IMO 1975 SL 10

Origin: GBR

Problem

The function f(x, y) is a homogeneous polynomial of the nth degree in x and y. If f(1, 0) = 1 and for all a, b, c, f(a + b, c) + f(b + c, a) + f(c + a, b) = 0, prove that f(x, y) = (x −2y)(x + y)n−1.

Solution

We shall prove that for all x, y with x+y = 1 it holds that f(x, y) = x−2y. In this case f(x, y) = f(x, 1 −x) can be regarded as a polynomial in z = x −2y = 3x −2, say f(x, 1 −x) = F(z). Putting in the given relation a = b = x/2, c = 1 −x, we obtain f(x, 1 −x) + 2f(1 −x/2, x/2) = 0; hence F(z) + 2F(−z/2) = 0. Now F(1) = 1, and we get that for all k,

F((−2)k) = (−2)k. Thus F(z) = z for infinitely many values of z; hence F(z) \equivz. Consequently f(x, y) = x −2y if x + y = 1. For general x, y with x+y ̸= 0, since f is homogeneous ,we have f(x, y) = (x + y)nf  x x+y, y x+y

= (x + y)n  x x+y −2 y x+y

= (x + y)n−1(x −2y). The same is true for x + y = 0, because f is a polynomial.