# All prime-squared Mersenne divisors are Wieferich

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 (mod*p*^{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*= 2*kq*+1 for some integer*k*. So2

^{q}≡ 2^{(p-1)/2k}≡ 1 (mod*p*^{2}).Raising this to the

*k*th 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 2*k*^{th}power to complete the proof. ∎

**Comment.** The only Wieferich primes less than 6,700,000,000,000,000
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 ϕ(*n*^{2}), where ϕ(*n*) is Euler's ϕ 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.