Best Perfect Number?

As pointed out by a previous poster, Euclid knew the relationship between what we now call Mersenne Primes (primes of the form 2^n - 1) and perfect numbers.

This page [utm.edu] has more info about Mersenne primes and how they can be discovered. There is a distributed project to find Mersenne primes [mersenne.org], which can even earn you a decent sum of money. A less selfish reason is that you can find the biggest known prime number, making you famous (among about 100 geeks).

Roughly 2000 years after Euclid, Euler proved that Mersenne Primes are responsible for ALL even perfect numbers. The proof isn't too complicated, so I'll reproduce it here for convenience. (Don't feel too bad if you don't understand it, though, since this wasn't found for 2000 years...)

First of all, we'll make a distinction between factors of a number, and "proper" factors of a number. A "proper" factor of a number is any factor of a number other than the number itself. So, for example, the factors of 6 are 1, 2, 3, and 6; the "proper" factors of 6 are 1, 2, and 3. A perfect number is defined to be a number which is the sum of all its "proper" factors; 6 = 1 + 2 + 3. This means the sum of all its factors is twice the number; 6 * 2 = 1 + 2 + 3 + 6. From here on out, we'll use this latter definition of perfect number, and we won't use the artificial concept of "proper" factors.

Consider the factorization of a number into prime powers. For example, 252 = 2^2 * 3^2 * 7^1.

Every single factor of 252 is of the form 2^a * 3^b * 7^c, where a and b are in {0, 1, 2}, and c is in {0, 1}; and all numbers of this form are factors of 252. Now, consider the product (2^0 + 2^1 + 2^2) * (3^0 + 3^1 + 3^2) * (7^0 + 7^1). If we expand this using the distributive property, we see that this is a sum of all numbers of the form 2^a * 3^b * 7^c described above.

Therefore, a number is perfect if and only if this product is twice the number. We can quickly check if 252 is perfect by seeing if (2^0 + 2^1 + 2^2) * (3^0 + 3^1 + 3^2) * (7^0 + 7^1) is equal to 252 * 2. (It isn't.) Equivalently, we can check if (2^0 + 2^1 + 2^2)/2^2 * (3^0 + 3^1 + 3^2)/3^2 * (7^0 + 7^1)/7^1 is equal to 2 (I just divided both sides of the equation by 252); and given the prime power factorization of any other number, we can check in the same manner. This last check is the one we'll analyze.

Any even number will have a (2^0 + 2^1 + ... + 2^n)/2^n term in this expression. If the next prime factor of the number is any less than (2^0 + 2^1 + ... + 2^n), the product is forced to be greater than 2. If the next prime factor of the number is any greater than (2^0 + 2^1 + ... + 2^n), the product cannot be 2 because the numerator of the first term has prime factors other than 2 that are not canceled out by any of the denominators.

Thus, all even perfect numbers are a product of a power of 2 and the associated Mersenne prime. (Yes, I left out a few minor details at the end, but if you understood the rest of the proof you can easily figure them out.)

However, the matter of whether odd perfect numbers exist or not is still an open question. I'm not aware of any interesting mathematics this has generated of late, but who knows what it will take to finally solve the problem?

Christopher's post [slashdot.org].

Note Chris's distinction between "proper" factors and factors. If you were allowed to count the original number, 1 would be the only perfect number. All others would have factors that add up to at least one more than themselves.