
Re: Does polynomial P need to be an affine mapping
Posted:
May 8, 2013 4:54 PM


In article <fc897489095248d297b6c5296d668dfd@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.

