2. Let $f(x)=x^{2}+6 x+6$. Solve the equation in real numbers

$$
\underbrace{f(f(\ldots f}_{2017}(x) \ldots))=2017
$$

(P. Alishev)
Answer: $x=-3 \pm \sqrt[2^{2017}]{2020}$.

Solution: Note that $f(x)=(x+3)^{2}-3$ or $f(x)+3=$ $=(x+3)^{2}$. Then

$$
\begin{gathered}
2020=\underbrace{f(f(\ldots f}_{2017}(x) \ldots))+3=(\underbrace{f(f(\ldots f}_{2016}(x) \ldots))+3)^{2}= \\
=(\underbrace{f(f(\ldots f}_{2015}(x) \ldots))+3)^{2^{2}}=\ldots=(f(x)+3)^{2^{2016}}=(x+3)^{2^{2017}} .
\end{gathered}
$$

In the end, we obtained the equation

$$
(x+3)^{2^{2017}}=2020 \Longrightarrow x_{1,2}=-3 \pm \sqrt[2^{2017}]{2020}
$$