IMO 1989 SL 4

Origin: BUL

Problem

Prove that for every integer n > 1 the equation xn n! + xn−1 (n −1)! + \cdot \cdot \cdot + x2 2! + x 1! + 1 = 0 has no rational roots.

Solution

First we note that for every integer k > 0 and prime number p, pk doesn’t divide k!. This follows from the fact that the highest exponent r of p for which pr|k! is r = k p

k p2

• \cdot \cdot \cdot < k p + k p2 + \cdot \cdot \cdot = k p −1 < k. Now suppose that \alpha is a rational root of the given equation. Then \alphan + n! (n −1)!\alphan−1 + \cdot \cdot \cdot + n! 2! \alpha2 + n! 1! \alpha + n! = 0, (1) from which we can conclude that \alpha must be an integer, not equal to \pm1. Let p be a prime divisor of n and let r be the highest exponent of p for which pr|n!. Then p | \alpha. Since pk|\alphak and pk ∤k!, we obtain that pr+1 | n!\alphak/k! for k = 1, 2, . . . , n. But then it follows from (1) that pr+1 | n!, a contradiction.