# Polynomial maps rational numbers onto itself

• May 9th 2010, 02:25 AM
james_bond
Polynomial maps rational numbers onto itself
Is there a polynomial of degree at least two that represents a one-to-one mapping of the set of rational numbers onto itself?
• May 9th 2010, 03:59 AM
roninpro
Here are some necessary conditions:

1. The polynomial $p(x)$ has odd degree
2. $p'(x)\geq 0$
3. $p(x)-r$ has exactly one root, and it is rational, for any rational number $r$

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 $p:\mathbb{R}\to \mathbb{R}$ satisfies the above conditions and $p|_\mathbb{Q}$ is a bijection to $\mathbb{Q}$. That way, I can use a little calculus.

Maybe we can see where $p(x)$ maps certain numbers. Write $p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$, where the $a_i$ are integers. For a moment, let us assume that $a_n=1$. We can find $r$ so that $a_0-r=1$. The only rational roots that can occur in $p(x)-r$ are $1$ and $-1$. If we suppose that $p(x)$ is injective / monotonic, then we can only take one of these as a root. A similar situation occurs when $a_0-r=-1$. We can make an analogous argument when $a_0-r$ is prime. So, $-p^{-1}(x)=p^{=1}(-x)$ if $x$ is prime or $\pm 1$. And, to preserve montonicity, we must have $p(x)=x$ or $p(x)=-x$.

The polynomial is beginning to look linear when you restrict to primes or $\pm 1$. You can then use the Mean Value Theorem to show that it really does take the form $p(x)=x$ or $p(x)=-x$ everywhere.

Hopefully you can generalize the argument for $a_n>1$.