 mersenneforum.org I think I found proof that no odd perfect numbers exist!
 Register FAQ Search Today's Posts Mark Forums Read 2014-11-16, 11:31 #1 Philly314   Nov 2014 1 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^(n-1))(2^n-1), 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^(n-1))(2^n-1) equals a perfect number, then both 2^(n-1) and (2^n)-1 are odd. For (2^n)-1, any number n will be odd, except for n=0, but (2^(0-1))(2^0-1) equals (1/2)(-1), or -1/2, which is not an odd number, or a perfect number. For 2^(n-1), only n=1 makes it odd, because 2^(1-1) = 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.   2014-11-16, 14:04 #2 legendarymudkip   Jun 2014 23×3×5 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.   2014-11-16, 14:04   #3
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2·23·137 Posts Quote:
 Originally Posted by Philly314 First, we can all agree that all perfect numbers are in the form (2^(n-1))(2^n-1), right?
That is a definition for an even perfect number.
Quote:
 Originally Posted by Philly314 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^(n-1))(2^n-1) equals a perfect number, then both 2^(n-1) and (2^n)-1 are odd. For (2^n)-1, any number n will be odd, except for n=0, but (2^(0-1))(2^0-1) equals (1/2)(-1), or -1/2, which is not an odd number, or a perfect number. For 2^(n-1), only n=1 makes it odd, because 2^(1-1) = 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.
Very clever, you just proved that an even perfect number can't be an odd perfect number.

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.   2014-11-16, 14:58   #4
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by retina That is a definition for an even perfect number.Very clever, you just proved that an even perfect number can't be an odd perfect number. 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.
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