- Group,

Q: Why are the number of them... finite?

A: They are bounded by two 'modulo' conditions,

namely...

G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1))

H(x): 2^(x+1) +1 == (2n+1) mod (x+1)

G. H. Hardy and E. M. Wright predicted the latter

condition and would have labeled the Fermat primes as

'pair-wise' when an F(odd x)/F(even x) gave the result

of [n=1]/[n=0], respectively. I noticed that this

'modulo' equation explains why they thought that the

Fermat primes were consecutively linked.

I conjecture that if two consecutive Fermat numbers

are 'pair-wise' by H(x) and also 'congruent' by G(x),

then F(odd x)/F(even x) are necessarily prime!

It's an idea that would undoubtedly be hard to prove,

but never-the-less, H(x) can be easily calculated, and

G(x) can be verified using the 'Try GMP!' evaluator on

the googled 'GNU Bignum Library' website.

That's why F(1)/F(2) and F(3)/F(4) are the only Fermat

primes in the sequence; the G(x) and H(x) 'modulo'

clocks are wound together so tightly.

Bill

____________________________________________________________________________________

Boardwalk for $500? In 2007? Ha! Play Monopoly Here and Now (it's updated for today's economy) at Yahoo! Games.

http://get.games.yahoo.com/proddesc?gamekey=monopolyherenow - --- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...>

wrote:>

Fermat or Fantasy ?????

> Group,

>

> Q: Why are the number of them... finite?

>

> A: They are bounded by two 'modulo' conditions,

> namely...

>

> G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1))

> H(x): 2^(x+1) +1 == (2n+1) mod (x+1)

>

> G. H. Hardy and E. M. Wright predicted the latter

> condition and would have labeled the Fermat primes as

> 'pair-wise' when an F(odd x)/F(even x) gave the result

> of [n=1]/[n=0], respectively. I noticed that this

> 'modulo' equation explains why they thought that the

> Fermat primes were consecutively linked.

>

> I conjecture that if two consecutive Fermat numbers

> are 'pair-wise' by H(x) and also 'congruent' by G(x),

> then F(odd x)/F(even x) are necessarily prime!

>

> It's an idea that would undoubtedly be hard to prove,

> but never-the-less, H(x) can be easily calculated, and

> G(x) can be verified using the 'Try GMP!' evaluator on

> the googled 'GNU Bignum Library' website.

>

> That's why F(1)/F(2) and F(3)/F(4) are the only Fermat

> primes in the sequence; the G(x) and H(x) 'modulo'

> clocks are wound together so tightly.

>

> Bill

>

> more info...

Let F(x)= 2^(2^x)+1 represent a Fermat number such that `x' is

a /whole number/.

The first few... F(0) = 3, F(1) = 5, F(2) = 17, F(3) = 257, F(4) =

65537... are all prime. F(5) has a factor of 641.

Here's evidence that "No more Fermat primes can exist!".

I have characterized the notion of `pair-wise'ness that G. H. Hardy

and E. M. Wright have conjectured about `F(odd x)/F(even x)' compan-

ions.

Fermat numbers are `pair-wise' iff... the condition H(x): 2^(x+1)+1 ==

(2n+1) (mod (x+1)) implies that [n=1]/[n=0], respectively and in suc-

cession.

Fermat numbers are `prime' iff... the `pair-wise' compan-ions both

satisfy the congruency G(x): F(x+1) == 2^q (mod (2^x +1)); the `pair-

wise' companions must imply an integral solution for `q'.

F(0) can't be evaluated under G(x), so it's considered to be similar

to a geometric point at/near infinity.

If F(1)= 5 and F(2)= 17, then we have...

F(1)=> H(0): 3 == (2n+1) (mod 1) implies [n=1], and...

F(2)=> H(1): 5 == (2n+1) (mod 2) implies [n=0].

F(1) & F(2) are `pair-wise', but are they `prime'???

G(0): 5==(1 = 2^0)(mod 2) and G(1): 17==(2 = 2^1)(mod 3); F(1) & F(2)

are considered `prime' because the congruency, G(x), has the integral

solutions 0 & 1, respectively.

Note: determining the primality of a Fermat number is possible using G

(x) because of its unusual `double-exponential' form.

If F(3)= 257 and F(4)= 65537, then we have...

F(3)=> H(2): 9 == (2n+1) (mod 3) implies [n=1], and...

F(4)=> H(3): 17 == (2n+1) (mod 4) implies [n=0].

F(3) & F(4) are `pair-wise', but are they `prime'???

G(2): 257==(2 = 2^1)(mod 5) and G(3): 65537==(8 = 2^3)(mod 9); F(3) &

F(4) are `prime' again because of the congruency G(x) has integral

solutions,... namely 1 & 3.

Someone may eventually notice the following:

[The `pair-wise' Fermat candidates] have the form...

If F(Mp)=> H(Mp-1): 2^(Mp)+1 == (2n+1) mod(Mp) and F(Mp+1)=> H(Mp): 2^

(Mp+1)+1 == (2n+1) (mod(Mp+1)) both imply [n=1], [n=0], respectively,

then the F(odd x)/F(even x) companions are `pair-wise', but are

they `prime'???

where Mp is the usual Mersenne prime.

They are `prime' iff both are congruent to 2^q, for some q.

F(Mp):G(Mp-1): 2^(2^(Mp))+1 == 2^q (mod(2^(Mp-1)+1)) and

F(2^y):G(2^y-1): 2^(2^(2^y))+1 == 2^q (mod(2^(2^y-1)+1))

using the `Try GMP!' feature on the GNU Bignum Library web-site to

evaluate the modulo equation, G(x), when it becomes too large.

Again, Mp is the usual Mersenne prime, and 2^y is the power of two

that is just one larger than a Mp.

Look at the `2^q' values in following table...

For those `x' values that passed the `pair-wise' condition:

(x+1) | (x) | (2^q)== (after applying the G(x) congruency)

1 0 1 == 2^0

2 1 2 == 2^1

They are `pair-wise' and `prime'!

3 2 2 == 2^1

4 3 8 == 2^3

They are `pair-wise' and `prime'!

...

7 6 62 <> 2^q

8 7 17 and so on...

They are `pair-wise' but `composite'(not congruent)

...

31 30 257

32 31 17

They are `pair-wise' but `composite'...

...

127 126 85070591730234615865843651857942052862 <> 2^q

128 127 17

They are `pair-wise' but `composite'...

...

8191 8190 (odd), however large it is...

8192 8191 17

They are `pair-wise' but `composite'...

...

131071 131072 (evaluator couldn't compute modulo)

131072 131073 (same)

...

Can someone compute 2^(2^(131071+1))+1 == 2^q (mod ((2^131071)+1))...

to further end the mystery ???

I believe that another Fermat prime will not be found.

Bill