Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively? 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?
 A: 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.
A: No, this can't happen. One way to prove this is via Hilbert irreducibility: The polynomial $p(x) - t$ is irreducible over $\mathbb Q[x,t]$, so there are infinitely many specializations $t = c$ with $c \in \mathbb Q$ such that $p(x) - c$ is irreducible in  $\mathbb Q[x]$. Since the degree of $p(x)$ is greater than 1, it follows that for each such $c$ the polynomial $p(x) - c$ has no rational roots.
A: John's answer is good, but an alternative, more elementary, method is to use the theory of height functions. Thus for a fraction in lowest terms $a/b$, we define the height of $a/b$ to be
$$ H(a/b) = \max\bigl\{ |a|,|b|\bigr\}. $$
One can show that if $p(x)\in\mathbb Q[x]$ is a polynomial of degree $d$, then there is a constant $C_1(p)>0$ depending only on $p$ such that
$$ H\bigl(p(a/b)\bigr) \ge C_1(p) H(a/b)^d. $$
From this one can prove a counting result:
$$ \# \bigl\{ a/b\in\mathbb Q : H\bigl(p(a/b)\bigr) < X \bigr\} \le C_2(p)\cdot X^{2/d}. $$
On the other hand, it's not hard to see that here is a constant $C_3>0$ such that
$$ \#\bigl\{ a/b\in\mathbb Q: H(a/b) < X\bigr\} \ge C_3\cdot X^2. $$
So the heights of the values of $p(a/b)$ grow too rapidly to cover all of $\mathbb Q$. (Indeed, part of the proof of Hilbert irreducibility uses an argument of this sort.)
