>One snag:

primes)

>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)

-Mike Oakes

[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: <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

> > true; and if it is satisfied then the second condition is trivially

primes)

> > 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

> 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 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.

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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