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Old 2014-11-16, 14:04   #3
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"The unspeakable one"
Jun 2006
My evil lair

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Originally Posted by Philly314 View Post
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.
Originally Posted by Philly314 View Post
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.
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