<s>[INST] How to solve this system of equations for $x^2+y^2+z^2$? For the complex numbers $x,y,z$, the system of equations

$x^2-yz=i~~~~~
y^2-zx=i~~~~~
z^2-xy=i$

It is not easy for me to get $x^2+y^2+z^2$ from the above. I don't need the values of $x,y,z$
I'm stuck in what to do at first. Any advice would be helpful.
My attempt : we can get $x^2+y^2+z^2-xy-yz-zx=3i$, 
so $(x-y)^2+(y-z)^2+(z-x)^2=6i$
I actually don't know whether this is a right way to solve the above.
 [/INST] Subtracting the equations in pairs we get
$$(x-y)(x+y+z) = 0$$
$$(x-z)(x+y+z) = 0$$
$$(y-z)(x+y+z) = 0$$
If $x+y+z \not = 0$ then $x=y=z$, but then $x^2-yz = 0 \not = i$ so we must have $x+y+z = 0$.
Next we can use the identity
$$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz)$$
togeather with the equation system to arrive at
$$x^2 + y^2 + z^2 = 2 i + \frac{(x+y+z)^2}{3} = 2i$$
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