Solve the system

$$
\begin{aligned}
& y=2 x^{2}-1 \\
& z=2 y^{2}-1 \\
& x=2 z^{2}-1
\end{aligned}
$$
From the system, it is clear that all unknowns are not less than -1. If one of them (for example, $x$) is greater than 1, then $y=2x^2-1=x^2+x^2-1>x+1-1=x$. Similarly, $z>y$, $x>z$. Contradiction: $x>z>y>x$.

Therefore, $|x| \leq 1$, and we can make the substitution $x=\cos \varphi, 0 \leq \varphi \leq \pi$. Then $y=\cos 2\varphi, z=\cos 4\varphi, x=\cos 8\varphi$. We obtain the equation

$\cos 8\varphi=\cos \varphi$, from which $8\varphi= \pm \varphi+2k\pi$, that is, $9\varphi=2k\pi$ or $7\varphi=2k\pi$.

## Answer

$(1,1,1);(-1/2,-1/2,-1/2);(\cos 2\pi/9, \cos 4\pi/9, -\cos \pi/9);(\cos 2\pi/7, -\cos 3\pi/7, -\cos \pi/7)$ and all sets obtained from the specified ones by cyclic permutations.