Prove that $2p+1$ is a divisor of $2^p-1$

Question

Suppose that $p$ is a Sophie Germain prime and $p=3 mod 4$. Want to prove that $2p+1$ is a divisor of $2^p-1$.

I got a hint that I should prove that $2$ is a square mod $2p+1$ along the way, but I don't really see the connection and wonder how it works.

Answer 1

Working mod $q:=2p+1$, If $0\not\equiv 2\equiv x^2$, then $2^p-1\equiv (x^2)^p-1\equiv x^{2p}-1$ and so $\equiv 0$ by Fermat's little theorem.

Now $q\equiv 7\pmod 8$ so $2$ is indeed a quadratic residue.

Answer 2

Using Fermat's little Theorem, for $q=2p+1$ prime and since $(2,q)=1$ we get
$2^{2p}\equiv 1 (q)$. So that $q|(2^p-1)(2^p+1)$. It can only happen one of the following: $q|(2^p-1)$, $q|(2^p+1)$, since $q$ is prime. The one that happens is the first one, since for the Legendre symbol we have $\big ( \frac{2}{q} \big )=(-1)^{\frac{q^2-1}{8}}=1$. Then $2^{\frac{q-1}{2}}=2^p\equiv 1 (q)$.