$f(x)^5-f(x)= (f(x)^2+1)(f(x)-1)f(x) (f(x)+1))$ has three real roots, so one of $f(x)-1,f(x),f(x)+1$ has exactly one root Case 1: $f(x)-1$ has one real root, than $f(x)=q(x)^2+1 \to f(x),f(x)+1$ don't have real roots, so in total there is one root Case 2: $f(x)$ has one real root, than $f(x)=q(x)^2$. $f(x)-1=q(x)^2-1$ has two real roots and $f(x)+1$ has no roots. So ordinate of vertex is $0$ Case 3: $f(x)+1$ has one real root, than $f(x)=q(x)^2-1, f(x)-1=q(x)^2-2$, every has two real roots, so in total there are $5$ roots.