3. Determine the complex numbers $x, y, z$, if $\left(\begin{array}{l}x y=z^{2}+2 z-x-y \\ x z=y^{2}+2 y-x-z \\ y z=x^{2}+2 x-y-z\end{array}\right.$.

Mariana-Liliana Popescu, Suceava
Solution. The relation $x y=z^{2}+2 z+1-1-x-y \Leftrightarrow(x+1)(y+1)=(z+1)^{2}$ suggests the substitutions $x+1=a$, $y+1=b, z+1=c$ and the system becomes: $\left(\begin{array}{l}a b=c^{2} \\ a c=b^{2} \\ b c=a^{2}\end{array}, a, b, c \in \mathbb{C}\right)$. We consider the cases:
I. $a=0 \Rightarrow b=c=0$.

II. $a \neq 0 \Rightarrow b, c \neq 0$. In this case, the system is equivalent to $a^{3}=b^{3}=c^{3}=a b c$.

We observe that the triplet $(a, a, a), a \in \mathbb{C}^{*}$ is a solution to the system.

If at least two of the numbers $a, b, c$ are distinct, then we deduce that all three numbers are distinct. In this case, $\left(\frac{b}{a}\right)^{3}=\left(\frac{c}{a}\right)^{3}=1 \Rightarrow \frac{b}{a}, \frac{c}{a}$ are the non-real complex cube roots of unity. Thus, $(a, b, c)=\left(z_{1}, z_{2}, z_{3}\right)$, where $z_{1} \neq z_{2} \neq z_{3} \neq z_{1}$ are complex numbers in the set $\left\{t, \omega t, \omega^{2} t\right\}, t \in \mathbb{C}^{*}$, and $\omega=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}$. Therefore, the solutions are $(x, y, z) \in\left\{(z, z, z),\left(z_{1}-1, z_{2}-1, z_{3}-1\right)\right\}, z \in \mathbb{C}$, $z_{1}, z_{2}, z_{3}$ as specified above.

## Scoring.

| ![](https://cdn.mathpix.com/cropped/2024_06_07_aa15727edf4b5156a708g-3.jpg?height=56&width=1358&top_left_y=716&top_left_x=182) | $2 \text{p}$ |
| :---: | :---: |
| $(z, z, z)$ is a solution to the system..... | $1 \text{p}$ |
| Determines the form of solutions $(x, y, z)$, with distinct components | $4 \text{p}$ |