Loading ...
Sorry, an error occurred while loading the content.

Fermat conjecture; please post it

Expand Messages
  • Bill Bouris
    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
    • 0 Attachment
      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.

      Thanks in advance,

      Bill Bouris
    • Jack Brennen
      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
      • 0 Attachment
        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.
        >
        > Thanks in advance,
        >
        > Bill Bouris
        >
        >
        >
        >
        >
        > ------------------------------------
        >
        > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
        > The Prime Pages : http://www.primepages.org/
        >
        > Yahoo! Groups Links
        >
        >
        >
        >
        >
      • Peter Kosinar
        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
        • 0 Attachment
          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
          your sample calculation suggests):

          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
        • djbroadhurst
          ... 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
          • 0 Attachment
            --- In primenumbers@yahoogroups.com,
            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
          • Bill Bouris
            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
            • 0 Attachment
              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.
              >
              > --- In primenumbers@yahoogroups.com,
              > Bill Bouris <leavemsg1@...> 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.
              >
              > 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.
              > >
              > > Thanks in advance,
              > >
              > > Bill Bouris
              > >
              >
              >
              >
            • maximilian_hasler
              ... 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
              • 0 Attachment
                > 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
              Your message has been successfully submitted and would be delivered to recipients shortly.