Existence of a surjective non-linear polynomial $P \in \Bbb Q[X]$

Does their exist a non-linear polynomial $P(x)$ such that for every rational number $y$ there exists a rational number $x$ such that $y=P(x)$?

Solution 1:

No. First of all, the coefficients of $P$ must be rational: taking $n+1$ different rational $x$'s ($n=\deg P$) with rational $y=P(x)$ will give $n+1$ independent linear equations for the $n+1$ coefficients of $P$, the equations have rational coefficients, so the unique solution (the coefficients of $P$) is rational. Or simply compute the coefficients of $P$ using Lagrange interpolation formula.

Now let me use a nuclear weapon: the polynomial in two variables $Q(x,y)=P(x)-y$ is certainly irreducible, so by Hilbert's irreducibility theorem there is a $a\in\mathbb Q$ such that $P(x)-a$ is irreducible in $\mathbb Q[x]$, in particular it has no rational root (if it is not of degree $1$). Hence $a$ cannot be represented as $P(x)$ for $x$ rational.

edit: since using nuclear weapons might be against international treaties, let me sketch a direct proof. $P$ must be of odd degree (else $P(x)$ can't be arbitrarily large both positive and negative). For $y$ big enough there is now a unique real $X(y)$ s.t. $P(X(y))=y$.

We can suppose that $P$ has integer coefficients (if not, multiply $P$ by common denominator of the coefficients). This implies that if $y\in \mathbb Z$ and if $b\in\mathbb Q$ such that $P(b)=y$ then $b\in\frac{1}{c_n}\mathbb Z$, where $c_n$ is the leading coefficient of $P$.

So suppose now, as you do, that $X(y)$ is rational for every $y\in\mathbb Z$ large enough, i.e. that $X(y)\in \frac{1}{c_n}\mathbb Z$. Given that $X(y)$ is strictly but slowly increasing (the derivative of $X$ (with respect to $y$) goes to zero as $y\to\infty$ if $n>1$), we get for $y\in\mathbb Z$ big enough (bigger than before) $0<X(y+1)-X(y)<1/c_n$, which is a contradiction with $X(y),X(y+1)\in \frac{1}{c_n}\mathbb Z$.