I would like to propose generalization of this question :
Let $p$ be a prime number such that : $p\equiv 1 \pmod 4$
Show that $~k\cdot p \pm a~$ is a primitive root modulo $p~$ iff
$a$ is a primitive root modulo $p$ , where $k \in \mathbb{Z}$ .
So we want to show that :
If $(kp\pm a) ^m \equiv 1 \pmod p ~$ then $(p-1) \mid m$
According to Freshman Dream Theorem it follows that :
$(kp+(\pm a)) ^p \equiv (kp)^p +(\pm a)^p \pmod p$
And from Fermat Little Theorem we know that :
$a^{p-1} \equiv 1 \pmod p$
How could we proceed from this point ?