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Here is the original problem :

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 $P(x)$.

Here is my solution, however i'm not sure of it , could you guys please check it ?

Solution :

It's easy to see that $P(P(x))=0$ for all $x=x_1,..,x_n$. The other roots that $P(P(x))$ may have are the eventual values $X_j$ for which $P(X_j)=x_j, j\neq 1$ (and those $X_j$ must be integers.) We have $P(0)=0$ so obviously $P(x)=a_nx^n+..+a_1x$. For all integers $a,b$ : $a-b\mid P(a)-P(b)$, since $P\in\mathbb Z[x]$. Taking $a=X_j$ and $b=x_j$ (for $i=2,..,n$, this is just : $X_j-x_j\mid x_j$. This means there exists some integer $k>1$ such that: $X_j=kx_j$, for all $2\le i\le n$. We have : $P(x_j)=a_nx_j^n+..+a_1x_j$. Thus : $P(X_j)=P(kx_j)=a_nk^nx_j^n+..+a_1kx_j=x_j$$\Longleftrightarrow$ $a_nk^nx_j^{n-1}+..+a_1k=1$. Thus : $a_nk^{n-1}x_j^{n-1}+..+a_1=\frac{1}{k}$. But $P$ has integer coefficients and $k,x_j$ are integers, so $\frac{1}{k}$ must be an integer, so $k=1\Longrightarrow X_j=x_j$, so the integer roots of $P(P(x))$ are the same as those of $P(x)$.

Thanks a lot !

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  • $\begingroup$ How do you conclude $X_j = kx_j$? From $X_j-x_j | x_j$ it follows that kX_j = (k+1)x_j$ for some integer $k$. Or did I misunderstand something? $\endgroup$ – MichalisN Jan 14 at 20:29
  • $\begingroup$ Oh i see, such a stupid error, thanks a lot ! :) $\endgroup$ – 1 2 3 Jan 14 at 20:50
  • $\begingroup$ Just editted my answer for clarity $\endgroup$ – Mike Jan 14 at 22:10
  • $\begingroup$ I actually, i think we could have used the fact that $\frac{k+1}{k}x_j=X_j$. For instance, we may write $P(x)=xQ(x)$ since $0$ is a root of $P$. It follows that $P(X_i)=x_i$ if and only if, $X_iQ(X_i)=x_i$ so $Q(X_i)=\frac{x_i}{X_i}$. $Q$ has integer coefficients, so the fraction must be an integer, so $\frac{k}{k+1}$ is an integer, so $k=0$. This means $X_j=x_j$. $\endgroup$ – 1 2 3 Jan 15 at 14:31
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So $P(x)=x\prod_{i=2}^n (x-x_i)$ where $x_2,\ldots, x_n$ are the nonzero integral roots of $P$. Now suppose there is a a nonzero root $y$ of $P(P(x))$ that is distinct from $0,x_2,\ldots, x_n$. Then $y\prod_{i=2}^n(y-x_i)$ must be in $\{x_2,x_3,\ldots, x_n\}$. We show that this is impossible for integral $y \not = 0,x_2,\ldots, x_n$ via Claim 1 below:

Claim 1: Let us use the notation as above, and let us write $x_n$ be the root of $P$ with the largest modulus i.e., $|x_n| \ge |x_i|$ for each $i=2,3,\ldots, n$. Let $y$ be a nonzero integer. Then $|P(y)| > |x_n|$.

Case 1: $0< |y| \le \frac{|x_n|}{2}$. Then $|y\prod_{i=2}^n (y-x_i)| > |x_n|$. Indeed, that $y$ is distinct from $0,x_2,\ldots, x_n$ implies $|y|$ and $|y-x_i|$ for each $i=2, \ldots, n-2$ is at least 1, and that at least 2 of $\{|y|, |y-x_i|; i=2,\ldots, n-1\}$ is at least 2, since the degree $n$ of $P$ is at least 5. So $|y\prod_{i=2}^{n-1}(y-x_i)|$ is at least 4. But then indeed as $|y| \le \frac{|x_n|}{2}$ it follows that $|y-x_n|$ is at least $\frac{|x_n|}{2}$, this implies that

$$|y\prod_{i=2}^{n} (y-x_i)| \ge \frac{|x_n|}{2} \times |y\prod_{i=2}^{n-1}(y-x_i)|$$

$$\ge \frac{|x_n|}{2} \times 4 > |x_n|.$$

So Claim 1 follws for Case 1.

Case 2: $|y| \ge \frac{|x_n|}{2}$. Then $|y\prod_{i=2}^n (y-x_i)| > |x_n|$. Indeed, that $y$ is distinct from $0,x_2,\ldots, x_n$ implies $|y-x_i|$ for each $i=2, \ldots, n$ is at least 1, and that $|y-x_i| \ge 2$ for at least 2 of the other $i$s in $2,\ldots, n$, since the degree $n$ of $P$ is at least 5. So $|\prod_{i=2}^{n}(y-x_i)|$ is at least 4. But then indeed as $|y|$ is at least $|x_n/2|$, this implies that

$$|y\prod_{i=2}^{n} (y-x_i)| \ge \frac{|x_n|}{2} \prod_{i=2}^{n}|(y-x_i)|$$

$$\ge \frac{|x_n|}{2} \times 4 > |x_n|.$$

So Claim 1 follows for the remaining Case 2 as well, thus Claim 1 follows.

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