The composition $f^2(x)$ has at least as many roots as $f(x)$

I have seen that a similar question has been asked, but that question was regarding polynomials of odd degree. My question is regarding single variable polynomials with integer coefficients in general.

I have read in a paper that the polynomial $f^2(x) = f(f(x))$ contains at least as many distinct roots as $f(x) \in \mathbb{Z}[x]$. It seems to me that, in order to prove this statement, one has to show that for every root $r \in \mathbb{Z}$ of $f(x)$ there exists a $q \in \mathbb{Z}$ such that $f(q) = r$. This is easy to show when the degree of the polynomial is odd, because the polynomial is surjective in that case. However, I do not see why this is necessarily the case when the degree is even.

Does the statement still hold for such polynomials? If it does, is this still the right strategy to prove it or should I approach this problem in a different way? If it does not, what is a counter example?

You would need, for instance, that the roots are all smaller than the minimum of the function. In degree $2$ this is satisfied for $$f(x)=(x+3)^2-1$$ which has roots at $-2, -4$ but no real solutions for $f(x)=-2$ and $f(x)=-4)$ resp. $$f(f(x))=(f(x)+2)(f(x)+4)=((x+3)^2+1)(x+3)^2+3)=(x+3)^4+4(x+3)^2+3$$ is always positive.
• Over the complex numbers the equation $f(z)=r_k$ has always at least one solution, and the solution sets are disjoint for different $r_k$. – LutzL May 16 '18 at 12:01
• Yes, as $f(z)-r_k$ is a polynomial with complex coefficients. – LutzL May 16 '18 at 12:13