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

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  • 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 1 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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