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Re: odd-perfect number don't exist

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  • djbroadhurst
    ... Here is the work: Tractatus de numerorum doctrina capita sedecim, quae supersunt http://eulerarchive.maa.org/pages/E792.html Here is the chapter: Cap. 3:
    Message 1 of 33 , Jul 13, 2011
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > As Peter remarked, Bill deals only with the case
      > 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.

      Here is the work:

      "Tractatus de numerorum doctrina capita sedecim, quae supersunt"
      http://eulerarchive.maa.org/pages/E792.html

      Here is the chapter:

      Cap. 3: "De summa divisorum cujusque numeri"
      http://www.math.dartmouth.edu/~euler/docs/originals/E792.chp3.pdf

      Euler's result is at the end of Section 109:

      "Sicque talis numerus perfectus hujusmodi habebit formam
      (4*n+1)^(4*lambda+1)*P*P,
      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)
    • Mathieu Therrien
      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.



        ________________________________
        From: Tom Hadley <kctom99@...>
        To: Mathieu Therrien <mathieu344@...>
        Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
        Sent: Thursday, July 14, 2011 4:41:26 PM
        Subject: Re: [PrimeNumbers] Post-Cartesian Puzzle





         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.
         
        Tom Hadley

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