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FLT

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  • paulmillscv@yahoo.co.uk
    Hello, Thanks to Chris Nash (!), David Broadhurst, Marcel Martin, Tom Hadley, Alan Wong and Michael Hartley for your comments on my FLT paper. Yes, I have
    Message 1 of 6 , Sep 4, 2001
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      Hello,
      Thanks to Chris Nash (!), David Broadhurst, Marcel Martin, Tom
      Hadley, Alan Wong and Michael Hartley for your comments on my FLT
      paper. Yes, I have found a bunker with my tee shot. X can be
      divisible by p in 3.37. But the news is that I have fixed it over
      the weekend and I am now writing up version 2. So the second shot is
      soaring towards the green.
      I have now put up some comments on version 1, at the link in my
      last post. I think you will enjoy the second shot. Stay tuned.

      Regards
      Paul Mills.
    • Hadley, Thomas H (Tom), NLCIO
      Paul, You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p (mod p) even when a and b are not rationals, given a definition of modularity
      Message 2 of 6 , Sep 4, 2001
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        Paul,
        You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p
        (mod p) even when a and b are not rationals, given a definition of
        modularity that says for example t = 1/2 mod p means that t = 1/2 + k.p for
        some integer k, which seems to me a reasonable definition.

        This is not true. A counter-example is easy:

        let a = 2, b = 1/5, p = 3
        (a+b)^p mod p would be (2+1/5)^3
        = (11/5)^3
        = 1331/125
        = 9 + 206/125
        = 3*3 + 206/125
        = 206/125 mod 3

        However,

        a^p + b^p would be 2^3 + (1/5)^3
        = 8 + 1/125
        = 3*2 + 2 + 1/125
        = 3*2 + 251/125
        = 251/125 mod 3

        These are not equal. Your "Binomial prime theorem" is false. Sorry, but
        it's back to the tee for you.

        I agree with Phil that you should probably be working with Usenet's sci.math
        group, not this list. I'm not going to spend any more time reviewing your
        stuff. I'm an amateur. Even if I _don't_ find a flaw, that would mean
        nothing.

        Tom Hadley
      • Hadley, Thomas H (Tom), NLCIO
        Typo: Second line should read even when a and b are not integers . Sorry, Tom ... From: Hadley, Thomas H (Tom), NLCIO Sent: Tuesday, September 04, 2001 10:29
        Message 3 of 6 , Sep 4, 2001
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          Typo: Second line should read "even when a and b are not integers".
          Sorry, Tom

          -----Original Message-----
          From: Hadley, Thomas H (Tom), NLCIO
          Sent: Tuesday, September 04, 2001 10:29 AM
          To: paulmillscv@...; primenumbers@yahoogroups.com
          Subject: RE: [PrimeNumbers] FLT


          Paul,
          You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p
          (mod p) even when a and b are not rationals, given a definition of
          modularity that says for example t = 1/2 mod p means that t = 1/2 + k.p for
          some integer k, which seems to me a reasonable definition.

          This is not true. A counter-example is easy:

          let a = 2, b = 1/5, p = 3
          (a+b)^p mod p would be (2+1/5)^3
          = (11/5)^3
          = 1331/125
          = 9 + 206/125
          = 3*3 + 206/125
          = 206/125 mod 3

          However,

          a^p + b^p would be 2^3 + (1/5)^3
          = 8 + 1/125
          = 3*2 + 2 + 1/125
          = 3*2 + 251/125
          = 251/125 mod 3

          These are not equal. Your "Binomial prime theorem" is false. Sorry, but
          it's back to the tee for you.

          I agree with Phil that you should probably be working with Usenet's sci.math
          group, not this list. I'm not going to spend any more time reviewing your
          stuff. I'm an amateur. Even if I _don't_ find a flaw, that would mean
          nothing.

          Tom Hadley


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        • paulmillscv@yahoo.co.uk
          Hello Tom, A nice observation and I will amend the paper so that the Binomial prime `theorem is a definition. But just a small comment. Your example seems to
          Message 4 of 6 , Sep 4, 2001
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            Hello Tom,
            A nice observation and I will amend the paper so that the
            Binomial prime `theorem' is a definition. But just a small comment.
            Your example seems to be a contradiction but in fact you are
            comparing a Binomial (2 + 1/5)^3 and a monomial (11/5)^3. :-). How
            does one expand a monomial? How long is a piece of string? What
            matters is that a consistent framework is developed and that is why
            maths is fun. If you think that Andrew Wiles et al have approached
            the problem of modulo rational arithmetic any other way you are in
            for a pleasant surprise. If it ain't broke, don't fix it. If it is
            broke, define! Stay tuned as you see the end of my proof in version
            2 and can you guess where it ends? No, it doesn't use Logical
            Resonance! Something much more visual.

            Regards,
            Paul Mills
          • John W. Nicholson
            http://primes.utm.edu/top20/page.php?id=2 Around 1825 Sophie Germain proved that the first case of Fermat s Last Theorem is true for such primes. Soon after
            Message 5 of 6 , Feb 7, 2005
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              http://primes.utm.edu/top20/page.php?id=2
              "
              Around 1825 Sophie Germain proved that the first case of Fermat's
              Last Theorem is true for such primes. Soon after Legendre began to
              generalize this by showing the first case of FLT also holds for odd
              primes p such that kp+1 is prime, k=4, 8, 10, 14 and 16. In 1991 Fee
              and Granville [FG91] extended this to k<100, k not a multiple of
              three."

              My question is: What about the converse? How many things does the
              FLT now prove about primes? Not just about SG primes and the others
              which were use to reduece the problem, I am thinking of what other
              ways it may be used. Like: Is there any type of primes of a form
              which includes x^n + y^n?
            • Joseph Moore
              So I was looking at the Mersenne numbers yesterday and found a curious property: 3^(2^(p-1)) = -3 (mod 2^p - 1) if (2^p - 1) divides S(p-1). For the sake of
              Message 6 of 6 , Mar 1, 2005
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                So I was looking at the Mersenne numbers yesterday and
                found a curious property: 3^(2^(p-1)) = -3 (mod 2^p -
                1) if (2^p - 1) divides S(p-1).

                For the sake of reference, I'm using S(1) = 4, S(n+1)
                = S(n)^2 - 2. Also, by 'Mersenne number' I mean "let
                p be prime, then 2^p - 1 is a Mersenne number, namely
                M_p".

                I got this result by taking the proof of the
                sufficiency of the Lucas-Lehmer test (see
                http://www.utm.edu/research/primes/notes/proofs/LucasLehmer.html)
                and doing some squaring and modding. I haven't been
                able to show (yet?) whether the property I found is a
                sufficient condition for primality, but so far it
                hasn't found any false positives or missed any actual
                primes (tested up through 521 at the time of this
                writing). It does seem to be (at least) a necessary
                condition.

                Of course... now that I look more closely at my code,
                this test appears to be one cycle longer than the
                Lucas-Lehmer test, so maybe it isn't really worth
                pursuing. Hard numbers: it was 6 seconds less time to
                check M_11213 (94 seconds for Lucas-Lehmer, 88 for my
                method, both using algorithms written in Lisp, run
                under SBCL 0.8.19 on a 566MHz Celeron), and a little
                less memory too...

                Joseph.



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