Diophantine equation using Pell's equation

I asked this question some days ago: Is there a way to find for which A the system

$X^2+Y^2=Z^2+T^2+1$

$XZ−YT=A$

has only one solution in positive integers?

Looking for the solution of the problem, I came upon this entry ,in which scientific value and usefulness I am not sure. That's why I will make a copy of it.

"Form: $x^2+y^2+z^2 = u^2+v^2$

One can set any of its terms equal to 1, but depending on which side of the equation it is chosen, it may need slightly different Pell equations. To set this form either as $p^2+q^2+1 = s^2+t^2$, or $p^2+q^2+r^2 = s^2+1$, can involve,

$x^2-(d+1)y^2 = ±1$; or $x^2-(d^2+1)y^2 = ±1$

respectively. To find the complete soln of this form, use the more general one for $x1^2+x2^2+x3^2 = y1^2+y2^2+y3^2$,

$(a-b)^2 + (c+d)^2 + (e-f)^2 = (a+b)^2 + (c-d)^2 + (e+f)^2$

where ab-cd+ef = 0. It is easy to make one of the terms vanish.

$(a-b)^2 + (c+d)^2 + 4n = (a+b)^2 + (c-d)^2$

where n = ab-cd. Let ${a,b,c} = {x+y, x-y, y^2}$ and this condition becomes,

$x^2-(d+1)y^2 = n$

where one can then set n = ±1. If d,x,y are odd, then all the terms above are even and can be reduced."

Can this help me to solve my problem? I did not manage to do it!

Thank you!

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The question referred to was math.stackexchange.com/questions/660634/… – Gerry Myerson Feb 16 '14 at 10:29
Yes,I know, I just add what I have found – Rally Feb 16 '14 at 10:44
I wasn't trying to tell you something you didn't know, I was trying to tell others something you should have told them. – Gerry Myerson Feb 16 '14 at 10:46
And he should have at least mentioned where he copied $x^2+y^2+z^2 = u^2+v^2$, Form 2 at sites.google.com/site/tpiezas/004. Sigh. – Tito Piezas III Dec 2 '14 at 2:23