IMO 1975 Problem B3 Find all polynomials P(x, y) in two variables such that: (1) P(tx, ty) = t n P(x, y) for some positive integer n and all real t, x, y; (2) for all real x, y, z: P(y + z, x) + P(z + x, y) + P(x + y, z) = 0; (3) P(1, 0) = 1.
(1) means that P is homogeneous of degree n for some n. Experimenting with low n, shows that the only solutions for n = 1, 2, 3 are (x - 2y), (x + y)(x - 2y), (x + y) 2 (x - 2y). It then obvious by inspection that (x + y) n (x - 2y) is a solution for any n. Taking x = y = z in (2) shows that P(2x,x) = 0, so (x - 2y) is always a factor. Taking x = y = 1, z = -2 gives P(1,-1) (2 n - 2) = 0, so (x + y) is a factor for n > 1. All this suggests (but does not prove) that the general solution is (x + y) n (x - 2y). Take y = 1 - x, z = 0 in (2) and we get: P(x, 1-x) = -1 - P(1-x, x). In particular, P(0,1) = -2. Now take z = 1 - x - y and we get: P(1-x, x) + P(1-y, y) + P(x+y, 1-x-y) = 0 and hence f(x+y) = f(x) + f(y), where f(x) = P(1-x, x) - 1. By induction we conclude that, for any integer m and real x, f(mx) = mf(x). Hence f(1/s) = 1/s f(1) and f(r/s) = r/s f(1) for any integers r, s. But P(0,1) = -2, so f(1) = -3. So f(x) = -3x for all rational x. But f is continuous, so f(x) = -3x for all x. So set x = b/(a+b), where a and b are arbitrary reals (with a+b non-zero). Then P(a,b) = (a+b) n P(1-x, x) = (a+b) n (-3b/(a+b) + 1) = (a+b) n-1 (a-2b), as claimed. [For a+b = 0, we appeal to continuity, or use the already derived fact that for n > 1, P(a,b) = 0.] s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 17th IMO 1975 © John Scholes jscholes@kalva.demon.co.uk 10 Oct 1998