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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.

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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

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$2^{p-1}-1 ≡0\mod p$$2^p-2≡0\mod p$$2^p-1≡1\mod p=kp+1$

$q|2^p-1$$q| kp+1$$kp|q-1$

$q-1$is even so p must divides the odd $\frac{q-1}{2}$

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