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Old 2015-06-23, 11:44   #5
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"Forget I exist"
Jul 2009

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Originally Posted by retina View Post
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

Last fiddled with by science_man_88 on 2015-06-23 at 11:47
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