Let $p$ be a prime number, and let $a$ be a primitive root $\mod p$.
Is it true that $a^m$ is a primitive root if and only if $\gcd(m,p-1)=1$?
One direction is correct: if $a^m$ is a primitive root, then let $d = gcd(m,p-1)$. Then $dq_1 = m, dq_2 = p-1$, where $q_1,q_2\in\mathbb{Z}$, and $q_2\leq p-1$. We then have that $$ \left(a^m\right)^{q_2} = \left(a^{dq_1}\right)^{q_2} = \left(a^{q_1}\right)^{dq_2} = \left(a^{q_1}\right)^{p-1} $$ From Fermat's little theorem, the rightmost element is congruent to $1\mod{p}$, since $a < p$. From this, we conclude that since $a^m$ is a primitive root, and $q_2\leq 18$, then necessarily $q_2 = 18$, which means $d=1$.
I am not sure about the other direction: does $gcd(m,p-1)=1$ imply $a^m$ is primitive, given $a$ is primitive?