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Topic: Consecutive Perfect Numbers (Read 626 times) |
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FiBsTeR
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Re: Consecutive Perfect Numbers
« Reply #1 on: Sep 26th, 2007, 8:57pm » |
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Isn't this just a restatement of "Are there any odd perfect numbers," which hasn't been shown yet?
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srn437
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Re: Consecutive Perfect Numbers
« Reply #2 on: Sep 26th, 2007, 9:44pm » |
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It's like trying to find an even non-integer.
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Aryabhatta
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Re: Consecutive Perfect Numbers
« Reply #3 on: Sep 27th, 2007, 12:47am » |
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on Sep 26th, 2007, 8:57pm, FiBsTeR wrote:| Isn't this just a restatement of "Are there any odd perfect numbers," which hasn't been shown yet? |
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It isn't. The statement in question, if true does imply the existence of odd perfect numbers, but not vice versa.
Basically, disproving that statement does not prove that there are no odd perfect numbers...
Note: I came across this in a math forum and this supposedly has a solution. Feel free to use the web for odd perfect number related theorems. If required we can move this into the easy/hard/putnam forum, as appropriate.
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Grimbal
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Re: Consecutive Perfect Numbers
« Reply #4 on: Sep 27th, 2007, 12:54am » |
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Given the sequence of known perfect numbers,
6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, 191561942608236107294793378084303638130997321548169216
I wouldn't seriously consider the possibility that 2 successive perfect numbers exist. It is not a proof, but I would bet my life on it.
PS: But the fact that there are so few perfect numbers should make it easier to show no 2 are consecutive. So the challenge is to find an elegant proof of that. It is much less to prove than to prove that no perfect number is odd.
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| « Last Edit: Sep 27th, 2007, 1:05am by Grimbal » |
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towr
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Re: Consecutive Perfect Numbers
« Reply #5 on: Sep 27th, 2007, 1:39am » |
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odd perfect numbers are of the form 12m + 1 or 36m + 9
even perfect numbers have the form 2n-1(2n - 1) where (2n - 1) is a prime
considering mod 4 means the odd perfect number would have to be one larger.
So look for
2n-1(2n - 1) = 12m
2n-1(2n - 1) = 36m + 8
For the first case we can reason that (2n - 1) is a prime other than 3 (we know there's no solution for n=2) and 2n-1 is indivisible by 3, so there can't be adjacent perfect numbers in that case.
For the second case if we look at even n, and consider modulo 3, we'd have (2n - 1) = 0 (mod 3), and so it can't work (0  2 (mod 3)). For odd n, we have 2n-1=1 (mod 3), so it reduces to (2*2n-1 - 1) = 1 2 (mod 3), so it patently fails as well.
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| « Last Edit: Sep 27th, 2007, 1:53am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Aryabhatta
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Re: Consecutive Perfect Numbers
« Reply #6 on: Sep 27th, 2007, 2:24pm » |
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Looks right to me. Nicely done.
Could you please provide a proof/(link to proof) to the odd perfect number being of the form 12m+1 or 36m+9, if not already on wikipedia?
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towr
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Re: Consecutive Perfect Numbers
« Reply #7 on: Sep 27th, 2007, 3:16pm » |
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on Sep 27th, 2007, 2:24pm, Aryabhatta wrote:| Could you please provide a proof/(link to proof) to the odd perfect number being of the form 12m+1 or 36m+9, if not already on wikipedia? |
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It is mentioned on the wiki page under "minor results" that odd perfect numbers should have that form, but it doesn't give a proof.
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Eigenray
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Re: Consecutive Perfect Numbers
« Reply #8 on: Sep 27th, 2007, 4:50pm » |
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on Sep 27th, 2007, 2:24pm, Aryabhatta wrote:| Could you please provide a proof/(link to proof) to the odd perfect number being of the form 12m+1 or 36m+9, if not already on wikipedia? |
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One can use the ideas here and here. Basically, (a) n can't be 3 mod 4, (b) n can't be 2 mod 3, and (c) 3 | n implies 9 | n.
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