Solve the following system of equations:

$$
x=2 y^{2}-1 ; \quad y=2 z^{2}-1 ; \quad z=2 x^{2}-1
$$
First, we show that the absolute values of $x, y$, and $z$ cannot be greater than 1. Suppose, contrary to our claim, that for example $|x|>1$. Since according to the first equation $x$ cannot be less than -1, this (for real numbers) is only possible if $x>1$. Then, according to the third equation,

$$
z=2 x^{2}-1=x^{2}+\left(x^{2}-1\right)>x>1
$$

thus, according to the second equation,

$$
y=2 z^{2}-1>z
$$

therefore, from the first equation,

$$
x=2 y^{2}-1>y
$$

The three inequalities derived from the assumption $x>1$:

$$
x<z<y<x
$$

lead to a contradiction.

By cyclic permutation of the variables, it can similarly be shown that $y$ and $z$ also cannot have an absolute value greater than 1.

Let $t$ be a real number between 0 and $\pi$ such that $x=\cos t$. Since $|x| \leq 1$, there is exactly one such $t$. Then, according to the third equation,

$$
z=2 x^{2}-1=2 \cos ^{2} t-1=\cos 2 t
$$

and according to the second equation,

$$
y=2 z^{2}-1=2 \cos ^{2} 2 t-1=\cos 4 t
$$

Substituting the obtained results into the first equation:

$$
\cos t=x=2 y^{2}-1=2 \cos ^{2} 4 t-1=\cos 8 t
$$

thus,

$$
\cos t=\cos 8 t
$$

Since for any real numbers $u$ and $v$, $\cos u=\cos v$ holds if and only if $u-v$ or $u+v$ is an integer multiple of $2 \pi$, this means that $7 t$ or $9 t$ is an integer multiple of $2 \pi$. There are eight such values of $t$ in the interval $[0, \pi]$:

$$
t_{1}=0 ; \quad t_{2}=\frac{2 \pi}{7} ; \quad t_{3}=\frac{4 \pi}{7} ; \quad t_{4}=\frac{6 \pi}{7} ; \quad t_{5}=\frac{2 \pi}{9} ; \quad t_{6}=\frac{4 \pi}{9} ; \quad t_{7}=\frac{6 \pi}{9} ; \quad t_{8}=\frac{8 \pi}{9}
$$

The corresponding solutions are:

$$
\begin{array}{lll}
x_{1}=\cos 0=1 ; & y_{1}=\cos 0=1 ; & z_{1}=\cos 0=1 \\
x_{2}=\cos \frac{2 \pi}{7} ; & y_{2}=\cos \frac{8 \pi}{7}=\cos \frac{6 \pi}{7} ; & z_{2}=\cos \frac{4 \pi}{7} ; \\
x_{3}=\cos \frac{4 \pi}{7} ; & y_{3}=\cos \frac{16 \pi}{7}=\cos \frac{2 \pi}{7} ; & z_{3}=\cos \frac{8 \pi}{7}=\cos \frac{6 \pi}{7} \\
x_{4}=\cos \frac{6 \pi}{7} ; & y_{4}=\cos \frac{24 \pi}{7}=\cos \frac{4 \pi}{7} ; & z_{4}=\cos \frac{12 \pi}{7}=\cos \frac{2 \pi}{7} \\
x_{5}=\cos \frac{2 \pi}{9} ; & y_{5}=\cos \frac{8 \pi}{9} ; & z_{5}=\cos \frac{4 \pi}{9} ; \\
x_{6}=\cos \frac{4 \pi}{9} ; & y_{6}=\cos \frac{16 \pi}{9}=\cos \frac{2 \pi}{9} ; & z_{6}=\cos \frac{8 \pi}{9} ; \\
x_{7}=\cos \frac{6 \pi}{9}=-\frac{1}{2} ; & y_{7}=\cos \frac{24 \pi}{9}=-\frac{1}{2} ; & z_{7}=\cos \frac{12 \pi}{9}=-\frac{1}{2} \\
x_{8}=\cos \frac{8 \pi}{9} ; & y_{8}=\cos \frac{32 \pi}{9}=\cos \frac{4 \pi}{9} ; & z_{8}=\cos \frac{16 \pi}{9}=\cos \frac{2 \pi}{9}
\end{array}
$$