Let $p$ be an odd prime and $q$ be a prime such that $q \mid 2^{p}-1.$ Prove that $p \mid \dfrac{q-1}{2}.$
My attempt:
By Euler's Theorem, $2^{q-1} \equiv 1 (\text{mod} \ q),$ so $2^{\frac{q-1}{2}} \equiv \pm 1(\text{mod} \ q).$ How do I relate this to the order of $2$ modulo $q?$
Appreciate any advice, thank you.
Best Answer
If ord$_q2=d>1$
$d$ must divide $g=(p,q-1)$
If $p\nmid (q-1),g=1\implies d|1,d=?$
Else $p|(q-1)$
As $q-1$ is even, odd $p$ will divide $(q-1)/2$ as well