3. (3 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}-4 x-5$

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# Solution:

It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:

$$
x^{2}-4 x-5=x \Leftrightarrow x^{2}-5 x-5=0 \Leftrightarrow x=\frac{1}{2}(5 \pm 3 \sqrt{5})
$$

Since the roots of the polynomial $f(x)-x=x^{2}-5 x-5$ are also roots of the polynomial $f(f(x))-x=x^{4}-8 x^{3}+2 x^{2}+55 x+40$, the first polynomial divides the second. The result of the division is $x^{2}-3 x-8$. Therefore, the original equation is equivalent to the system of equations:

$$
\left[\begin{array} { l } 
{ x ^ { 2 } - 5 x - 5 = 0 } \\
{ x ^ { 2 } - 3 x - 8 = 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\frac{1}{2}(5 \pm 3 \sqrt{5}) \\
\frac{1}{2}(3 \pm \sqrt{41})
\end{array}\right.\right.
$$

Answer: $\frac{1}{2}(5 \pm 3 \sqrt{5}), \frac{1}{2}(3 \pm \sqrt{41})$.