How to solve this symmetric system of equations?

Question

How many solutions are there to this equation?

$$\begin{align*} x^2-y^2&=z\\ y^2-z^2&=x\\ z^2-x^2&=y \end{align*}$$

Answer 1

Sum both sides and get $x+y+z=0$, so you're on a plane.

Further hint:

$$z=(x^2-y^2)=(x-y)(x+y)=-z(x-y)\implies \ldots$$

Answer 2

We have \begin{align} x^2 - y^2 & = z\\ y^2 - z^2 & = x\\ z^2 - x^2 & = y \end{align} Adding the first two equations, we get that $$x^2-z^2 = z+x \implies z=-x \text{ or }x-z=1$$ Similarly, we get that $$x=-y \text{ or }y-x=1$$ $$y=-z \text{ or }z-y=1$$ Hence, there are $8$ possible choices of getting $3$ equations. But the choice $$x-z=1 \text{ and }y-x=1 \text{ and }z-y=1$$ gives no solution. The rest of the seen choices give the following solutions $$(0,0,0); (-1,0,1); (0,1,-1); (1,-1,0)$$