Proof of Form of Divisors of Mersenne Numbers
Any prime divisor of the Mersenne Primes
\[M_p=2^p-1\]
where \[p\]
is an odd prime, is off the form \[2kp+1\]
for some positive integer \[k\]
<,br /> ProofLet
\[q\]
be a prime divisor of \[M_p=2^p-1\]
. Then \[2^p \equiv 1 \; (mod \; q)\]
.From Fermat's Little Theorem
\[2^{q-1} \equiv 1 \; (mod \; q)\]
. Let the order of 2 modulo \[q\]
be \[c\]
so the \[q, \; q-1\]
are multiples of \[c\]
. As \[p\]
is prime and \[c \gt 1\]
(since \[2^1 \equiv 1 \; (mod \; q))\]
is not possible, it follows that \[c =p\]
. Hence \[q-1\]
is a multiple of \[p\]
.Writing
\[q-1=tp\]
for some integer \[t\]
gives \[q=tp+1\]
. If \[t\]
is odd then \[q\]
is even which cannot be the case. Hence \[t\]
is positive and even so we can write \[p=2k\]
then \[q=2kp+1\]
. 
