Quote:
Originally Posted by retina
What about when n=0?
What about the more general k^n?
A much more interesting proof would be to show that no odd number can be a perfect number.

For n=0 k^n is odd but a list of proper divisors doesn't exist, the second statement can be partially worked out since odd +odd =even only odd numbers with an odd number of proper divisors can be perfect. Which also means that for k=odd only n=even need be considered, and yes I know this was likely all rhetorical