Re: odd-perfect number don't exist
- --- In email@example.com,
"leavemsg1" <leavemsg1@...> wrote:
> You'll like my traditional approach!You seem to have arrived at the following conclusion:
Proposition 1: If N is an odd perfect number,
then bigomega(N) cannot be even.
That is indeed true and is a trivial consequence of
this stronger result, from Euler:
Proposition 2: If N is an odd perfect number,
then there exists a prime p = 1 mod 4 and an
exponent e = 1 mod 4 such that N/p^e is a square.
I remark that there is now a proof tree claimed for the following.
Proposition 3: If N is an odd perfect number, then
(a) N > 10^1500,
(b) bigomega(N) > 100,
(c) the largest prime power dividing N exceeds 10^62.
- Thx, I will rework it.
From: Tom Hadley <kctom99@...>
To: Mathieu Therrien <mathieu344@...>
Cc: "firstname.lastname@example.org" <email@example.com>
Sent: Thursday, July 14, 2011 4:41:26 PM
Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle
Mathieu Therrien <mathieu344@...> wrote:
>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.
>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
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.
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