How to find all the complex solutions to $ X^2= Y Z , Y^2 =X Z, Z^2= X Y$?

How do I solve the following system to find the all the complex solutions:

$$ X^2= Y Z\\ Y^2 =X Z\\ Z^2= X Y$$

I know about the obvious solution $(0,0,0)$. When I try to compute it with Mathematica, it yields

$$\begin{array}{cc} Y\to X & Z\to \frac{Y^2}{X} \\ Y\to -\sqrt[3]{-1} X & Z\to \frac{Y^2}{X} \\ Y\to (-1)^{2/3} X & Z\to \frac{Y^2}{X} \\ \end{array} $$

And I know how to reach to each of these expressions by hand, I just don't know what to do with them later in order to actually obtain solutions.

Solution 1:

Setting apart the case $x=y=z=0$, in all the other cases, $x,y,z$ are different from zero, allowing to write the conditions under the equivalent form:

$$\dfrac{x}{z}=\dfrac{y}{x}=\dfrac{z}{y} =: a\tag{1}$$

where $a$ is defined as the common value of these ratios. Then:

$$x=za, \ \, y=xa, \ \ z=ya \tag{2}$$

As a consequence :

$$xyz=a^3 xyz$$

giving

$$a^3=1 \ \iff \ a=\omega^k \ for \ \omega:=e^{2i k\pi/3}\tag{3}$$

(for $k=0,1,2$) implying that the general solution is, by taking an arbitrary value $x=x_0$ in (2):

$$y=ax_0, \ \ \ \ z=ya=a^2x_0, \ \ \ \ \underbrace{x=za=s^3x_0=x_0}_{\text{sanity check...}}$$

for any of the three values given by (2). Otherwise said, set apart the solution $(0,0,0)$, the nonzero solutions are:

$$(x_0,x_0,x_0), \ \ \ \ (x_0,\omega x_0,\omega^2 x_0), \ \ \ \ (x_0,\omega^2 x_0,\omega x_0)$$

Edit: (1) can be treated in a somewhat more sophisticated way.

Indeed, it can be written under the form:

$$\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z \end{pmatrix}=a \begin{pmatrix}x\\y\\z \end{pmatrix}\tag{4}$$

where the matrix is the rotation matrix of $\alpha := \frac {2 \pi}{3}$ around vector with coordinates $1,1,1$. Please note that the case $(x,y,z)=(0,0,0)$ is included under the form (4).

We recognize in (4) an eigenvalue/eigenvector formulation with, in particular, eigenvalues

$$a \ = \ 1, \ \ e^{i \alpha}, \ \ e^{-i \alpha},$$

the value of the corresponding eigenvectors being readily computed.

Solution 2:

If any of $X, Y, Z$ are zero they all are all zero, so assume they are all nonzero in the following.

Dividing the first equation by the second leads to $X^3 = Y^3$, and the second and third similarly lead to $Y^3 = Z^3$. So the three variables can only differ by a factor of a cube root of unity. From here it's not to hard to use the original equations to show that either the three variables are equal or one has $(X,Y,Z) = (X, X\omega, X\omega^2)$ or $(X, X\omega^2, X\omega)$ for $\omega$ a complex cube root of unity.