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• ... 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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