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Re: [PrimeNumbers] Re: No more Fermat primes :)!

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  • Joseph Moore
    I have to say, I have a lot of trouble following any argument that uses a modulus of 1. Could you use more words, more generalized equations, describe more
    Message 1 of 3 , Aug 1, 2007
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      I have to say, I have a lot of trouble following any
      argument that uses a modulus of 1. Could you use more
      words, more generalized equations, describe more
      specifically where your special cases arise, and maybe
      outline the overall idea? The conjecture itself is
      not at all clear to me. I don't understand the /
      notation you're using.

      Joseph.


      --- leavemsg1 <leavemsg1@...> wrote:

      > --- In primenumbers@yahoogroups.com, "leavemsg1"
      > <leavemsg1@...>
      > wrote:
      > >
      > > Hi, Group, et. al.
      > >
      > > I have a rather simple observation that may imply
      > that the number
      > of
      > > Fermat primes is finite.
      > >
      > > Conjecture: If a Fermat number of the form F(k,m)=
      > 2^ (2^ ((2^k)
      > *m))
      > > +1 is prime, then [2^ (2^ ((2^k)*m)) +1 == 2^q
      > (mod 2^(m-k) +1)]
      > > where 'k' is /-1 or a whole number/ and 2^k <= 'm'
      > which is
      > a /whole
      > > number/ for some /odd number/ 'q'.
      > >
      > > Since 'q' is /odd/, this argument further reduces
      > to...
      > > [((2^k)*m) == (2n+1) mod (m-k)] and is only
      > satisfied when k= -1
      > and
      > > m= 0, 2; or k= 0 and m= 1, 3; or k= 1 and m= 2.
      > For all other 'k's
      > &
      > > whole numbers 'm' > 3, the last equation gives the
      > incompatible re-
      > > sult of 2n == some /odd number/ or...
      >
      > >>>
      > ((would imply that 'n' > 1 (now, I think my argument
      > is preserved.))
      > >>>
      >
      > >
      > > Just examine the following equations which imply
      > that n= 0 or 1
      > from
      > > the reduced modulo equation/argument...
      > >
      > > k=-1, m=0;
      > > [0 == (2n+1) mod 1]=> [-0 == (2n) mod 1]=> [2n=
      > 0]=> [n= 0]; F0=3,
      > >
      > > k=0, m=1;
      > > [1 == (2n+1) mod 1]=> [0 == (2n) mod 1]=> [2n=
      > 0]=> [n= 0]; F1=5,
      > >
      > > k=-1, m=2;
      > > [1 == (2n+1) mod 3]=> [0 == (2n) mod 3]=> [2n=
      > 0]=> [n= 0]; F2=17,
      > >
      > > k=0, m=3;
      > > [3 == (2n+1) mod 3]=> [2 == (2n) mod 3]=> [2n=
      > 2]=> [n= 1]; F3=257,
      > >
      > > k=1, m=2;
      > > [2 == (2n+1) mod 1]=> [1 == (2n) mod 1]=> [2n=
      > 0]=> [n= 0];
      > F4=65537.
      > >
      > > Other values for 'k' and 'm' would make 2n== some
      > /odd number/ or
      >
      > >>>
      > ((would imply that 'n' > 1 (now, I think my argument
      > is preserved.))
      > >>>
      >
      > > Now, if the observation held for only one or two
      > of the Fermat
      > prime
      > > numbers, then I would've been hesitant to imply
      > that the modulo ar-
      > > gument is, in fact, a reliable method for
      > determining whether F(k,
      > m)
      > > = 2^ (2^ ((2^k)*m)) +1 IS prime or not.
      > >
      > > This is just a rough draft... but I believe it's a
      > great indicator.
      > >
      > > Could someone work a few other modulo sentences to
      > convince me that
      > > it's not just a hoax?
      > >
      > > Respectfully,
      > >
      > > Bill Bouris
      > >
      > It's very difficult to express one's ideas via
      > e-mails; final draft.
      >
      >




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