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Old 2014-11-16, 11:31   #1
Philly314
 
Nov 2014

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Default 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.
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Old 2014-11-16, 14:04   #2
legendarymudkip
 
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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.
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Old 2014-11-16, 14:04   #3
retina
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Quote:
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.
Quote:
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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Old 2014-11-16, 14:58   #4
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by retina View Post
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.
You know that cranks prefer to shoot their mouth off, rather than READ,
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