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.

I believe the general solution to $x^2+y^2-z^2=1$ is $x=(rs+tu)/2$, $y=(rt-su)/2$, $z=(rs-tu)/2$, where $rt+su=2$. 
EDIT: Solutions to $x^2+y^2+1=z^2$ can be obtained by choosing $a$, $b$, $c$, $d$ such that $ad-bc=1$ and then letting $x=(a^2+b^2-c^2-d^2)/2$, $y=ac+bd$, $z=(a^2+b^2+c^2+d^2)/2$, though I'm not sure you get all the integer solutions this way. 
The rational solutions are a bit easier. $(0,0,1)$ is a (rational) point on the surface. The line $(0,0,1)+t(a,b,c)$ through that point intersects the surface again at $x=2ac/(a^2+b^2-c^2)$, $y=2bc/(a^2+b^2-c^2)$, $z=(a^2+b^2+c^2)/(a^2+b^2-c^2)$, giving all the rational points on the surface. 
