- --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:>

of

> 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...

>>>

>>>

from

>

> 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

>>>

>>>

prime

> 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?

>

> Respectfully,

>

> Bill Bouris

>

- 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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