Count the ordered triplets

In general the equation $x^2+y^2-z^2=w$ has infinitely many solutions. Rewrite the equation as $(x-z)(x+z)=w-y^2$ Now if $w-y^2=ab$ then $(x, z)=\left(\frac{a+b}{2}, \frac{a-b}{2}\right)$ will be a solution.

It's enough that $a$ and $b$ have same parity, for $x$ and $z$ being integers. Whatever the parity of $w$ is you can find infinitely many $y$ such that $w-y^2$ can be written as product of two odd numbers ( $1$ is allowed).

$x^2+y^2-z^2=1997$ $(y-z)(y+z)=1997-x^2$ Let $x=2t$ for any integer $t$ . We may set $y-z=1$ and $y+z=1997-4t^2$

Addition and subtraction yields $2y=1998-4t^2$ and $2z=1996-4t^2$ respectively.

Hence $(2t,999-2t^2,998-2t^2)$ is a solution of $x^2+y^2-z^2=1997$ .

Since $t$ is an arbitrary integer, there are infintely many solutions.

So the answer is $\boxed{777}$