IMO 1992 SL 9

Origin: IRN

Problem

Let f(x) be a polynomial with rational coefficients and \alpha be a real number such that \alpha3 −\alpha = f(\alpha)3 −f(\alpha) = 331992. Prove that for each n \geq1, (f (n)(\alpha))3 −f (n)(\alpha) = 331992, where f (n)(x) = f(f(. . . f(x))), and n is a positive integer.

Solution

Since the equation x3 −x −c = 0 has only one real root for every c > 2/(3 \sqrt 3) , \alpha is the unique real root of x3 −x −331992 = 0. Hence f n(\alpha) = f(\alpha) = \alpha. Remark. Consider any irreducible polynomial g(x) in the place of x3 − x −331992. The problem amounts to proving that if \alpha and f(\alpha) are roots

of g, then any f (n)(\alpha) is also a root of g. In fact, since g(f(x)) vanishes at x = \alpha, it must be divisible by the minimal polynomial of \alpha, that is, g(x). It follows by induction that g(f (n)(x)) is divisible by g(x) for all n \inN, and hence g(f (n)(\alpha)) = 0.