theory of Numbers

Exercise 2.8.18 ($gg'$ is not a primitive root of $p$)

Show that if $g$ and $g^{'}$ are primitive roots modulo an odd prime $p$ , then $g g^{'}$ is not a primitive root modulo $p$ .

Answers

Proof. Since $g$ is a primitive root modulo $p$ , there is some integer $k$ such that $g^{'} \equiv g^{k} (\mod p)$ .

By Lemma 2.33, $g^{k}$ is a primitive root if and only if $k \land (p - 1) = 1$ . Therefore $k$ is odd.

Then $g g^{'} = g^{k + 1}$ , where $2 ∣ (k + 1) \land (p - 1)$ . Hence, by the same lemma, $g g^{'}$ is not a primitive root modulo $p$ . □

2024-09-13 09:17

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