6th Australian 1985 Problem B3 Find all real polynomials p(x) such that p(x 2 + x + 1) = p(x) p(x + 1).
0, 1, (x 2 + 1) n for n = 1, 2, 3, ... Solution Suppose p(x) has a real root. Then take k to be the largest real root. But k 2 + k + 1 > k and is also a real root. Contradiction. So p(x) has no real roots. So p(x) must have even degree. Note also that it is immediate from the relation that the leading coefficient is 1. It is easy to check that x 2 + 1 is a solution and the only quadratic solution. Also notice that if q(x) and r(x) are solutions, then so is q(x)r(x), so we might conjecture that the solutions are (x 2 + 1) n . Let p(x) be a solution with degree 2n. Put q(x) = p(x) - (x 2 + 1) n . Suppose q(x) ≠ 0. It has no term in x 2n , so let deg q = m < 2n. Hence q(x 2 + x + 1) - q(x)q(x+1) has degree < 4n. But it equals p(x)((x+1) 2 +1) n + p(x+1)(x 2 +1) n , which has degree 4n. Contradiction. So we must have q(x) = 0, which shows that the only solution of degree 2n is (x 2 + 1) n . Finally notice that in addition to the solution 1 of degree 0, we also have the solution 0. Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered by email. 6th Aus 1985 © John Scholes jscholes@kalva.demon.co.uk 4 Jan 2004