Functional Equation of iterations 
Problem: Let $f : \mathbb{Q} \to \mathbb{Q}$ satisfy  $$f(f(f(x)))+2f(f(x))+f(x)=4x$$ and $$f^{2009}(x)=x$$ ($f$ iterated
  $2009$ times).
Prove that $f(x)=x$.
   This is a contest type problem so it is supposed to have an elegant solution.

My approach: I considered $f^{(k)}(x)=a_k$ then we get,
$a_n+2a_{n-1}+a_{n-2}-4a_{n-3}=0$ with $f(a_{n-1})=a_n$ and we need to show that the sequence is constant, $a_i=a_0$. 
The characteristic root of that recurrence are not good.
 Also I observed that f must be a bijection
Someone please help.

I'm not sure that my solution is elegant.
So let $g(x) = f(x) - x$. We may rewrite the first equation from task as
$$
f(f^2(x)) - f^2(x) + 3f^2(x)-3f(x) + 4f(x) - 4x = 0,
$$
or using $g(x)$
$$
g(f^2(x)) + 3g(f(x)) + 4g(x) = 0.
$$
Let $a_n = g(f^n(x))$. Then $a_n + 3a_{n-1} + 4a_{n-2} = 0$. We may make this recurrent relation simpler: let $b_n = a_n + \frac{3 - i\sqrt{7}}{3}a_{n-1}$. Then $b_n + \frac{3+i\sqrt{7}}{2}b_{n-1} = a_n + 3a_{n-1} + 4a_{n-1} = 0$ or $$b_n = -\frac{3+i\sqrt{7}}{2}b_{n-1} = \left(-\frac{3+i\sqrt{7}}{2}\right)^{n-1} b_1.$$
Now using relation $f^{2009}(x) = x$ we get that $b_{2010} = b_1$. Thus
$$
b_1 = \left(-\frac{3+i\sqrt{7}}{2}\right)^{2009} b_1 \; \Rightarrow \; b_1 = 0\; \Rightarrow \; b_n = 0.
$$
Thus $a_n + \frac{3 - i\sqrt{7}}{2}a_{n-1} = 0$ or $$a_n = \left(-\frac{3 - i\sqrt{7}}{2}\right)a_{n-1} = \left(-\frac{3 - i\sqrt{7}}{2}\right)^{n}a_0$$
and again, as $f^{2009}(x) = f(x)$ we get $a_{2009} = a_0$ or
$$
a_0 = \left(-\frac{3 - i\sqrt{7}}{2}\right)^{2009}a_0 \;\Rightarrow\; a_0 = 0.
$$
As $a_0 = g(x) = f(x) -x$ we get $$f(x) - x = 0$$ QED.
