It has often been asked if all Mersenne
numbers (with prime exponents) are square-free. The theorem we prove
below makes this very likely because Wieferich primes
are rare! But before we explain this, lets pause for a short proof break.

Theorem.

Let p and q be primes. If p^{2}divides M_{q},
then 2^{(p-1)/2} = 1 (modp^{2}), so in particular, p is a Wieferich prime.

Proof.

First note that p and q must be odd. Elsewhere
we have shown that if p divides M_{q}, then
p = 2kq+1 for some integer k. So

2^{q} = 2^{(p-1)/2k}
= 1 (mod p^{2}).

Raising this to the kth power gives the first result of the theorem.
Recall that the Wieferich primes are the primes p for which
2^{p-1} = 1 (mod p^{2}), so we can raise
the modular equation above to the 2kth power to complete the proof.

Comment. The only Wieferich primes less than 4,000,000,000,000
are 1093 and 3511. The first of these does not satisfy the full force
of theorem and the second never divides an M_{q} (with
q prime), so M_{q} is square-free for all primes
less than 4^{.}10^{12}.

If we allow composite
exponents, then every odd square n^{2} divides infinitely
many "Mersennes" 2^{m}-1; just make m any multiple
of phi(n^{2}), where phi(n) is Euler's phi function.
Then we know n^{2} divides 2^{m}-1 (and
indeed b^{m}-1 for all brelatively
prime to n) by Euler's Theorem.