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 p2 divides Mq, then 2(p-1)/2 ≡ 1 (modp2), 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 Mq, then p = 2kq+1 for some integer k. So

2q ≡ 2(p-1)/2k ≡ 1 (mod p2).

Raising this to the kth power gives the first result of the theorem. Recall that the Wieferich primes are the primes p for which 2p-1 ≡ 1 (mod p2), so we can raise the modular equation above to the 2kth 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 Mq (with q prime), so Mq is square-free for all primes less than 4⋅1012.

If we allow composite exponents, then every odd square n2 divides infinitely many "Mersennes" 2m-1; just make m any multiple of ϕ(n2), where ϕ(n) is Euler's ϕ function. Then we know n2 divides 2m-1 (and indeed bm-1 for all b relatively prime to n) by Euler's Theorem.

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