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On square divisors of Mersenne numbers, II

By John Blythe Dobson (j.dobson@uwinnipeg.ca)

Although the author is employed by the University of Winnipeg, he has no affiliation with the Mathematics Department or with any other academic department.


It has often been asked whether Mersenne numbers, i.e. numbers of the form Mp = 2p − 1 with p prime, can have square divisors. In a previous note, we reviewed the pertinent literature and demonstrated various relations which must hold between any such square divisor and its cofactor. Although the present discussion of the same subject is quite brief, it seemed better to issue it separately rather than disturbing with new content the earlier piece, which appeared nearly a year ago. This note thus refines a result given in the previous one, but does not supersede it entirely.

Refining the conditions on the cofactor of the square divisor

Conditions involving the modulus 3·16 = 48

As before, we write 2p − 1 = q2·c (where the cofactor c is not necessarily assumed to be squarefree or free of the divisor q, although its squarefree part is certainly greater than 1). We remind the reader of Euler’s result that the divisors of Mp are all of the form 2kp + 1. We previously showed, by a somewhat complicated argument, that c ≡ −1 (mod 16). While much of that argument still seems unavoidable in the context of the earlier note, so far as the linear form of c is concerned it is possible to obtain a somewhat sharper result by a much simpler line of reasoning, which we regret not having previously noticed.

As is well known, Mersenne numbers greater than M3 = 7 are all congruent to 31 (mod 48). This is easily seen, because modulo 48, 24 ≡ 16 and 162 ≡ 16, so 16k = 24k ≡ 16 for all k > 0. Thus 24k + 1 − 1 ≡ 16·2 − 1 ≡ 31, and 24k + 3 − 1 ≡ 16·8 − 1 ≡ 31, so 2p − 1 ≡ 31 for all odd p > 3.

Now any divisor q of M which is congruent to ±1 (mod 8) has its square congruent to 1 or 33 (mod 48). We may disregard the case of the residue 33, as it results from values of q divisible by 3, and 32 cannot divide any Mersenne number (since 3 divides 22 − 1 only linearly, and cannot divide any other Mersenne number of prime exponent as such numbers are relatively prime to one another). Thus, q ≡ 1, 7, 17, 23, 25, 31, 41, or 47 (mod 48), and since in every case we have

q2 ≡ 1 (mod 48),


we must therefore have

c ≡ 31 (mod 48).


This consideration establishes nothing new about q, merely emphasizing the trivial fact that it is not divisible by 3, but it narrows the number of possible residue classes to which c could belong from 3 of 48 down to 1 of 48. Specifically, recalling the fact that all divisors of Mersenne numbers are congruent to 1 modulo the exponent p, it shows that any Mersenne number M with a square divisor has also a non-square divisor, not necessarily prime or squarefree, of the form 6kp + 1 for some odd integer k congruent to 5/p ≡ 5p (mod 8).

While we have, of course, no prior knowledge as to whether or not the squarefree part of Mp is a prime, it is nonetheless instructive to consider the consequences if it is assumed to be. Then the prime c would divide 2(c − 1)/6k − 1, so it would have 2 as a cubic as well as a quadratic residue, and a result of Gauss in the theory of cubic residuacity would guarantee the existence of the (unique) representation

c = a2 + 27b2


For some coprime integers a and b. Since this form is congruent to a2 − b2 (mod 7) we cannot have a2 ≡ b2 or a ≡ ±b (mod 7), otherwise the expression would be divisible by 7. In the present case, the limitation to c ≡ 31 (mod 48) requires that a ≡ ±2 (mod 12) and b ≡ ±1 (mod 8), as is readily verified.

Still assuming c to be prime, the quadratic form expressed in (3) would of course be reflected in M itself, since we would have

M = (qa)2 + 27(qb)2.


However, it is not obvious what such a representation would allow us to infer concerning M or q.

Conditions involving the modulus 5·16 = 80

Mersenne numbers Mp with p > 3 are all congruent to 31 or 47 (mod 80) according as p ≡ 1 or 3 (mod 4). Now it happens that the squares of integers congruent to ±1 (mod 8) and which are not divisible by 5 are congruent to 1 or 49 (mod 80). This makes for a fairly simple table of solutions for the residue classes to which c could belong:

 M ≡ 3147
q2 ≡ 13147

[Note added 12 Feb. 2011: fix of an error in the bottom row, in which the two values for c were reversed.]

Combining the criteria

Combining (1) and the statement respecting the form of q2 (mod 80), we have:

q2 ≡ 1 or 49 (mod 240),


which is, after all, only a statement about squares of the required linear form. The conditions on c (mod 80) fail to improve on what is already known from (2).

With the caviat that they are of no possible value for computational purposes, we introduce some conditions on the expressions q2 + c and q2 − c, which arise naturally from the ideas developed in our first note and sharpen the results of (5) and (2). It is obvious from the definitions of q and c that

q2 + c ≡ 2 (mod 2p).


In the first note, we showed that

q2 + c ≡ 0 (mod 32).


From (1) and (2) above, we have

q2 + c ≡ 32 (mod 48).


Finally, it can be seen from the table above of solutions for c (mod 80) that both sets of solutions satisfy one of the congruences

q2 + c ≡ 32 or 48 (mod 80).


The product of the highest powers of primes in the moduli in (7) through (9) is 480, leading to the congruences

q2 + c ≡ 32 or 128 (mod 480),


which can also be expressed as

(q2 + c)/32 ≡ 1 or 4 (mod 15).


Now to consider the corresponding congruences for the difference of q2 and c. First, from the definition:

q2c ≡ 0 (mod 2p).


From the first note,

q2 − c ≡ 2 (mod 256).


From (1) and (2) above, we have

q2c ≡ 18 (mod 48).


Finally, from the table above of solutions for c (mod 80), both sets of solutions satisfy one of the congruences

q2c ≡ 34 or 50 or 66 (mod 80).


The product of the highest powers of primes in the moduli in (13) through (15) is 3840, leading to the congruences

q2c ≡ 1026 or 1794 or 3330 (mod 3840),


which can also be expressed as

(q2c)/2 ≡ 513 or 897 or 1665 (mod 1920).


If it is possible for a Mersenne number of prime exponent to have a square divisor q, such a divisor must satisfy the set of simultaneous congruences {(6), (10), (12), (16)}. This provides some hint as to why none have yet been found, for if any exist they must be quite thinly distributed, apart from the fact (discussed in our previous note) that q must possess the extremely restrictive Wieferich property.


We have shown that a Mersenne number M of prime exponent with a square divisor q must also have a non-square divisor, not necessarily prime or squarefree, congruent to 31 (mod 48) and hence of the form 6kp + 1 (k odd). While no amount of information concerning the form of this cofactor can be of any computational value in testing known divisors of Mersenne numbers for multiplicity, our hope is that, in combination with other criteria, it might help to settle theoretically the question of whether such multiple divisors can exist.

First published 22 June 2008
With minor revisions through 17 January 2016