## Proof Of Condition for Prime Dividing Mersenne Number

Theorem
If
$p$
and
$2p+1$
are prime numbers, where
$p \equiv 3 \; (mod \; 4)$
then
$2p+1$
divides the Mersenne Number
$M_p=2^p-1$
.
Since
$2p+1$
is prime,
$2^{(2p+1)-1} \equiv 1 \; (mod \; 2p+1) \rightarrow 2^{2p} -1 \equiv 0 \; (mod \; 2p+1)$
by Fermat's Little Theorem/
$2^{2p}-1=(2^p-1)(2^p+1)$
so by Euclid's Lemma either
$(2p+1) | (2^p-1)=M_p$
or
$(2p+1) | (2^p+1)$
.
$p \equiv 3 \; (mod \; 4) \rightarrow p=4k+3$
for some integer
$k$
, then
$2p+1=2(4k+3)+1=4k+7$
.
Suppose
$(2p+1) | (2^p+1)$
.
$2^{4k+3-1}+1=2^{4k+2}+1 \equiv 0 \; (mod \; 4k+7)$
which is impossible since
$4^{4k+2} \equiv 0 \; (mod \; 4)$
.
Hence
$(2p+1) | M_p$
.

Alternative Proof
Let
$p, \; q=2p+1$
be prime.
$p=4k+3 \rightarrow q= 2p+1=8k+7 \equiv 7 \; (mod \; 8)$
. This Theorem on The Quadratic Character of 2 gives
$(2/q)-1$
so that 2 is a quadratic residue of
$q$
. Euler's Criterion gives
$2^{\frac{q-1}{2}} \equiv 1 \; (mod \; q)$
, so that
$q=2p+1$
divides
$2^{\frac{q-1}{2}}-1=2^p-1=M_p$
.