5. (BUL 5) Solve the system
$$
\begin{array}{l}
x^{2}+x-1=y \\
y^{2}+y-1=z \\
z^{2}+z-1=x .
\end{array}
$$
5. If one of $x, y, z$ is equal to 1 or -1 , then we obtain solutions $(-1,-1,-1)$ and $(1,1,1)$. We claim that these are the only solutions to the system. Let $f(t)=t^{2}+t-1$. If among $x, y, z$ one is greater than 1 , say $x>1$, we have $xf(x)=y>f(y)=z>f(z)=x$, a contradiction. This proves our claim.