Roots of a polynomial that is composed n times with itself

Let $f(x)=x(4x^2-3)(64x^6-96x^4+36x^2-3)$ and $f^{(n)}=f(f(f(\cdots f(x))\cdots)$ (composed with itself $n$ times). Prove that for all positive integers $n$, $f^{(n)}(x)=x$ has $9^n$ distinct real solutions $x$.

I could only manage to prove this for $n=1$. For that i put $2x=t$ and found the roots of $t(t^2-3)(t^6-6t^4+9t^2-3)=t$

$O$ is an obvious root. So, i proceeded to show that $(m-3)(m^3-6m^2+9m-3)=1$ has $4$ distinct positive roots.

Expanding i got $m^4-9m^3+27m^2-30m+8=0$. $2, 4$ are natural to guess and easy to verify and the other two roots by dividing out $(m-2)$ and $(m-4)$.

Now, someone please help me to prove for $n>1$.

• Use induction and count preimages. – Avi Steiner Oct 12 '15 at 4:25
• Chebyshev polynomials have a composition property... – DVD Oct 15 '15 at 17:40

$$f(x)=x(4x^2-3)(64x^6-96x^4+36x^2-3)=256x^9-576x^7+432x^5-120x^3+9x=T_9(x)$$ Where $T_n(x)$ is the n-th Čebyšëv polynomials of the first kind. It's known this nesting propriety: $$T_n(T_m(x))=T_{nm}(x)$$ In addition, $T_n(x)$ has exatcly n reals roots.
So, using induction, soppose that the following holds: $$f^{n}(x)=T_{9^n}(x)$$ So It has $9^n$ reals roots, Then $$f^{n+1}(x)=f^n(f(x))=T_{9^n}(T_9(x))=T_{9^{n+1}}(x)$$ So It has $9^{n+1}$ roots.
In your question you said that you have proved that f^{(2)} holds. So by induction and by the proprieties of Čebyšëv polynomials $f^{(n)}(x)$ has exatcly $9^n$ reals roots $\forall n$