Two or three days ago, someone found this site by Googling odd perfect numbers. I intended to do a little write-up on the subject then, but I’m a fairly chronic procrastinator (don’t ask me about the Singapore article I promised 3 months ago or else I might change my banning policy…). So in case that person is still here, here’s the shebang:
Every integer has divisors. For example, the divisors of 4 are 1, 2, and 4. It makes sense to talk about the sum of all the divisors, in this case 7. We can define a function that returns for every integer the sum of its divisors. It’s usually denoted with a lowercase Sigma, but I’m going to use s for reasons of convenience. We get, for example, s(4) = 7, s(2) = 3, s(5) = 6, s(10) = 18…
The case we’re really interested in is when s(n) = 2n. That means that n is the sum of all of its divisors, including 1 but not itself. For instance, 6 = 1 + 2 + 3, 28 = 1 + 2 + 4 + 7 + 14, 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248. We call these numbers perfect.
Now, here’s the thing: all known perfect numbers are even. There’s a formula that generates all even perfect numbers: if for some k, 2^k – 1 is prime (in which case k has to be prime), then (2^k – 1) * 2^(k-1) is perfect.
One feature of the s function is that if a and b have no factors in common except 1, then s(ab) = s(a)s(b). There’s a fairly easy intuitive proof: every factor of ab is a factor of a times a factor of b because no prime appears in the prime factorization of both a and b, and similarly every factor of a times a factor of b is a factor of ab. So in a way, you can factor the sum of all factors of ab as the sum of all factors of a times the sum of all factors of b.
Now, if 2^k – 1 is prime, then its only factors are itself and 1, so s(2^k – 1) = 2^k. And you can check that every power of 2 is almost perfect, in the sense that s(n) = 2n – 1, so s(2^(k-1)) = 2^k – 1. We can multiply them together and get exactly twice (2^k – 1)(2^(k-1)).
Actually finding prime numbers of the form 2^k – 1 is exceedingly difficult. If k is composite, then 2^k – 1 won’t be prime; even if k is prime, chances are 2^k – 1 won’t be prime. There are 43 known primes of this form, the largest of which has around 9 million digits. On the other hand, it’s possible to prove that all even perfect numbers are of that form.
Still, there are no analogous proofs about odd perfect numbers. For example, there’s no upper bound on the number of primes dividing an odd perfect number; there are lower bounds, but these can be found with simple trial and error. There’s no proof that shows that an odd perfect number can’t be divisible by a specific prime, except of course 2.