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Topic: Does polynomial P need to be an affine mapping
Replies: 7   Last Post: May 10, 2013 6:45 AM

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Ken.Pledger@vuw.ac.nz

Posts: 1,374
Registered: 12/3/04
Re: Does polynomial P need to be an affine mapping
Posted: May 8, 2013 4:54 PM
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In article <fc897489-0952-48d2-97b6-c5296d668dfd@googlegroups.com>,
steinerartur@gmail.com wrote:

> Let P be a polynomial with real coefficients. Suppose there are non empty
> intervals I and J such that P maps surjectively the rationals of I into the
> rationals of J. Does this imply P is an affine mapping ?
>
> Why or why not? So far, I couldn't come to a conclusion....



A fascinating problem! Intuitively it seems obvious, but I can't
prove it either. I've managed a first stage of proof (as you may have
done yourself): P must have rational coefficients.

Proof of that bit.
Let n be the degree of P. Choose any n+1 rationals in I. Their
images under P must be rationals in J. Then the Lagrange interpolation
formula gives a polynomial which matches P for more than n values of x,
so must be P itself. Therefore all coefficients in P are rational.

After that, it seems natural to take an arbitrary rational a/b in
J, so then some rational c/d in I satisfies P(c/d) = a/b. Clearing
the fractions and thinking about divisibility properties of the first
and last coefficients seems a good way to go, but I haven't been able to
make it work. :-(

Good luck!

Ken Pledger.



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