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19053Re: No more Fermat primes :)!

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  • leavemsg1
    Aug 1, 2007
      --- 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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