Sum of Divisors Equals a Power of 2

 

In 1959 the Polish mathematician Waclaw Sierpinski noted that the sum of the divisors of an integer N>1 is a power of 2 if and only if N is a product of distinct Mersenne primes. (See "Sur les nombres dont la somme de diviseurs est une puissance du nombre 2" Calcutta Math. Soc. Golden Jubilee Commemoration 1958/59, Part I. Calcutta: Calcutta Math. Soc., pp. 7-9, 1963.) The Problems section of a 1990 issue of Mathematics Magazine challenged itís readers to prove this assertion.

 

Let N > 1 be an integer with the prime factorization

 

 

where the pj are distinct primes. Since the sum of divisors of N, denoted by σ(N), is multiplicative, we have

 

 

from which it follows that σ(N) is a power of 2 if and only if each of σ(pjaj) is a power of 2. Now, for any prime power pa we have

 

 

For this sum to be even, both p and a must be odd. Thus there is a non-negative integer m such that a = 2m + 1, so we can factor (1+p) out of equation (1) to give

 

 

Clearly σ(pa) is a power of 2 if and only if all of its factors are powers of 2, so we have

 

 

for integers u > 1 and v ≥ 0. Equation (2) proves that p must be a Mersenne prime, because it has the form p = 2u − 1, where u is obviously a prime because 2ab − 1 has the non-unit factor 2a − 1 for any integers a,b > 1. Thus a necessary condition for σ(N) to be a power of 2 is that N be the product of Mersenne primes.

 

To prove that N is necessarily the product of distinct Mersenne primes, suppose that equation (3) were satisfied for some positive integer m (corresponding to an exponent aj > 1 in the prime factorization of N). Then v > 0, and m must be odd, so we have m = 2r+1 for some non-negative integer r. We could then factor (1 + p2) from the left side of (3) to give

 

 

which implies that (1 + p2) is a power of 2. However, from equation (2) we also have

 

 

proving that 1 + p2 has an odd factor greater than 1, contradicting the previous equation. It follows that the only solution of equation (3) (if p is a Mersenne prime) is the trivial solution with m = v = 0. Therefore, each non-zero exponent in the prime factorization of N must equal 1, so N must be a product of distinct Mersenne primes.

 

Finally, we observe that the stated condition is also sufficient, since σ(p) = 1 + p is (by definition) a power of 2 if p is a Mersenne prime.

 

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