10th Irish 1997 Problem A3 Find all polynomials p(x) such that (x-16) p(2x) = (16x-16) p(x).
k(x-2)(x-4)(x-8)(x-16) Solution There is no loss of generality in taking the leading coefficient to be 1. Since we have a factor 2 4 on the rhs, p(x) must have degree 4. (x-16) is a factor of the lhs, so it must also be a factor of p(x), so one root is 16. Similarly x-1 must be a factor of p(2x) so p(x) must have a root 2. Thus p(x) = (x-2)(x-16)(x-a)(x-b), p(2x) = 16(x-1)(x-8)(x-a/2)(x-b/2) and we have (x-8)(x-a/2)(x-b/2) = (x-2)(x-a)(x-b). Hence one of a, b is 8. wlog a = 8. Hence b = 4. So p(x) = (x-2)(x-4)(x-8)(x-16). 10th Irish 1997 © John Scholes jscholes@kalva.demon.co.uk 17 Dec 2003 Last updated/corrected 17 Dec 03