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Math Help - Polynomial maps rational numbers onto itself

  1. #1
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    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?
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  2. #2
    Senior Member roninpro's Avatar
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    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.
    Last edited by roninpro; May 9th 2010 at 05:47 AM.
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