Re: [PrimeNumbers] The Function y^2 - x^2
>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
which implies the functional equation
thinking(Jens) > thinking(Mike)
[Non-text portions of this message have been removed]
- Fermat realized this 300 years earlier.
And, this is the basis of his factoring algorithm.
Milton L. Brown
miltbrown at earthlink.net
> [Original Message]conjecture is
> From: Jens Kruse Andersen <jens.k.a@...>
> To: <email@example.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
> > true; and if it is satisfied then the second condition is triviallyprimes)
> > 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
> The second part means x=y-1 is a solution if 2y-1 is a product of 2
> As example, let y=11. Then 2y-1 = 21 = 3*7, so x=10 is a solution.it is
> There are of course also Goldbach solutions for y=11.
> This makes our problem slightly weaker than Goldbach's Conjecture, i.e.
> conceivable (but highly unlikely) that Goldbach is false but there alwaysis x
> for our problem. That would be the case iff 2y-1 is a product of 2 primesfor
> all Goldbach counter examples 2y.unproved.
> As to why Goldbach remains unproved?
> Well, lots and lots of conjectures with strong heuristic support are
> Heuristics are usually useless for proofs.prove:
> A probably weak explanation for why many prime conjectures are hard to
> 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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