19053Re: No more Fermat primes :)!
- Aug 1, 2007--- In email@example.com, "leavemsg1" <leavemsg1@...>
> Hi, Group, et. al.
> I have a rather simple observation that may imply that the number
> Fermat primes is finite.*m))
> Conjecture: If a Fermat number of the form F(k,m)= 2^ (2^ ((2^k)
> +1 is prime, then [2^ (2^ ((2^k)*m)) +1 == 2^q (mod 2^(m-k) +1)]a /whole
> where 'k' is /-1 or a whole number/ and 2^k <= 'm' which is
> number/ for some /odd number/ 'q'.and
> Since 'q' is /odd/, this argument further reduces to...
> [((2^k)*m) == (2n+1) mod (m-k)] and is only satisfied when k= -1
> 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-((would imply that 'n' > 1 (now, I think my argument is preserved.))
> sult of 2n == some /odd number/ or...
> Just examine the following equations which imply that n= 0 or 1
> the reduced modulo equation/argument...F4=65537.
> 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];
>((would imply that 'n' > 1 (now, I think my argument is preserved.))
> Other values for 'k' and 'm' would make 2n== some /odd number/ or
> Now, if the observation held for only one or two of the Fermat
> numbers, then I would've been hesitant to imply that the modulo ar-m)
> gument is, in fact, a reliable method for determining whether F(k,
> = 2^ (2^ ((2^k)*m)) +1 IS prime or not.It's very difficult to express one's ideas via e-mails; final draft.
> 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?
> Bill Bouris
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