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#1 |
Nov 2014
1 Posts |
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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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#2 |
Jun 2014
23×3×5 Posts |
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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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#3 | ||
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·32·7·53 Posts |
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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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#4 | |
"Bob Silverman"
Nov 2003
North of Boston
2×33×139 Posts |
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