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From: alainverghote on 20 May 2010 13:48
Bonsoir,

Be x,y,z,t four positive integer numbers.

Let us consider the form x^2 + 2y^2 = z^2 + 2t^2 ,
and suppose {x,t} given , example x=2 , t=3
which solution will you propose?

Can we get a general formula?

Best regards,

Alain
From: Peter on 20 May 2010 14:28
On May 20, 1:48 pm, "alainvergh...(a)gmail.com"
<alainvergh...(a)gmail.com> wrote:
> Bonsoir,
>
> Be x,y,z,t  four  positive integer numbers.
>
> Let us consider the form  x^2 + 2y^2 = z^2 + 2t^2 ,
> and suppose {x,t} given , example x=2 , t=3
> which solution will you propose?
>
> Can we get a general formula?
>
> Best regards,
>
> Alain

Re-arranging your equation:
(i) x^2 - z^2 = 2*(t^2 - y^2)
shows that
1) both y and t must always be even
2) z is odd iff x is odd

which doesn't tell you a great deal, but it's a start ;>)

Good luck,

Peter
From: William Elliot on 21 May 2010 05:22
On Thu, 20 May 2010, alainverghote(a)gmail.com wrote:
> Bonsoir,
> Be x,y,z,t four positive integer numbers.
>
> Let us consider the form x^2 + 2y^2 = z^2 + 2t^2 ,
> and suppose {x,t} given , example x=2 , t=3
> which solution will you propose? Can we get a general formula?

Given x,t and
z^2 - 2y^2 = x^2 - 2t^2

solve for z and y. Clearly z = +-x, y = +-t are solutions.

>
From: alainverghote on 21 May 2010 05:40

Yes, William

Form x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given

In case we wish four distinct squares, example for {2,3}
2^2 + 2*13^2 = 18^2 + 2*3^2
we need another solution,
a simple one
could be {x,y,z,t} = {x,3t+2x,4t+3x,t}

Alain
From: William Elliot on 22 May 2010 04:27
On Fri, 21 May 2010, alainverghote(a)gmail.com wrote:

> Form x^2 + 2y^2 = z^2 + 2t^2 , {x,t} given
>
> In case we wish four distinct squares, example for {2,3}
> 2^2 + 2*13^2 = 18^2 + 2*3^2
> we need another solution,
> a simple one
> could be {x,y,z,t} = {x,3t+2x,4t+3x,t}
>
y = 2x + 3t; z = 3x + 4t is a solution to x^2 + 2y^2 = z^2 + 2t^2?

x^2 + 2(4x^2 + 12xt + 9t^2) = 9x^2 + 24xt + 16t^2 + 2t^2
x^2 + 8x^2 + 24xt + 18t^2) = 9x^2 + 24xt + 16t^2 + 2t^2

Ok, that is correct. So is y = t, z = x a solution.

You are looking for elements r,s in the ring R = Z[sqr 2]
for which norm(r) = n(r) = n(s).
n(r) = rr* where (n + m.sqr 2)* = n - m.sqr 2 for n,m in Z.

There should be an infinite sequence of such numbers.
Find some u /= +-1 in R with n(u) = 1.
For example 3 + 2.sqr 2 or 3 - 2.sqr 2.
Then for all j in N, n(r) = n(u^j r).



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