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?



Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively?
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.


