<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] No solutions exist for real, distinct $(x,y,z)$.
Writing $f(x)=x^2+x-1$, the existence of such a solution would mean that $f$ has a real point of least period 3; that is, $f^3(x)=x$ for some real $x$ with $f(x)\neq1$ i.e. $x\neq \pm 1$. Sarkovsky's theorem then implies that $f$ has points of arbitrary least period. In particular, there would exist $x\in\mathbb{R}-\{-1,1\}$ such that $f^2(x)=x$. But $$f^2(x) - x = (x^2+x-1)^2+(x^2+x-1)-1-x = x^4+2x^3-2x-1 = (x-1)(x+1)^3.$$ and so there are no points of least period 2. Hence $f$ has no real points of least period 3.
P.S.: I'm actually only using the simplest case of Sarkovsky's theorem, namely that 3-cycles imply period 2-cycles. But the full theorem implies that the same conclusion follows for every least period other than one (that is, $f$ has fixed points but not cycles.) This means that we would be no no better off if we increased the number of variables and equations beyond 3: we would still find no nontrivial real solutions. (Though this can be proven much more simply by the same RMS-GM inequality stated by cirpis in his answer...)
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