All prime-squared Mersenne divisors are Wieferich

By Chris Caldwell

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 M_q, 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 M_q, then p = 2kq+1 for some integer k. So

2^q ≡ 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 2^p-1 ≡ 1 (mod p²), so we can raise the modular equation above to the 2k^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⋅10¹².

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