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From: alainverghote on 29 Jul 2010 13:24
The idea is quite simple , for squares and other
powers we may build such things.

I here , just want considering second degree forms.

x(n)^2 + y(n)^2 = z(n)^2
with a seed {x0,y0,z0}
the following linear relations work:

x(n+1) = x(n) +2y(n) +2z(n)
y(n+1) = 2x(n)+y(n) +2z(n)
z(n+1) = 2x(n)+2y(n)+3z(n) .

Can we find likewise relations to generate new
forms like x(n)^2+y(n)^2 ) = 2z(n)^2+2t(n)^2
say from {x0,y0,z0,t0} = {7,1,3,4}
No need to say we're looking for solutions other than
{x(n+1),y(n+1),z(n+1),t(n+1)} = c*{x(n),y(n),z(n),t(n)}

Is this a well known problem?

Thanks for help,

Alain
From: alainverghote on 2 Aug 2010 12:47
On 29 juil, 19:24, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
wrote:
> The idea is quite simple , for squares and other
> powers we may build such things.
>
> I here , just want considering second degree forms.
>
> x(n)^2 + y(n)^2 = z(n)^2
> with a seed {x0,y0,z0}
> the following linear relations work:
>
> x(n+1) = x(n) +2y(n) +2z(n)
> y(n+1) = 2x(n)+y(n) +2z(n)


Good evening,

Getting back to propose a generic solution to
x^2+y^2 = 2z^2+2t^2
We may start with equality
x^2 + (x +2z)^2 = 2z^2 + 2(z+x)^2 and
x^2 + {y(n)(x,z)}^2 = 2z^2 + 2{t(n)(x,z)}^2
with passage rules:
y(n+1)(x,z) = {y(n)(3x+4z,2x+3z)
t(n+1)(x,z) = {t(n)(3x+4z,2x+3z)

Then the third term will be
x^2 + {41x+58z)}^2 = 2z^2 + 2{29x+41z}^2
for {x;z} = {2;1}
2^2 + 140^2 = 2*1^2+ 2*99^2


I still do not know how generalizable is
this process,to other powers?! ;..

Alain


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