Roots of $P(P(x))$ where 5th degree (or higher) $P(x)$ has all its coefficients and roots in $\mathbb Z$

Question

A polynomial $P(x)$ of degree $n \geq 5$ with integer coefficients and $n$ distinct integer roots is given. Find all integer roots of $P(P(x))$ given that $0$ is a root of $P(x)$.

This is an old Czech math olympiad problem (1998) that I cannot find a solution to. Any ideas?

Answer 1

If $p(p(x))=0$ then $p(x)=n$ where $n$ is a root of $p$. Thus we have $$n=x(x-n)q(x)$$ where $q(x)$ is the product of at least three integers. Since $x|n-x+x$ we have $x|n-x$ and since $n-x|n-x+x$ we have $n-x|x$ thus $n-x=\pm x$ the plus sign must hold since otherwise $n=0$. And the equation becomes $$2x=\pm x^2 q(x)$$ or if $x\neq 0$, $$2=\pm xq(x)$$ where $q(x)$ is a product of at least three distinct integers. This is impossible. Thus we must have either $x=0$ or $n=0$, in the latter case we see that $p(x)=0$ so $x$ is a root of the original polynomial, this includes $x=0$.

Answer 2

Use the rational roots theorem to extract all rational roots (which includes integer roots). If $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x^1$, then

$$P(P(x))=(a_n)^{n+1}x^{2n}+\dots+a_1x^1$$

Factor out $x$ and then apply the rational roots theorem.

Thus, all rational roots are of the form

$$r=\pm\frac pq$$

where $p$ are the factors of $a_1$ and $q$ are the factors of $(a_n)^{n-1}$

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To extract the trivial ones,

$$P(r)=0\tag{given roots}$$

$$P(P(r))=P(0)=0\tag{also given}$$

So all the original roots are still roots.

Answer 3

Let's say $P(x) = Ax\prod\limits_{k=2}^n(x-a_k)$ where $a_k$ are distinct non-zero integers. It is clear if $P(P(x)) = 0$, then $P(x)$ either vanishes or equal to one of the $a_i$. We claim that the second case is impossible for any integer $x$.

Assume the contrary, let says $P(m)$ equal to some $a_k$, say $a_2$ for integer $m \ne 0$.

Since $a_2 = A m\prod\limits_{k=2}^n (m - a_k)$ and the RHS (excludes $A$) is a product of $n \ge 5$ distinct integers,

$$|a_2| \ge |\pm 1|^2|\pm 2|^2|\pm 3| = 12$$

Rewrite the RHS as $m (a_2 - m) A \times\prod_{k=3}^n (m-a_k)$, the part after $A$ consists of a product of at least 3 distinct integers. this implies

$$\left|\prod_{k=3}^n (m-a_k)\right| \ge |\pm 1|^2|\pm 2| = 2$$

and leads to

$$|a_2| = |LHS| = |RHS| \ge 2|m(m-a_2)| \ge 2||a_2| - 1|$$

It is easy to see above equation doesn't have any solution when $|a_2| \ge 12$. As a result, the second case is impossible. So the only integer roots of $P(P(x))$ are those of $P(x)$.