How to solve equation system with 3 parameters

Question

How to solve this problem?

$$y=2x^2-1$$ $$z=2y^2-1$$ $$x=2z^2-1$$

x,y,z - ?

I am not good at math. Can you help me? Thanks in advance

Answer 1

Hint:

$$\begin{align}y &= 2x^2 - 1\tag{1}\\z &= 2y^2 - 1\tag{2}\\x &= 2z^2 - 1\tag{3}\end{align}$$

Substitute $(2)$ into $(3)$ to get $$x = 2\left(2y^2 - 1\right)^2 - 1\tag{4}$$

You now have two simultaneous equations, i.e., $(1)$ and $(4)$, that you can solve to find the values of the variables $x$ and $y$. Once you have either of the two, finding the value of variable $z$ is trivial with either $(2)$ or $(3)$.

Answer 2

Let $x=\cos A$ $$y=\cos2A,z=\cos4A,x=\cos8A$$

$$\cos8A=\cos A$$

$$8A=360^\circ n\pm A$$

Either $$(8+1)A=360^\circ n\implies A=40^\circ n;0\le n<180/40$$

What if $(8-1)A=360^\circ n?$