IMO 1993 SL 11

Origin: IRE

Problem

Let n > 1 be an integer and let f(x) = xn + 5xn−1 + 3. Prove that there do not exist polynomials g(x), h(x), each having integer coefficients and degree at least one, such that f(x) = g(x)h(x).

Solution

Due to the extended Eisenstein criterion, f must have an irreducible factor of degree not less than n −1. Since f has no integral zeros, it must be irreducible. Second solution. The proposer’s solution was as follows. Suppose that f(x) = g(x)h(x), where g, h are nonconstant polynomials with integer coefficients. Since |f(0)| = 3, either |g(0)| = 1 or |h(0)| = 1. We may assume |g(0)| = 1 and that g(x) = (x−\alpha1) \cdot \cdot \cdot (x−\alphak). Then |\alpha1 \cdot \cdot \cdot \alphak| =

1. Since \alphan−1 i (\alphai + 5) = −3, taking the product over i = 1, 2, . . ., k yields |(\alpha1 + 5) \cdot \cdot \cdot (\alphak + 5)| = |g(−5)| = 3k. But f(−5) = g(−5)h(−5) = 3, so the only possibility is deg g = k = 1. This is impossible, because f has no integral zeros. Remark. Generalizing this solution, it can be shown that if a, m, n are positive integers and p < a −1 is a prime, then F(x) = xm(x −a)n + p is irreducible. The details are left to the reader.