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 2⁶⁴ (the search limit as of February 2026) are 1093 and 3511. A short calculation shows that the squares of these primes divide no M_q. The next possible p with p² dividing some M_q is the next possible Wieferich prime, which means p > 2⁶⁴. Since p² divides M_q, we have (2⁶⁴)² < p² ≤ M_q = 2^q-1, so 2¹²⁸ < 2^q-1 and q > 128, that is, M_q is square-free for all q < 128. This is not very helpful. For q somewhat larger than 128, M_q has been factored completely. One can look at the factors and see that no prime is repeated. This works for all q < 1213 (as of February 2026), so M_q is square-free for all q < 1213.

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.