Is there a polynomial of degree at least two that represents a one-to-one mapping of the set of rational numbers onto itself?
Here are some necessary conditions:
- The polynomial $\displaystyle p(x)$ has odd degree
- $\displaystyle p'(x)\geq 0$
- $\displaystyle p(x)-r$ has exactly one root, and it is rational, for any rational number $\displaystyle r$
It seems that the third condition will be difficult to satisfy. I haven't thought about it completely - it is a little too early in the morning for me to concentrate.
Any ideas?
EDIT:
I'd like to assume that $\displaystyle p:\mathbb{R}\to \mathbb{R}$ satisfies the above conditions and $\displaystyle p|_\mathbb{Q}$ is a bijection to $\displaystyle \mathbb{Q}$. That way, I can use a little calculus.
Maybe we can see where $\displaystyle p(x)$ maps certain numbers. Write $\displaystyle p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$, where the $\displaystyle a_i$ are integers. For a moment, let us assume that $\displaystyle a_n=1$. We can find $\displaystyle r$ so that $\displaystyle a_0-r=1$. The only rational roots that can occur in $\displaystyle p(x)-r$ are $\displaystyle 1$ and $\displaystyle -1$. If we suppose that $\displaystyle p(x)$ is injective / monotonic, then we can only take one of these as a root. A similar situation occurs when $\displaystyle a_0-r=-1$. We can make an analogous argument when $\displaystyle a_0-r$ is prime. So, $\displaystyle -p^{-1}(x)=p^{=1}(-x)$ if $\displaystyle x$ is prime or $\displaystyle \pm 1$. And, to preserve montonicity, we must have $\displaystyle p(x)=x$ or $\displaystyle p(x)=-x$.
The polynomial is beginning to look linear when you restrict to primes or $\displaystyle \pm 1$. You can then use the Mean Value Theorem to show that it really does take the form $\displaystyle p(x)=x$ or $\displaystyle p(x)=-x$ everywhere.
Hopefully you can generalize the argument for $\displaystyle a_n>1$.