Ahh, this will be long. Here goes...

Proof (<==)

n = 2^(m - 1)*(2^m - 1) where m >= 2, 2^m - 1 is prime. Show n is an even perfect number.

Note that since 2^m - 1 is prime implies m is prime.

Let sigma(n) = sum of divisors.

sigma(n) = 2n

m >= 2 implies m - 1 >= 1 implies 2|n implies n is even.

From 2|n we know 2^{exponent}*(2^m - 1)

sigma(n) = sigma(2^(m - 1)*(2^m - 1)) = (fill in here after noting the following-------- that something: sigma(2^(m - 1))*sigma(2^m - 1)

gcd(2^(m - 1), 2^m - 1) = 1, and therefore we can separate sigma over each piece since sigma is multiplicative.

We know 2^m - 1 is odd; (there are no odd divisors of 2^(m - 1)

From the red we know (2^m - 1)/2 - 1 * {note sigma(p) = 1 + p} and therefore (2^m -1 + 1) .. since 2^m - 1 is prime, remember.

= 2^m*(2^m - 1)

= 2*2^(m - 1)*(2^m - 1) = 2n (do you see how I got this?)

See next post for (==> part of proof)