Fermat primes

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• 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):
Message 1 of 2 , Sep 25, 2007
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

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

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• ... Fermat or Fantasy ????? 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,
Message 2 of 2 , Oct 14, 2007
--- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...>
wrote:
>
> 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
>

Fermat or Fantasy ?????

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
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