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Re: [PrimeNumbers] Re: odd-perfect number don't exist

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  • Peter Kosinar
    Bill, You seem to have missed Jack s point -- he wasn t talking about 22021 as being the odd-prime candidate . It was the number N defined as ... which your
    Message 1 of 33 , Jul 12, 2011
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      Bill,

      You seem to have missed Jack's point -- he wasn't talking
      about 22021 as being the odd-prime "candidate". It was the
      number N defined as

      > N = 3^2*7^2*11^2*13^2*22021

      which your alleged "proof" might get fooled by. There are
      two main sources of confusion. Before getting to the "hard"
      one which Jack had in mind; would you care to explain how
      your proof deals with number which are not square-free (e.g. 9,
      25, 45, ...)?

      Following your line of reasoning, when looking at number 9,
      one could either use f1 = 1 and f2 = 9, or f1 = f2 = 3.
      The problem is that in neither case, the LHS (1 + f1 + f2)
      manages to express the sum of divisors of number 9 -- in
      the first case, it equals 1+1+9 = 11, in the other, it's
      1+3+3 = 7. So yes, the proof can correctly conclude that
      LHS cannot be equal to RHS... but it since LHS wasn't equal
      to the sum of divisors of the product f1*f2 in the first
      place, this inequality says nothing about the perfect-ness
      of the number f1*f2.

      Peter
    • 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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