5. (BUL 5) Solve the system  $$ \begin{aligned} & x^{2}+x-1=y \\ & y^{2}+y-1=z \\ & z^{2}+z-1=x \end{aligned} $$
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.