Is there a polynomial of degree at least two that represents a one-to-one mapping of the set of rational numbers onto itself?

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- May 9th 2010, 02:25 AMjames_bondPolynomial 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 AMroninpro
Here are some necessary conditions:

- The polynomial has odd degree
- has exactly one root, and it is rational, for any rational number

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 satisfies the above conditions and is a bijection to . That way, I can use a little calculus.

Maybe we can see where maps certain numbers. Write , where the are integers. For a moment, let us assume that . We can find so that . The only rational roots that can occur in are and . If we suppose that is injective / monotonic, then we can only take one of these as a root. A similar situation occurs when . We can make an analogous argument when is prime. So, if is prime or . And, to preserve montonicity, we must have or .

The polynomial is beginning to look linear when you restrict to primes or . You can then use the Mean Value Theorem to show that it really does take the form or everywhere.

Hopefully you can generalize the argument for .