## The Function y^2 - x^2

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• Hi All! Could someone out there explain to me in simple terms, why, with all your research into prime numbers, it is not possible to prove When y = ( 2,3,4...)
Message 1 of 6 , Feb 5, 2005
Hi All!

Could someone out there explain to me in simple terms, why, with all your research into prime numbers, it is not possible to prove

When y = ( 2,3,4...) and x = ( 0,1,2,3...),

the function y^2-x^2 inevitably returns a solution, which is the product of just two prime factors? (Heuristically, of course, the evidence is overwhelming)

I just want to know why such a proof is so difficult - what are the stumbling blocks?

Many thanks

Bob Gilson

PS Please don't tell me this is just another way of re-stating Goldbach's Conjecture, that explanation does not explain anything.

---------------------------------
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[Non-text portions of this message have been removed]
• In a message dated 05/02/2005 08:12:13 GMT Standard Time, bobgillson@yahoo.com writes: Could someone out there explain to me in simple terms, why, with all
Message 2 of 6 , Feb 5, 2005
In a message dated 05/02/2005 08:12:13 GMT Standard Time,
bobgillson@... writes:

Could someone out there explain to me in simple terms, why, with all your
research into prime numbers, it is not possible to prove

When y = ( 2,3,4...) and x = ( 0,1,2,3...),

the function y^2-x^2 inevitably returns a solution, which is the product of
just two prime factors? (Heuristically, of course, the evidence is
overwhelming)

Your notation is a little ambiguous.

After a bit of experimentation I have found one interpretation which seems
both to be non-trivial and to hold up (for small values of y, at least).

If we re-state it as follows:-

Let y be any integer > 1, then there exists at least one integer x, 0 <= x
<= y, such that (y^2-x^2) is the product of just two prime factors

then it seems indeed to be true:-

y y^2 x y^2-x^2
2 4 0 2*2
3 9 0 3*3
4 16 1 3*5
5 25 0 5*5
6 36 1 5*7
7 49 0 7*7
8 64 3 5*11
9 81 2 7*11
10 100 3 7*13
11 121 0 11*11
12 144 5 7*17
13 169 0 13*13

Remarkable!

Is this what you meant?

-Mike Oakes

[Non-text portions of this message have been removed]
• ... 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
Message 3 of 6 , Feb 5, 2005
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.

-Mike Oakes

[Non-text portions of this message have been removed]
• ... 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
Message 4 of 6 , Feb 5, 2005
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
• ... primes) which implies the functional equation thinking(Jens) thinking(Mike) -Mike Oakes [Non-text portions of this message have been removed]
Message 5 of 6 , Feb 5, 2005
>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)

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 ... conjecture is ...
Message 6 of 6 , Feb 5, 2005
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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> The Prime Pages : http://www.primepages.org/
>
>