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Is there more than one Descartes number?

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  • djbroadhurst
    ... Bill did get one bit of maths right: 2011 - 373 = 1638 ...Mais je pense pouvoir démontrer qu il n y en a point de pairs qui soient parfaits, excepté
    Message 1 of 33 , Jul 14, 2011
      --- In primenumbers@yahoogroups.com,
      Bill Bouris <leavemsg1@...> wrote:

      > overlooked for 373 years

      Bill did get one bit of maths right:
      2011 - 373 = 1638

      "...Mais je pense pouvoir démontrer qu'il n'y en a point de
      pairs qui soient parfaits, excepté ceux d'Euclide; & qu'il
      n'y en a point aussi d'impairs, si ce n'est qu'ils soient
      composés d'un seul nombre premier, multiplié par un carré
      dont la racine soit composée de plusieurs autres nombres
      premiers. Mais je ne vois rien qui empêche qu'il ne s'en
      trouve quelques uns de cette sorte: car, par exemple, si
      22021 était nombre premier, en le multipliant par 9018009,
      qui est un carré dont la racine est composée des nombres
      premiers 3, 7, 11 & 13, on aurait 198585576189, qui serait
      nombre parfait..."

      Réne Descartes, Letter to Mersenne, November 15, 1638

      See Chapter 8, on "Descartes numbers", in the thesis
      http://tinyurl.com/6yssqa6
      of C. Wesley Nevans, for bounds on the existence
      of a second odd number of the form n = k*m
      for two integers k, m > 1 such that
      sigma(k)*(m + 1) = 2*n
      whether or not m is prime.
      This work also appears in
      http://tinyurl.com/6ec9n9x

      David
    • Mathieu Therrien
      Thx, I will rework it. ________________________________ From: Tom Hadley To: Mathieu Therrien Cc:
      Message 33 of 33 , Jul 14, 2011
        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

        [Non-text portions of this message have been removed]
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