Sufficient condition for a polynomial $f$ over $\mathbb Q$ to be linear is $f(\mathbb Q)=\mathbb Q$.

Question

I am reading a book (Polynomial Mappings by W. Narkiewicz) on invariance of polynomial maps where I found the following result along with a proof:

Theorem: Let $f\in\mathbb Q[x]$ be a polynomial such that $f(\mathbb Q)=\mathbb Q$. Then $f$ must be linear.

Proof. If such an $f$ exists with degree at least 2, then there must also be such a polynomial over $\mathbb Z$. Let $$f(x)=a_Nx^N+\cdots+a_0$$ be a polynomial over $\mathbb Z$ satisfying the condition that $f(\mathbb Q)=\mathbb Q$ with $a_N\neq 0$ and $N\ge 2$. By possibly replacing $f$ by $-f$ one can assume that $a_N>0$. Then for some $a$ large enough, $f$ is increasing on $[a,\infty)$. Let $x_n=p_n/q_n$, $(p_n,q_n)=1$, be positive rational numbers such that $f(x_n)=n$ for $n\in\mathbb N$. For sufficiently large $n$ one has $x_{n+1}-x_n\ge 1/a_N$, so that $1=f(x_{n+1})-f(x_n)$ tends to infinity, an impossibility.

I have the following queries.
Q1) Why can we always choose positive $x_n$'s such that $f(x_n)=n$?

Q2) Why is it true that for sufficiently large $n$ $x_{n+1}$ is bigger than $x_n$? I know how to show that this would lead to $x_{n+1}-x_n\ge 1/a_N$.

Q3) Also, I fail to understand where exactly was the condition of non-linearity being used here?

Answer 1

(1)/(2) Even when $f$ is linear, it is not generally possible to pick a sequence $(x_n)$ of positive values $x_n$ for which $f(x_n)=n$ for all naturals $n\in\Bbb N$. Indeed, $f(x)=x+C$ makes this false for large enough $C\in\Bbb N$. Instead, this statement itself should be "for all sufficiently large" $n$. Plus, it is generally possible to find values $x$ outside of $[a,\infty)$ for which $f(x)\in\Bbb N$, so the proof also needs to specify the $x_n$s are chosen from the interval $[a,\infty)$, and (without loss of generality) $a>0$.

Suppose $f$ is increasing on the interval $[a,\infty)$ with $a>0$. The image of this interval under $f$ is $[f(a),\infty)$, and the inverse image of $[f(a),\infty)\cap\Bbb Q$ must be $[a,\infty)\cap\Bbb Q$. Then we can pick unique values $x_n\in[a,\infty)$ for which $f(x_n)=n$ for all sufficiently large naturals, i.e. $n\ge N$ where $N\in[a,\infty)\cap\Bbb N$ is arbitrary. The values $x_n$ are unique because $f$ increasing implies bijective on this interval $[a,\infty)$, they are positive since $x_n\ge a>0$, and they are increasing ($x_{n+1}>x_n$) since $f$ is increasing.

(3) The assumption $f$ is nonlinear implies $\lim\limits_{n\to\infty}[f(x_{n+1})-f(x_n)]=+\infty$, which is what contradicts the other observation that $f(x_{n+1})-f(x_n)=1$ for all $n$. Indeed,

$$f(x_{n+1})-f(x_n) ~>~ f(x_n+\tfrac{1}{a_N})-f(x_n)$$

for large enough $n$ since $f$ is increasing and $x_{n+1}-x_n>\frac{1}{a_N}$ for large enough $n$, and the RHS above has positive nonconstant leading term as a polynomial in $x_n$ (and $x_n\to\infty$ with $n$).

Answer 2

I'll address some of the more terse parts of the proof. Also, I'll write the polynomial as \begin{equation*} f(x)=a_kx^k+\cdots+a_0 \end{equation*} to avoid confusion between $N$ and $n$.

0. As you mentioned, $x_{n+1} - x_n \geq 1/a_k $ because $q_n \mid a_k$ for every $n$. This can be seen as $f(x_n) = n \implies a_k (p_n / q_n)^k + a_{k-1} (p_{n} / q_n)^{k-1} + \cdots + a_0 = n$. Now multiply $q_n^k$ to both sides of the equation, modulo $q_n$, and use the fact that $a_i \in \mathbb{Z}$, $(p, q) = 1$ to deduce this fact.

1. Why is it true that for sufficiently large $n$ $x_{n+1}$ is bigger than $x_n$?

I think what the author is trying to say is that for large enough $a$, the polynomial is increasing on $[a, \infty)$. Choose rational points $x_n \in [a, \infty)$ such that $f(x_n) = n$. From the fact that the polynomial is increasing it is clear that $x_{n+1} > x_n$.

2. Why can we always take positive $x_n$'s?

This is because $a_k > 0$. So one can choose $[a, \infty)$ with $a > 0$ and large enough such that the $a_k x^k$ term dominates.

3. Also, I fail to understand where exactly was the condition of non-linearity being used here?

Non-linearity is used here because we have a sequence of points $\{ x_n \}$ that are `sufficiently spaced' apart ($\geq 1/a_k$ spacing between them), yet the difference in the polynomial evaluation of those points are evenly spaced. This is impossible unless $f$ is linear.