## Task 3 - 191243

Determine all triples $(x, y, z)$ of real numbers for which

$$
2 x + x^{2} y = y \quad ; \quad 2 y + y^{2} z = z \quad ; \quad 2 z + z^{2} x = x
$$

holds. Here, $x, y$, and $z$ should be expressed using given real numbers through repeated application of operations $+,-, \cdot, :$, real-valued power functions, exponential functions, trigonometric functions, or their real-valued inverse functions.
Considering the last equation $2 z+z^{2} x=x$, it can also be written in the form $2 z=x\left(1-z^{2}\right)$, where $1-z^{2} \neq 0$ because $1-z^{2}=0$ would immediately lead to the contradiction $\pm 2=0$. By cyclically permuting the variables $x, y, z$, which the system of equations is clearly invariant under, we immediately obtain the three additional equations

$$
x=\frac{2 z}{1-z^{2}}, \quad y=\frac{2 x}{1-x^{2}}, \quad z=\frac{2 y}{1-y^{2}}
$$

which, in their form, immediately remind one of the formula

$$
\tan (2 \varphi)=\frac{2 \tan \varphi}{1-\tan ^{2} \varphi}
$$

from trigonometry. Therefore, the ansatz

$$
x=\tan \alpha, \quad y=\tan \beta, \quad z=\tan \gamma \quad\left(\alpha, \beta, \gamma \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)
$$

is very natural, which allows us to write $\left({ }^{*}\right)$ simply as

$$
\alpha \equiv 2 \gamma \quad \bmod \pi, \quad \beta \equiv 2 \alpha \quad \bmod \pi, \quad \gamma \equiv 2 \beta \quad \bmod \pi \quad(* *)
$$

Substitution then leads to

$$
7 \alpha=8 \alpha-\alpha \equiv 4 \beta-\alpha \equiv 2 \gamma-\alpha \equiv 0 \quad \bmod \pi
$$

i.e., in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\alpha$ has a total of 7 solutions

$$
\alpha \in\left\{0, \pm \frac{\pi}{7}, \pm 2 \frac{\pi}{7}, \pm 3 \frac{\pi}{7}\right\}
$$

which, in conjunction with $\left({ }^{* *}\right)$, allows us to easily provide the complete solution to the given system of equations:

$$
x=\tan \alpha, \quad y=\tan (2 \alpha), \quad z=\tan (4 \alpha) \quad\left(\alpha \in\left\{0, \pm \frac{\pi}{7}, \pm 2 \frac{\pi}{7}, \pm 3 \frac{\pi}{7}\right\}\right)
$$