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

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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 ... 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

> [Original Message]
> From: Jens Kruse Andersen <jens.k.a@...>
> 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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