NMO Selection Tests for the Junior Balkan Mathematical Olympiad

We have $y + 2 = \frac{(x+2)^3}{3x^2+4}$, $y - 2 = \frac{(x-2)^3}{3x^2+4}$ and the analogous relations.
If $y = 2$, then $x = z = 2$. If $y \ne 2$, then $\frac{y+2}{y-2} = \left(\frac{x+2}{x-2}\right)^3$, $\frac{z+2}{z-2} = \left(\frac{y+2}{y-2}\right)^3$ and $\frac{x+2}{x-2} = \left(\frac{z+2}{z-2}\right)^3$. We obtain $\frac{x+2}{x-2} = \left(\frac{x+2}{x-2}\right)^{27}$, which leads to $\frac{x+2}{x-2} \in \{-1, 0, 1\}$. We find two more solutions, $x = y = z = -2$ and $x = y = z = 0$.

Let $f : \mathbb{R} \to \mathbb{R}$, $f(x) = \frac{x^3 + 12x}{3x^2 + 4}$. We have $f(a) - f(b) = \frac{3(ab - 4)^2 + 4(a-b)^2}{(3a^2 + 4)(3b^2 + 4)} \cdot (a-b)$, so $f$ is strictly increasing.

Our equalities become $y = f(x)$, $z = f(y)$, $x = f(z)$. Without loss of generality, we may assume that $x = \min\{x, y, z\}$. Then $f(x) \le f(y)$, so $y \le z$, which leads to $f(y) \le f(z)$ or $z \le x$. Hence $x \le y \le z \le x$, so $x = y = z$. As the equation $f(x) = x$ has three solutions, namely $x = -2$, $x = 0$ and $x = 2$, we find that $(x, y, z) \in \{(-2, -2, -2); (0, 0, 0); (2, 2, 2)\}$.