odd perfects
ok we shall start at the very beginnning (a very good place to start) when you read you begin with abc, when you deal with number theory you begin with the primes. (as opposed to do rae me) A prime number is a number that is such that it has only two integer numbers which evenly divide it. These two divisors of course being 1 and the number itself. The fundamental theorem of arithmetic is that every counting number has an unique prime factorization, or in other words a unique set of primes which multiply to it.
The second beastie we shall meet on the way to odd perfects is the divisor sigma function. The divisor sigma function is merely the sum of all the divisors of a number. The divisor sigma is usually denoted with a lower case sigma but since this is an e-mail I will just dennote it S(n) Looking at a few small numbers we can easily see that S(2) = 2+ 1 = 3 S(3) = 3 + 1= 4 S(4) = 1+ 2+ 4 = 7 S( 5) = 5+ 1 = 6 S(6)= 1+2+ 3 + 6 = 12
The most obvious way to obtain S(n) is to simply check if each number up to n divides it and take the sum of the ones that do. But this gets rather intensive rather quickly and there is a much easier and more useful way of doing things. We take a moment to note that S(a * b) = S(a) * S(b) if and only if a and b share no common factors. (a common factor by the way would be a number other than 1 that divides both a and b if two numbers share no common factors they are said to be co-prime, for instance 15 and 28 are co-prime because even though neither 15 nor 28 are prime they share no primes in their factorizations 3 * 5 and 2^2 * 7 which implies they have no common factor)
ok now for the perfects part, a perfect number is a number which satisfies 2n = S(n) such a number is 6 since 12 = 2*6 = S(6) ta da. a number whose S(n) is greater thant 2n is called abundant while a number whos S(n) is between n+1 and 2n is called deficient. Of course it is a little more natural to just ignore n when finding the sum of the divisors of n in which case these classifications make a little more sense since then a number is deficient if its sum is less than itself and abundant if it is greater than itself and of course perfect if the sum is equal to itself. Obviously a number of perfect numbers exist, in fact at least 40+ perfect numbers exist one for each mersenne prime. It is not known whether there is an infinite number of perfects but that is likely the case. Anyway it is not known at all however whether even a single perfect number is odd.
obviously an odd perfect number cant have a 2 in its factorization or it wouldn't be odd! Since all the numbers in its factorization are odd we have some product of odd primes n = A1^a1 A2^a2 A3^a4..... Aj^aj and so on consider for a moment that each prime is coprime with the others so we can split the determination of S(n) into S(A1^a1) * S(A2^a2) * .... * S(Aj^aj). All of these sigmas are of the form of the sum of a number which is a power of a prime number. The only proper divisors the power of a prime number P^a are all the powers of the prime P^b with b<=a if the number werent a power of the prime then it wouldn't divide P^a and so wouldn't contribute to S(P^a). Meaning that S(P^a) consists of a sum of a+1 odd numbers (since of course p^0 or 1 is in S(P^a) too!) and so if a is even then S(P^a) will be a sum of an odd number of odds which will itself be odd. But if a is odd then S(P^a) will be even.
Going back and taking this piece of information to look at the odd perfect again we see that if more than one of the Aj's had an odd power then it would mean that S(n) would be divisible by 4. But that would mean that S(n)/2 would be even. Which is of course a contradiction since S(n) / 2 = n and we started with the stipulation that n be odd. Therefore n has a single "special" prime in its factorization whose exponent is odd. The rest of the exponents are even giving us the rather interesting insight that n = P^(2m -1) * K^2