From: John Rickard Subject: Re: Surprisingly tricky Putnam-type number theory/polynomial problems Date: 17 Feb 1999 13:47:57 +0000 (GMT) Newsgroups: sci.math Keywords: Integral polynomials with the same range Larry Penn wrote: : (A) Let p(x),q(x) be two polynomials with integral coefficients. : : Restrict their domains to the integers Z. : : If they have the same ranges, then prove that they must be related via: : p(x) = q(a*x + b), with a= 1 or -1, b in Z. Not true: x^2 + x and 4x^2 + 2x have the same ranges. I think I can prove it if you also allow a to be 2 or 1/2 (and b to be a half-integer if a = 1/2). -- John Rickard ============================================================================== From: "Larry Penn" Subject: Re: Surprisingly tricky Putnam-type number theory/polynomial problems Date: Wed, 17 Feb 1999 09:59:40 -0500 Newsgroups: sci.math Well done! I know that the rational case is definitely true, and I thought that it would carry over to the integer case, but I was wrong! I had problems with the even-degree case that I thought could be overcome, but clearly they can't be! Do you think that the integer case is true if you restrict the polynomials to having the same leading coefficient? John Rickard wrote in message ... > >Not true: x^2 + x and 4x^2 + 2x have the same ranges. I think I can >prove it if you also allow a to be 2 or 1/2 (and b to be a >half-integer if a = 1/2). ============================================================================== From: John Rickard Subject: Re: Surprisingly tricky Putnam-type number theory/polynomial problems Date: 17 Feb 1999 18:49:54 +0000 (GMT) Newsgroups: sci.math Larry Penn wrote: : Do you think that the integer case is true if you restrict the polynomials : to having the same leading coefficient? Yes. In fact, I believe that the only exceptions to the original statement are pairs of polynomials of the form f(x) = p((x+a)(x+a+1)) g(x) = p((2x+b)(2x+b+1)) where p is a polynomial with integer coefficients, and a and b are integers. -- John Rickard