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Topic: Odd perfect number (Read 451 times) |
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kaushiks.nitt
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Odd perfect number
« on: Jun 13th, 2009, 10:27am » |
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This is a partial proof for the famous odd perfect number conjecture .
Consider any number N = pa1 * pa2 * ... pak . ( Note a1,...ak are the indices for prime numbers , Actually N can be of the form pa1^r1 * pa2^ r2 *.... pak^rk . However i am interested only in those numbers which can be expressed as N = pa1 * pa2 * ... pak ) . Thus the factors would be 1,pa1 , pa2 , .. pak , pa1*pa2,.. pa1*pak, pa2*pa3 ,.. Similarly taken three at a time and so on till taken all at a time.
A number is perfect if all the factors expect the one taken all at a time i.e N ( or all the prime factors taken all at a time ) should be equal to N.
If the above statement is not clear then let me explain it with a number which has just three factors i.e. N =pa*pb*pc.
Then N is perfect if 1+ pa+pb+pc+ pa*pb + pb*pc + pa*pc = pa*pb*pc.
i.e 1/ pa*pb*pc + 1/ pb*pc + 1/ pa*pc + 1/pa*pb + 1/pa + 1/pb + 1/pc =1.
i.e (pa+1)* (pb+1) * ( pc +1) / pa *pb*pc =2 . ( This is just rearrangement of the above equation).
The similar thing can be expanded for k prime factors also.
Thus we have (pa1 + 1)* ( pa2 +1) *..... *( pak +1) / pa1*pa2*... *pak = 2.
I have written pa1, pa2 ,.. pak in ascending order .
Thus if N is odd then none of the pak's = 2.
Also all the number in the numerator can be can be expressed as 2^k * pbk . It is trivial that pbk < pak.
(Note if pa1 is 3 . then it becomes 2^2 * 1 . No pbk term this is an exception).
We note that there is no term in the numerator which will cancel pak( largest prime number) . Thus the ratio is never an integer thus it is not equal to 2 also . Hence there doesn't exist any odd perfect number N = pa1 * pa2 * ... pak .
Also from the above statement we can infer two things.other than 6 there can be no even perfect number also of the form N = pa1 * pa2 * ... pak .
all even perfect numbers are of the form 2^n * (2^n+1 - 1) .
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Obob
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Re: Odd perfect number
« Reply #1 on: Jun 13th, 2009, 12:10pm » |
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Everything you've said looks correct. I wouldn't call it a "partial proof" of the odd perfect number conjecture, though; all you've shown is that the only squarefree perfect number is 6.
You've probably already seen it, but the Wikipedia page for perfect numbers has some partial results towards the odd perfect number conjecture. One condition stated there is that all the prime factors except for one of them must occur with an even power.
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| « Last Edit: Jun 13th, 2009, 12:16pm by Obob » |
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