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

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• Right. I suspect Bill might have the basic outline of a proof that no square-free odd number can be perfect. But even there, he makes it way too complicated;
Message 1 of 33 , Jul 12, 2011
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Right. I suspect Bill might have the basic outline of a proof that
no square-free odd number can be perfect. But even there, he
makes it way too complicated; it's basically trivial to see that
if N is a square-free odd composite, sigma(N) is divisible by 4,
while 2*N is not divisible by 4, therefore sigma(N) != 2*N.

On 7/12/2011 8:29 PM, Peter Kosinar wrote:
> Bill,
>
> You seem to have missed Jack's point -- he wasn't talking
> about 22021 as being the odd-prime "candidate". It was the
> number N defined as
>
>> N = 3^2*7^2*11^2*13^2*22021
>
> which your alleged "proof" might get fooled by. There are
> two main sources of confusion. Before getting to the "hard"
> one which Jack had in mind; would you care to explain how
> your proof deals with number which are not square-free (e.g. 9,
> 25, 45, ...)?
>
> Following your line of reasoning, when looking at number 9,
> one could either use f1 = 1 and f2 = 9, or f1 = f2 = 3.
> The problem is that in neither case, the LHS (1 + f1 + f2)
> manages to express the sum of divisors of number 9 -- in
> the first case, it equals 1+1+9 = 11, in the other, it's
> 1+3+3 = 7. So yes, the proof can correctly conclude that
> LHS cannot be equal to RHS... but it since LHS wasn't equal
> to the sum of divisors of the product f1*f2 in the first
> place, this inequality says nothing about the perfect-ness
> of the number f1*f2.
>
> Peter
>
>
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>
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>
>
• 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.