Re: odd-perfect number don't exist
- --- In firstname.lastname@example.org,
"djbroadhurst" <d.broadhurst@...> wrote:
> As Peter remarked, Bill deals only with the caseHere is the work:
> of an odd square-free number that is the product
> of an even number of prime divisors.
> It's hard to imagine a greater degree of ignorance
> of Euler's result.
"Tractatus de numerorum doctrina capita sedecim, quae supersunt"
Here is the chapter:
Cap. 3: "De summa divisorum cujusque numeri"
Euler's result is at the end of Section 109:
"Sicque talis numerus perfectus hujusmodi habebit formam
existente P numero impari, et 4*n+1 primo."
As I remarked to Warren, we are indeed privileged to be able
to access Euler's original work, on-line.
David (as ever, in awe of Leonhard)
- Thx, I will rework it.
From: Tom Hadley <kctom99@...>
To: Mathieu Therrien <mathieu344@...>
Cc: "email@example.com" <firstname.lastname@example.org>
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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