Symmetric System of Equations

Question

I'm new on studying Systems of equations. I just want to know the number of real solutions of this system of equations:

$$x^2-y^2=z$$ $$y^2-z^2=x$$ $$z^2-x^2=y$$

I also want to know how was your solution and explanation on how did you find your answer.

Answer 1

Add $x^2 - y^2 = z$ and $y^2 - z^2 = x$ to get $x^2 - z^2 = x + z$. This means either $x + z = 0$, or $x - z = 1$. Similarly, either $y + x = 0$ or $y - x = 1$, and either $z + y = 0$ or $z - y = 1$.

Now there are a total of $8$ different combinations of equations to deal with.

1. $x + z = 0$, $y + x = 0$, $z + y = 0$
2. $x + z = 0$, $y + x = 0$, $z - y = 1$
3. $x + z = 0$, $y - x = 1$, $z + y = 0$
4. $x + z = 0$, $y - x = 1$, $z - y = 1$
5. $x - z = 1$, $y + x = 0$, $z + y = 0$
6. $x - z = 1$, $y + x = 0$, $z - y = 1$
7. $x - z = 1$, $y - x = 1$, $z + y = 0$
8. $x - z = 1$, $y - x = 1$, $z - y = 1$

And then you just proceed to solve for $x$, $y$, and $z$, and find which cases make sense and which ones don't.