I am having difficulty with the following proof. A prime $p$ is said to be a Sophie Germain prime if $2p+1$ is also prime. It is known that if $p$ is a sophie Germain prime, then $2p+1$ is a factor of the Mersenne number $M_p$. For example, $23$ is a sophie Germain prime since $47$ is also prime, so 47 divides $M_{23}$ and thus $M_{23}$ is not a Mersenne prime.
Given the result that $(\frac{2}{47})=1$ (this is Legendre's symbol), prove that $47$ is indeed a factor of $M_{23}$.
I know Legendre's symbol means that $2^{23}\equiv1\mod 47 $ and that $M_{23}$ is of the form $2^{23}-1$. So can I say:
$2^{23}\equiv1\mod 47$
$2^{23}-1\equiv0\mod 47$
$M_{23}\equiv0\mod 47$ so $47$ is indeed a factor.
This question is worth a few marks so I can't help but feel it isn't as simple as this. Can anyone tell me where/if I may have gone wrong?
Let we consider the prime $p=47$. Is $2$ a quadratic residue $\!\!\pmod p$?
Of course, since $47$ is a number of the form $8k-1$. That ensures: $$ \left(\frac{2}{p}\right) \equiv 2^{\frac{p-1}{2}} \equiv +1\pmod{47} $$ hence $47$ is a divisor of $2^{23}-1$, since $2^{\frac{p-1}{2}}-1\equiv 0\pmod{47}$.
Consequence: if $q>3$ is a prime and $2q+1$ is a prime of the form $8k- 1$, $$2^q-1\text{ is not a prime}.$$