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• ## Re: [PrimeNumbers] Re: odd-perfect number don't exist

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• ________________________________ I guess its the proof for just 1 piece out of a couple then? From: Jack Brennen To: Prime Numbers
Message 1 of 33 , Jul 11, 2011
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________________________________

I guess its the proof for just 1 piece out of a couple then?

From: Jack Brennen <jfb@...>
Sent: Monday, July 11, 2011 9:41:12 PM
Subject: Re: [PrimeNumbers] Re: odd-perfect number don't exist

Right. I'm mentioned this to him just about every time that
he posts a "proof" -- and he has never addressed the issue.

Every one of his proofs -- as far as I can tell -- claims that
such a construct can't exist. My counterexample shows that
such a construct can in fact exist, and the only reason it's
not an odd perfect number is because 22021 is divisible by 19.
But none of his arguments ever explain why such a construct
can exist with one of the factors composite, but cannot exist
with all of the factors prime.

On 7/11/2011 5:44 PM, djbroadhurst wrote:
>
>
> Jack Brennen<jfb@...> wrote:
>
>> According to your proof, why isn't N = 3^2*7^2*11^2*13^2*22021
>> an odd perfect number?
>
> Jack's point is well made; he has shown that
> sigma(N)/(2*N) = sigma(22021)/(1+22021).
> If 22021 were prime, N would be an odd perfect number.
> Of course, 22021 = 19^2*61 is not prime.
> However Bill did not mention decomposition into /primes/
> in his vague remarks; so he might not detect that N is abundant.
>
> David
>
>
>
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[Non-text portions of this message have been removed]
• Thx, I will rework it. ________________________________ From: Tom Hadley To: Mathieu Therrien Cc:
Message 33 of 33 , Jul 14, 2011
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Thx, I will rework it.

________________________________
To: Mathieu Therrien <mathieu344@...>
Sent: Thursday, July 14, 2011 4:41:26 PM

Mathieu Therrien <mathieu344@...> wrote:

>If I understood the purpose of Sigma(N/M) = N/P_1 = M*P_2   for P = P_1 * P_2 correctly,
>
>then Many solution are possibles as long as (M+1) is divided by 2 only once
>
>for example u have m=5 ; N = P_2 * m * 3 = 15*P_2  and as long that P_2 is odd
>
>So m=5 and N=45 is 1 solution
>
>
I think you have misunderstood the sigma() function.  In Pari-GP, sigma(x) is the sum of the divisors of x.  So sigma(9) = 1+3+9 = 13.

The puzzle is: Find a pair of odd integers (N,m) with m|N,
sigma(N/m)*(1+m) = 2*N, and bigomega(m) = 1.

The proposed solution, N=45, m=5 doesn't work, since
sigma(45/5)*(1+5) = sigma(9)*6 = 13*6 = 78, which is not 2*N=2*45=90.