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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.

Martin Sleziak
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Alexy Vincenzo
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2 Answers2

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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

lab bhattacharjee
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  • Thank you for your help! How do I deduce that $q \equiv \pm 1 (\text{mod} \ 8) ?$ – Alexy Vincenzo Oct 30 '18 at 05:27
  • @Alexy $p\equiv\pm1\pmod4,$ right? – lab bhattacharjee Oct 30 '18 at 05:37
  • Hi, it is $q \equiv \pm 1( \text{mod} \ 8)$ – Alexy Vincenzo Oct 30 '18 at 07:15
  • @AlexyVincenzo, See https://math.stackexchange.com/questions/1274743/when-is-2-a-quadratic-residue-mod-p?rq=1 or https://math.stackexchange.com/questions/1670849/when-is-2-a-quadratic-residue-in-a-finite-field – lab bhattacharjee Oct 30 '18 at 07:19
  • So it suffices to prove that $2$ is quadratic residue mod $q \ ?$ – Alexy Vincenzo Oct 30 '18 at 07:54
  • Since $2^{p} \equiv 1 (\text{mod} \ q), \ $ we have $2^{p+1} = (2^{\frac{p+1}{2}})^2\equiv 2 (\text{mod} \ q). $ Hence, $2$ is quadratic residue modulo $q.$ So, $\Big(\frac{2}{q}\Big)= (-1) ^{\frac{q^2-1}{8}}=1 \implies 16 \mid (q+1)(q-1).$ Since $q+1-(q-1) = 2,$ we cannot have $4 \mid q+1$ and $4 \mid q-1.$ It follows that either $8 \mid q+1$ or $8 \mid q-1.$ – Alexy Vincenzo Oct 30 '18 at 08:02
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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 $\frac{q-1}{2}$

sirous
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