## odd-perfect number don't exist

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• Hi, Group. I had someone contact me anonymously about my proof. We argued back and forth in e-mails for over a week. Finally, he convinced me that my work was
Message 1 of 33 , Jul 10 3:55 PM
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Hi, Group.

We argued back and forth in e-mails for over a week.
Finally, he convinced me that my work was incorrect.
So, I quickly corrected my proof and posted it; It's
better than I EVER imagined; I was able to complete
it in only a few minutes.

Please visit www.oddperfectnumbers.com to view it at

Thanks,

Bill Bouris
Aurora, IL
• Thx, I will rework it. ________________________________ From: Tom Hadley To: Mathieu Therrien Cc:
Message 33 of 33 , Jul 14 1:55 PM
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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.