Roots of composite function

By completing the square, we can write $f(x) = (x+5)^2 - 5$ so that $f(f(x)) = \Big( \big((x+5)^2 - 5\big) + 5\Big)^2 - 5 = \Big( (x+5)^2 \Big)^2 - 5 = (x+5)^4 - 5 \\ f(f(f(f(x)))) = \Big( \big( (x+5)^4 - 5\big) + 5\Big)^4 - 5 = \Big((x+5)^4\Big)^4 - 5 = (x+5)^{16} - 5$

In other words, we're just asking how many real solutions $(x+5)^{16} - 5 = 0$ has. This can be solved directly as $\begin{aligned} (x+5)^{16} - 5 &= 0 \\ (x+5)^{16} &= 5 \\ x+5 = \pm \sqrt[16]{5} \\ x = -5 \pm \sqrt[16]{5} \end{aligned}$

showing there are $\boxed{2}$ real solutions.

Note that $f(x)$ has two zeros: $x=-5\pm\sqrt{5}$ . Furthermore, the minimum value of $f(x)$ is $f(-5)= -5$ . Thus, there is no $x$ for which $f(x)=-5-\sqrt{5}$ , but there are values of $x$ for which $f(x)=-5+\sqrt{5}$ . Specifically, since $f(x)$ is monic, the values $x=-5\pm\sqrt[4]{5}$ satisfy $f(x)=-5\pm\sqrt{5}$ and thus are the only real zeros of $f(f(x))$ .

Similarly, the minimum value of $f(f(x))$ occurs at the value(s) of $x$ for which $f(x)=-5$ , which is itself $x=-5$ . As before, there are no values of $x$ for which $f(f(x)) = -5-\sqrt[4]{5}$ and there are two for which $f(f(x))=-5+\sqrt{5}$ : $x=-5\pm\sqrt[8]{5}$ . These are the only real zeros of $f(f(f(x)))$ . In general, the only roots of $f^n(x)$ (where $f^n$ denotes $f$ composed with itself $n$ times) are $x=-5\pm\sqrt[2^n]{5}$ . Thus there are always precisely $\boxed{2}$ solutions to the equation $f^n(x)=0$ (in our case $n=4$ ).