IMO 1992 SL 12

Origin: NET

Problem

Let f, g, and a be polynomials with real coefficients, f and g in one variable and a in two variables. Suppose f(x) −f(y) = a(x, y)(g(x) −g(y)) for all x, y \inR. Prove that there exists a polynomial h with f(x) = h(g(x)) for all x \inR.

Solution

Let us set deg f = n and deg g = m. We shall prove the result by induction on n. If n < m, then degx[f(x) −f(y)] < degx[g(x) −g(y)], which implies that f(x)−f(y) = 0, i.e., that f is constant. The statement trivially holds. Assume now that n \geqm. Transition to f1(x) = f(x) −f(0) and g1(x) = g(x) −g(0) allows us to suppose that f(0) = g(0) = 0. Then the given condition for y = 0 gives us f(x) = f1(x)g(x), where f1(x) = a(x, 0) and deg f1 = n −m. We now have a(x, y)(g(x) −g(y)) = f(x) −f(y) = f1(x)g(x) −f1(y)g(y) = [f1(x) −f1(y)]g(x) + f1(y)[g(x) −g(y)].

Since g(x) is relatively prime to g(x)−g(y), it follows that f1(x)−f1(y) = b(x, y)(g(x)−g(y)) for some polynomial b(x, y). By the induction hypoth- esis there exists a polynomial h1 such that f1(x) = h1(g(x)) and con- sequently f(x) = g(x) \cdot h1(g(x)) = h(g(x)) for h(t) = th1(t). Thus the induction is complete.