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Re: [PrimeNumbers] The Function y^2 - x^2

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  • Jens Kruse Andersen
    ... One snag: y^2-x^2 = (y-x)*(y+x) is a product of 2 primes if and only if (y-x and y+x are both primes) _or_ (y-x=1 and y+x is a product of 2 primes) The
    Message 1 of 6 , Feb 5, 2005
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      Mike Oakes realized y^2-x^2 = (y-x)(y+x) and wrote:

      > If y^2-x^2 = p*q, where y > x >= 0, and p and q are prime, with p >= q,
      > then
      > y + x = p
      > y - x = q
      > so
      > 2*y = p + q
      > 2*x = p - q
      >
      > The first condition is always satisfiable ONLY if Goldbach's conjecture is
      > true; and if it is satisfied then the second condition is trivially
      > satisfied.

      One snag:
      y^2-x^2 = (y-x)*(y+x) is a product of 2 primes if and only if
      (y-x and y+x are both primes) _or_ (y-x=1 and y+x is a product of 2 primes)

      The second part means x=y-1 is a solution if 2y-1 is a product of 2 primes.
      As example, let y=11. Then 2y-1 = 21 = 3*7, so x=10 is a solution.
      There are of course also Goldbach solutions for y=11.

      This makes our problem slightly weaker than Goldbach's Conjecture, i.e. it is
      conceivable (but highly unlikely) that Goldbach is false but there always is x
      for our problem. That would be the case iff 2y-1 is a product of 2 primes for
      all Goldbach counter examples 2y.

      As to why Goldbach remains unproved?
      Well, lots and lots of conjectures with strong heuristic support are unproved.
      Heuristics are usually useless for proofs.

      A probably weak explanation for why many prime conjectures are hard to prove:
      Primes are defined by a property with multiplication.
      Most conjectures involve addition (or subtraction).
      Connecting these is easy in examples but often hard in proofs.

      --
      Jens Kruse Andersen
    • mikeoakes2@aol.com
      ... primes) which implies the functional equation thinking(Jens) thinking(Mike) -Mike Oakes [Non-text portions of this message have been removed]
      Message 2 of 6 , Feb 5, 2005
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        >One snag:
        >y^2-x^2 = (y-x)*(y+x) is a product of 2 primes if and only if
        >(y-x and y+x are both primes) _or_ (y-x=1 and y+x is a product of 2
        primes)

        which implies the functional equation
        thinking(Jens) > thinking(Mike)

        -Mike Oakes


        [Non-text portions of this message have been removed]
      • Milton Brown
        Fermat realized this 300 years earlier. And, this is the basis of his factoring algorithm. Milton L. Brown miltbrown at earthlink.net ... conjecture is ...
        Message 3 of 6 , Feb 5, 2005
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          Fermat realized this 300 years earlier.
          And, this is the basis of his factoring algorithm.

          Milton L. Brown
          miltbrown at earthlink.net

          > [Original Message]
          > From: Jens Kruse Andersen <jens.k.a@...>
          > To: <primenumbers@yahoogroups.com>
          > Date: 2/5/2005 6:52:19 AM
          > Subject: Re: [PrimeNumbers] The Function y^2 - x^2
          >
          >
          > Mike Oakes realized y^2-x^2 = (y-x)(y+x) and wrote:
          >
          > > If y^2-x^2 = p*q, where y > x >= 0, and p and q are prime, with p >= q,
          > > then
          > > y + x = p
          > > y - x = q
          > > so
          > > 2*y = p + q
          > > 2*x = p - q
          > >
          > > The first condition is always satisfiable ONLY if Goldbach's
          conjecture is
          > > true; and if it is satisfied then the second condition is trivially
          > > satisfied.
          >
          > One snag:
          > y^2-x^2 = (y-x)*(y+x) is a product of 2 primes if and only if
          > (y-x and y+x are both primes) _or_ (y-x=1 and y+x is a product of 2
          primes)
          >
          > The second part means x=y-1 is a solution if 2y-1 is a product of 2
          primes.
          > As example, let y=11. Then 2y-1 = 21 = 3*7, so x=10 is a solution.
          > There are of course also Goldbach solutions for y=11.
          >
          > This makes our problem slightly weaker than Goldbach's Conjecture, i.e.
          it is
          > conceivable (but highly unlikely) that Goldbach is false but there always
          is x
          > for our problem. That would be the case iff 2y-1 is a product of 2 primes
          for
          > all Goldbach counter examples 2y.
          >
          > As to why Goldbach remains unproved?
          > Well, lots and lots of conjectures with strong heuristic support are
          unproved.
          > Heuristics are usually useless for proofs.
          >
          > A probably weak explanation for why many prime conjectures are hard to
          prove:
          > Primes are defined by a property with multiplication.
          > Most conjectures involve addition (or subtraction).
          > Connecting these is easy in examples but often hard in proofs.
          >
          > --
          > Jens Kruse Andersen
          >
          >
          >
          >
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          > The Prime Pages : http://www.primepages.org/
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