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Re: odd-perfect number don't exist

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  • leavemsg1
    sorry that I ve disappointed you in the past. the fact that 22021 isn t an odd-perfect number is because it has four factors total including the 1 . the
    Message 1 of 33 , Jul 12, 2011
      sorry that I've disappointed you in the past. the fact that 22021 isn't an odd-perfect number is because it has four factors total including the '1'.
      the summation would be even-- 23622; not worring about it being abundant, it has to have and 'even' number of factors plus 1 so that 'odd' summation equals 'odd' product. 22021 disobeys the definition from that standpoint.

      --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
      >
      > Right. I'm mentioned this to him just about every time that
      > he posts a "proof" -- and he has never addressed the issue.
      >
      > Every one of his proofs -- as far as I can tell -- claims that
      > such a construct can't exist. My counterexample shows that
      > such a construct can in fact exist, and the only reason it's
      > not an odd perfect number is because 22021 is divisible by 19.
      > But none of his arguments ever explain why such a construct
      > can exist with one of the factors composite, but cannot exist
      > with all of the factors prime.
      >
      >
      >
      > On 7/11/2011 5:44 PM, djbroadhurst wrote:
      > >
      > >
      > > --- In primenumbers@yahoogroups.com,
      > > Jack Brennen<jfb@> wrote:
      > >
      > >> According to your proof, why isn't N = 3^2*7^2*11^2*13^2*22021
      > >> an odd perfect number?
      > >
      > > Jack's point is well made; he has shown that
      > > sigma(N)/(2*N) = sigma(22021)/(1+22021).
      > > If 22021 were prime, N would be an odd perfect number.
      > > Of course, 22021 = 19^2*61 is not prime.
      > > However Bill did not mention decomposition into /primes/
      > > in his vague remarks; so he might not detect that N is abundant.
      > >
      > > David
      > >
      > >
      > >
      > > ------------------------------------
      > >
      > > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
      > > The Prime Pages : http://www.primepages.org/
      > >
      > > Yahoo! Groups Links
      > >
      > >
      > >
      > >
      > >
      >
    • 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

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