20141116, 11:31  #1 
Nov 2014
1_{8} Posts 
I think I found proof that no odd perfect numbers exist!
First, we can all agree that all perfect numbers are in the form (2^(n1))(2^n1), right?
So, if a number is odd, then it has two factors that are both odd. I can prove this by drawing a multiplication table. I'm just going to do 2 & 3, but you can check for yourself. 2 3 2 4 6 3 6 9 The only pair of factors that have an odd product are the factors that are both odd; 3 and 3. If this is the case, then if (2^(n1))(2^n1) equals a perfect number, then both 2^(n1) and (2^n)1 are odd. For (2^n)1, any number n will be odd, except for n=0, but (2^(01))(2^01) equals (1/2)(1), or 1/2, which is not an odd number, or a perfect number. For 2^(n1), only n=1 makes it odd, because 2^(11) = 1, but that would mean that the other factor would = (2^1)1 = 1. 1*1 = 1. Does this mean that 1 is an odd perfect number? Otherwise, there is no odd perfect number. 
20141116, 14:04  #2 
Jun 2014
1111000_{2} Posts 
It is only even perfect numbers that have that form. As well as this, one of the factors is even, because it is a power of 2.

20141116, 14:04  #3  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5×1,201 Posts 
Quote:
Quote:
Now there is just the small problem remaining to prove that an odd number can't be a perfect number. BTW: Even the Wikipedia page could have saved you all this embarrassment. 

20141116, 14:58  #4  
Nov 2003
2^{2}·5·373 Posts 
Quote:


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