number of solutions of $f(f(f(f(x))))$

Hint : $$ f(x) = x^2+10x+20 = (x+5)^2 - 5 $$

so $$ f(f(x)) = ((x+5)^2 - 5 + 5)^2 - 5 = (x+5)^4 - 5 $$ $$f(f(f(x))) = (x+5)^8 - 5 $$ ... and it should be easier to analyse $f^{(4)} (x)$ now.

An alternative method to those already presented:

Consider first a simpler case: when is $f^2(x)=0$? (I let $f^2(x)$ denote $f(f(x))$, or, equivalently, $(f \circ f)(x)$.)

It will be whenever $f(x) \in f^{-1}(0)$. (Recall that $f^{-1}(a) = \{ x \mid f(x)=a \}$, i.e. this is a set.)

A similar idea applies to your case. $f^4(x) = 0$ if and only if $f(x) \in (f^{-1})^3(0)$. This is something we can proceed through iteratively.

• First, what elements are in $f^{-1}(0)$? That is, which elements does $f$ map to $0$? As you have determined, these are $-5 \pm \sqrt 5$.

• Next, what elements are in $(f^{-1})^2(0) = f^{-1}(-5 \pm \sqrt 5)$? This means that $$x^2 + 10x + 20 = -5 \pm \sqrt 5$$ so just solve for $x$. You get $-5 \pm \sqrt[4]5$ for the positive root, and $-5 \pm i \sqrt[4]5$ for the negative root. Only the former is relevant.

• Iterate one more time. Then $$x^2 + 10x + 20 = -5 \pm \sqrt[4]5$$ and we solve for $x$. The negative solution gives us more complex numbers, but the positive gives us $-5 \pm \sqrt[8]5$.