General integer solution for $x^2+y^2-z^2=\pm 1$ How to find general solution (in terms of parameters) for diophantine equations
$x^2+y^2-z^2=1$ and $x^2+y^2-z^2=-1$?
It's easy to find such solutions for $x^2+y^2-z^2=0$ or $x^2+y^2-z^2-w^2=0$ or $x^2+y^2+z^2-w^2=0$, but for these ones I cannot find anything relevant.

Of course for the equation $X^2+Y^2=Z^2+t$
There is a particular solution:
$X=1\pm{b}$
$Y=\frac{(b^2-t\pm{2b})}{2}$
$Z=\frac{(b^2+2-t\pm{2b})}{2}$
But interessuet is another solution:
$X^2+Y^2=Z^2+1$
If you use the solution of Pell's equation: $p^2-2s^2=\pm1$
Making formula has the form:
$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$
$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$
$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$
number $L$  and any given us.
The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple.
$a^2+b^2=c^2$
This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$
Allows you to find Pythagorean triples with a given difference.
$a=2s(p+s)$
$b=p(p+2s)$
$c=p^2+2ps+2s^2$
$b-a=\pm{k}$
Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.
