<s>[INST] Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
                   y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if  $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
                   y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get  $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
                y^2=z+1-y\\
                z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$    $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..   
 [/INST] $\textbf{Hint:}$Note that $\displaystyle x^2-x-1=y$ is equal $\displaystyle (x-\frac{1}{2})^2-\frac{5}{2}=y-\frac{1}{2}$, so if you substitute $x_1=x-\frac{1}{2}$, $y_1=y-\frac{1}{2}$ and $z_1=z-\frac{1}{2}$ you get:
$$x_1^2-\frac{5}{2}=y_1$$
$$y_1^2-\frac{5}{2}=z_1$$
$$z_1^2-\frac{5}{2}=x_1$$
Substitute (1) to (2), next (2) to (3), you get polynomial equation to calculate $x_1$. The same with $y_1$ and $z_1$.
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