## Fermat conjecture; please post it

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• Hello, Professor Caldwell. I like your website.  Could you please post the following conjecture: if n = 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod (2^(n-1)+1)
Message 1 of 6 , Nov 11, 2009
Hello, Professor Caldwell.

if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod (2^(n-1)+1) == -1]

(or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is prime.

it requires someone as formidable as Lucas, Lehmer, etc. to prove it.

examples...
n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!

n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!

n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!

n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!

I wish someone had the technical expertise to prove it; it's valid, and
I've studied it... trying to come up with a proof. Share it with a close
colleague, if you like.

Bill Bouris
• A couple of easy counterexamples can be found, for n = 16 and n = 36, both of which meet your hypothesis, but both of which are known not to be prime.
Message 2 of 6 , Nov 11, 2009
A couple of easy counterexamples can be found, for n = 16 and n = 36,
both of which meet your hypothesis, but both of which are known not
to be prime.

Bill Bouris wrote:
> Hello, Professor Caldwell.
>
> I like your website. Could you please post the following conjecture:
>
> if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod (2^(n-1)+1) == -1]
>
> (or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is prime.
>
>
> it requires someone as formidable as Lucas, Lehmer, etc. to prove it.
>
> examples...
> n= 2; F(n)= 17; 17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!
>
> n= 3; F(n)= 257; 257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!
>
> n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!
>
> n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!
>
> I wish someone had the technical expertise to prove it; it's valid, and
> I've studied it... trying to come up with a proof. Share it with a close
> colleague, if you like.
>
>
> Bill Bouris
>
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
>
>
• Hi Bill, ... I believe you were trying to say something like this (at least that s what your sample calculation suggests): If n = 2 and F(n) = 2^(2^n) + 1.
Message 3 of 6 , Nov 11, 2009
Hi Bill,

> if n>= 2 and F(n)= 2^(2^n)+1, then iff
> [F(n) mod (2^(n-1)+1) == -1] (or) [F(n) mod (2^(n-1)+1) == -1],
> then F(n) is prime.
>
> n= 2; F(n)= 17; 17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!
> n= 3; F(n)= 257; 257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!
> n= 4; F(n)= 65537; 65537 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!
> n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!

I believe you were trying to say something like this (at least that's what

If n >= 2 and F(n) = 2^(2^n) + 1. then, if either of the following two
conditions holds, F(n) is a prime.
Condition 1: F(n) mod (2^(n-1)+1) == -1
Condition 2: F(n) mod (2^(n-1)-1) == -1

> I wish someone had the technical expertise to prove it; it's valid, and
> I've studied it... trying to come up with a proof.

> it requires someone as formidable as Lucas, Lehmer, etc. to prove it.
> examples...

... and it takes a few milliseconds in Pari to prove it... wrong.

Let n = 16. Then, we have
2^(n-1)+1 = 32769,
F(n) = 2^(2^16) + 1,
F(n) mod 32769 = -1.

Yet, F(n) mod 825753601 = 0, thereby proving compositeness of F(n) (in
order to avoid nitpickers, note that F(n) > 825753601).

Myth^H^H^H^HConjecture busted!

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• ... It also fails for n = 256: {if(Mod(2,36986355*2^258+1)^(2^256)+1==0 &&Mod(2,2^255+1)^(2^256)+2==0,print(fail))} fail David
Message 4 of 6 , Nov 11, 2009
Jack Brennen <jfb@...> wrote:

> A couple of easy counterexamples can be found,
> for n = 16 and n = 36

It also fails for n = 256:

{if(Mod(2,36986355*2^258+1)^(2^256)+1==0
&&Mod(2,2^255+1)^(2^256)+2==0,print(fail))}
fail

David
• whoa, I got a lot of responses!; didn t have time to look at any of them. I made the correction to the typo earlier, but sent the e-mail to myself; this proof
Message 5 of 6 , Nov 12, 2009
whoa, I got a lot of responses!; didn't have time to look at any of them.
I made the correction to the typo earlier, but sent the e-mail to myself;
this proof makes it rock solid with the typo displayed and supplying the
proof... hope no one's mad; do you like it PROFESSOR Caldwell ??? your
website is immense if nothing else... done with maths for a while...

Professor Caldwell, et. al.

I couldn't resist... here's my idea for proof:

theorem: iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime!

if F(n)== -1 mod (r +1), then r= 2^h >1 and (r+1)*b= F(n) +1;

so, if (r +1)*b = 2^(2^n)+2, then b must equal 2; so, r +1=

2^(2^n-1) +1 implies... 2^h= 2^(2^n -1) implies... h= 2^n -1;

and if 2^n is replaced by n, then 2^n -1 is rel. prime to n-1.

now, (2^(n-1) +1) is mutually exclusive with (2^(2^n -1) +1),

and thus, if both modulators produce the same result, then

iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime; the other

case of (r -1) would be proved similarly.

*QED

I noticed it in an hour the night before & proved it last night.

Bill

--- On Wed, 11/11/09, leavemsg1 <leavemsg1@...> wrote:

> From: leavemsg1 <leavemsg1@...>
> Subject: Re: Fermat conjecture; please post it
> To: "Bill Bouris" <leavemsg1@...>
> Date: Wednesday, November 11, 2009, 9:16 AM
> small typo... look below.
>
> Bill Bouris <leavemsg1@...> wrote:
> >
> > Hello, Professor Caldwell.
> >
> following conjecture:
> >
> > if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod
> (2^(n-1)+1) == -1]
> >
> > (or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is
> prime.
>
> the second half should read ... F(n) mod (2^(n-1)-1) == -1;
> sorry.
>
> >
> >
> > it requires someone as formidable as Lucas, Lehmer,
> etc. to prove it.
> >
> > examples...
> > n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is
> prime!
> >
> > n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1;
> 257 is prime!
> >
> > n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 ==
> +3; 65537 is prime!
> >
> > n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15
> = +2; composite!
> >
> > I wish someone had the technical expertise to prove
> it; it's valid, and
> > I've studied it... trying to come up with a
> proof.  Share it with a close
> > colleague, if you like.
> >
> >
> > Bill Bouris
> >
>
>
>
• ... did you mean if ? But it s still wrong for n=16 (and n=36) : F(16) = -1 (mod 2^15+1) but F(16) is known to be composite, as Peter wrote, it is divisible
Message 6 of 6 , Nov 12, 2009
> theorem: iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime!

did you mean "if" ?

But it's still wrong for n=16 (and n=36) :
F(16) = -1 (mod 2^15+1)
but F(16) is known to be composite,
as Peter wrote, it is divisible by 825753601.

> so, if (r +1)*b = 2^(2^n)+2, then b must equal 2; so, r +1=

I think here you use that
2^(2^n-1)+1 has no factor of the form 2^h+1,
but this is not true in general...
For example with n=16 you have
2^65535+1 which is divisible by 2^3+1 = 9.

[Sorry, in my previous mail I misread the definition of r vs 2^h, please ignore...]

Maximilian
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