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Unsure if this is relevant: In the previous part of the question I deduced that if a number of the form $2^n-1$ is prime (a Mersenne prime), $n$ must also be prime.

Question: For primes $p$ and $p'$, where $p'=2p+1$, does $p\equiv 3\pmod{4}$ imply that $2^p\equiv 1\pmod{p'}$?

Attempt: By Euler's theorem, since $(2,p)=(2,p')=1$ we have $2^{\phi(p')}\equiv 2^{p'-1}\equiv 1\pmod{p'}$. Substituting $p'=2p+1$, we obtain $2^{p'-1}\equiv2^{2p}\equiv 1\pmod{p'}$, and thus $2^p\equiv\pm 1\pmod{p'}$.

Letting $p\equiv 4p_0+3$, then $p'\equiv 2(4p_0+3)+1 \equiv 8p_0+7$. So $p'\equiv 7\pmod{8}$.

I am stuck as to how to link this last bit to show that $2^p\equiv 1\pmod{p'}$, not $2^p\equiv -1\pmod{p'}$.

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As you pointed out, $p'$ is of the form $8k+7$. You have probably already seen the proof of the fact that $2$ is a quadratic residue of the odd prime $q$ if and only if $q\equiv \pm 1\pmod{8}$.

So $2$ is a quadratic residue of $p'$, say $2\equiv x^2\pmod{p}$. But $x^{p'-1}=x^{2p}\equiv 1\pmod{p'}$ and therefore $2^{p}\equiv 1\pmod{p'}$.

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    $\begingroup$ Yes, thank you, fixed. I am typo prone, particularly when I am typing outside in bright sunshine while smoking. $\endgroup$ – André Nicolas Mar 22 '15 at 18:07

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