Let $P(x) \in R[x]$ be a polynomial with real coefficients such that $$(\forall n \in \mathbb N)(\exists q \in \mathbb Q)(P(q)=n)$$
What's the biggest possible value of $\deg P$? ($\deg P$ is the degree of polynomial $P$)
I have literally no clue as where to start, and a hint would be more then enough.
Thanks in advance.
An initial observation: If $P(x_n) = n $ for each $n$, then the sequence $(x_n)$ can not have an accumulation point, or else $P$ would have a singularity there, a contradiction (since $P$ is a polynomial).
Let's try $P(x) =ax+2+bx+c $.
If $P(x) = n$, then $ax^2+bx+c=n$, so $x =\dfrac{-x\pm\sqrt{b^2-4a(c-n)}}{2a} $, so $b^2-4a(c-n) =r^2 $ or $n =\dfrac{r^2-b^2}{4a}+c $. For $n+1$, $n+1 =\dfrac{s^2-b^2}{4a}+c $ for some rational $s$.
Subtracting, $1 =\dfrac{s^2-r^2}{4a} $ or $4a =s^2-r^2 $ or $s-r =\dfrac{4a}{s+r} $.
Since $(x_n)$ does not have an accumulation point, we can make $|s|$ and $|r|$ arbitrarily large.
But then we can make $s-r$ arbitrarily small, so $(x_n)$ would have an accumulation point at zero.
This is a contradiction.
Therefore $P$ can not be of degree 2.
I'm not sure how to generalize this, since it depends on the explicit solution for quadratics.
I'll leave it at this and hope that someone can use the ideas in this answer.
