- I wrote:
>Remarkable!

Of course, I should have done a bit more thinking and less experimentation

before posting, as I'm sure you have realized by now.

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.

Does this answer the question in your original post?

-Mike Oakes

[Non-text portions of this message have been removed] - 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,

One snag:

> 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.

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 >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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