IMO 1993 · Problem 1

Let \(f(x) = x^n + 5x^{n-1} + 3\), where \(n > 1\) is an integer. Prove that \(f(x)\) cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Official text of the problem. Official problem paper (PDF)

Solution

Suppose, for a contradiction, that \(f=gh\), where \(g,h\in\mathbb Z[x]\) are nonconstant. Their leading coefficients multiply to \(1\), so, changing both signs if necessary, we may assume that both are monic.

Since \[ g(0)h(0)=3, \] we may interchange the factors so that \(g(0)\in\{1,-1\}\).

Write \(\overline g,\overline h\) for their reductions modulo \(3\). In \(\mathbb F_3[x]\), \[ \overline g(x)\,\overline h(x) =x^n+2x^{n-1} =x^{n-1}(x-1). \] Because \(\overline g(0)\ne0\), the polynomials \(\overline g\) and \(x^{n-1}\) are relatively prime. Hence \(\overline g\) divides \(x-1\).

As \(g\) is monic, reduction modulo \(3\) preserves its degree and leading coefficient. Thus \(\overline g\) is nonconstant and monic, forcing \[ \overline g=x-1,\qquad \deg g=1. \] Consequently \(g=x+c\), where \(c=g(0)\in\{1,-1\}\). Its reduction gives \(c\equiv-1\pmod3\), so \(c=-1\).

Therefore \(x-1\) divides \(f\), contradicting \[ f(1)=1+5+3=9\ne0. \] This proves the assertion for every integer \(n>1\). \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).