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Re: odd-perfect number don't exist

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  • djbroadhurst
    ... 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
    Message 1 of 33 , Jul 11, 2011
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      --- In primenumbers@yahoogroups.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.

      http://tinyurl.com/5styuvd

      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.

      http://www.lri.fr/~ochem/opn/opn.pdf

      David
    • 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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