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Let $p=2q+1$ and $q$ be prime numbers, with $q\equiv 3\:(4)$. Show that $2^q\equiv 1\:(p)$.

So far I've tried to write $q$ as an expression of $p$ and I am assuming that I have to use little Fermat somehow, since $p$ is prime. But I can't figure out how to combine those two and I fail to see an alternative. I also don't see how to use the fact that $q\equiv 3\:(4)$.

I've also tried rewriting the equation to $2^q-1\equiv 0\:(p)$ from which I need to show that $p=2q+1|2^q-1$. But I also got stuck there.

Any help is appreciated.

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  • Since $q\equiv 3\pmod{4}\implies p\equiv 7\pmod{8}$

  • You have to show $2^{\frac{p-1}{2}}\equiv 1\pmod{p}$

  • Note that $\left(\frac{2}{p}\right)\equiv 2^{\frac{p-1}{2}}\pmod{p}$ is $1$ if $p\equiv \pm{1}\pmod{8}$ and $-1$ if $p\equiv \pm{3}\pmod{8}$. Here $\left(\frac{a}{p}\right)$ denotes the Legendre Symbol and i have used Euler's criterion which says $\left(\frac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\pmod{p}$.

  • From Apostol's Analytic Number Theory Book this proof

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  • $\begingroup$ Is there a way to prove this without the use of the Legendre Symbol? I understand what you did and what it means, just wondering if there's a way without it. $\endgroup$ – SirCircle Oct 8 at 13:16
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    $\begingroup$ I have added the screen shot. Hope it helps :) $\endgroup$ – crskhr Oct 8 at 15:22

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