Does a solution exist?

$y=2-x$

$x(2-x)-z^2=1$

$(-x^2+2x-1)=z^2$

$-(x-1)^2=z^2$

LHS is zero or negative, RHS is zero or positive, therefore both sides must be zero.

$x=1$ , $z=0$ and $y=1$

Answer = $\boxed{0}$

As described by @Alak Bhattacharya ,

From $x+y=2 \implies y = 2-x$ . Putting in

$\begin{aligned} xy - z^2 & = 1 \\ x(2-x) - z^2 & = 1 \\ 2x-x^2 - z^2 & = 1 & \small \blue{\text{Rearrange}} \\ x^2-2x + 1 + z^2 & = 0 \\ (x-1)^2 + z^2 & = 0 \end{aligned}$

Note that $(x-1)^2 \ge 0$ and $z^2 \ge 0$ . Since the RHS $=0$ , the LHS must be $=0$ . And for the LHS $=0$ , the only solution is when both $x-1=0$ and $z=0$ . Therefore the solution is $x=1, y=1, z=0$ and $x-y+z = \boxed 0$ .

Given $x + y = 2$ , let $\delta = x - 1$ . (Note that \delta need not be positive.) Then $x = 1 + \delta$ and $y = 1 - \delta$ , so $xy = 1 - \delta^2$ .

The second equation tells us that $1 - \delta^2 - z^2 = 1$ . Thus $\delta = z = 0$ and $x = y = 1$ . As a result, $x - y + z = 0$ .