TITLE: Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively?
QUESTION [11 upvotes]: Clearly this is impossible for $p$ of even degree, and I imagine that Cardano’s formula quickly reveals it to be impossible in the cubic case, although I have not checked in detail. My guess is that no such $p$ exists. Does one exist? If so, is there an explicit example?
Failing a general yes or no answer, are there sufficient conditions to identify a non-surjective polynomial function?

REPLY [8 votes]: I claim that no polynomial $q$ of degree greater than $1$ and rational coefficients can be a surjective mapping from $\mathbb{Q}$ to $\mathbb{Q}$.
Suppose that a polynomial $q$ of degree greater than $1$ is surjective from $\mathbb{Q}$ to $\mathbb{Q}$. For simplicity, by replacing $q(x)$ with $p(x)=\alpha(q(\beta x)-\gamma)$ where $\alpha,\beta,\gamma$ are rational with $\alpha,\beta\neq 0$, we can assume that $p(x)$ is a surjective monic polynomial with constant term $0$ and integer coefficients. Suppose now that $p(x)=x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x$ where the coefficients $a_{1},\dots,a_{n-1}$ are integers.
If $\alpha,\beta$ are integers with $p(x)=\frac{\alpha}{\beta}$, then $\beta x^{n}+\dots+\beta a_{1}x=\alpha$, so by the rational root theorem, $x$ must be of the form $\frac{r}{s}$ where $r$ is a factor of $\alpha$ and $s$ is a factor of $\beta$. In particular, in the case where $\beta=1$, if $p(x)=\alpha$, then $x$ must be a factor of $\alpha$. Therefore,
$p$ must restrict to a surjective function from $\mathbb{Z}$ to $\mathbb{Z}$. This is impossible.