

It has often been asked if all Mersenne
numbers (with prime exponents) are squarefree. 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.
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. So2^{q} = 2^{(p1)/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^{p1} = 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 squarefree 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 b relatively prime to n) by Euler's Theorem. 
Another prime page by Chris K. Caldwell <caldwell@utm.edu> 