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Re:Prime Question

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  • asen@theory.saha.ernet.in
    ... Now,if n=m what do we have? 2^n=1+n*p This implies n can never be even...except of course n=0...then it is trivial. If n is not even...cann t say offhand!
    Message 1 of 5 , May 30, 2001
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      > > Hi!
      > > First of all...the two n's on either side do not seem to be the same.So
      > > you basically say
      > > 2^n-1=m*p.Right?We'll say about n=m in a while.
      > > Now it seems trivial.
      > > 2^n-1 is odd and we know any odd number is factorisable into odd integers.
      > > Now taking one of the prime factors as p and the rest as m the result
      > > follows.
      > > Did I clear the point or did I miss it altogether?
      > > Again,for Mersenne primes m=1.
      Now,if n=m what do we have?
      2^n=1+n*p
      This implies n can never be even...except of course n=0...then it is
      trivial.
      If n is not even...cann't say offhand!


      > >
      > > On Wed, 30 May 2001 paulmillscv@... wrote:
      > >
      > > > Hi to all,
      > > > Apologies for not being on the list recently but the local TV
      > > > station did a rerun of the Pink Panther movies. So, I wish to
      > > > have "speaks" with the group.
      > > >
      > > > I have 'good reason to believe' that
      > > > 2^n - 1 = n*p for some integer n, p a prime.
      > > >
      > > > n is odd, "I know that, I know that.."
      > > >
      > > > Can you prove me wrong, right!
      > > >
      > > > regards
      > > > Paul Mills
      > > > Keniworth,
      > > > England.
      > > >
      > > >
      > > > Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
      > > > The Prime Pages : http://www.primepages.org
      > > >
      > > >
      > > >
      > > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
      > > >
      > > >
      > >
      > >
      >
      >
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