IMO 2002 SL A3

Origin: POL | Category: Algebra

Problem

Let P be a cubic polynomial given by P(x) = ax3+bx2+cx+ d, where a, b, c, d are integers and a ̸= 0. Suppose that xP(x) = yP(y) for infinitely many pairs x, y of integers with x ̸= y. Prove that the equation P(x) = 0 has an integer root.

Solution

Let x, y be distinct integers satisfying xP(x) = yP(y); this is equivalent to a(x4 −y4) + b(x3 −y3) + c(x2 −y2) + d(x −y) = 0. Dividing by x −y we obtain a(x3 + x2y + xy2 + y3) + b(x2 + xy + y2) + c(x + y) + d = 0. Putting x + y = p, x2 + y2 = q leads to x2 + xy + y2 = p2+q , so the above equality becomes apq + b 2(p2 + q) + cp + d = 0, i.e. (2ap + b)q = −(bp2 + 2cp + 2d). Since q \geqp2/2, it follows that p2|2ap + b| \leq2|bp2 + 2cp + 2d|, which is possible only for finitely many values of p, although there are infinitely many pairs (x, y) with xP(x) = yP(y). Hence there exists p such that xP(x) = (p −x)P(p −x) for infinitely many x, and therefore for all x. If p ̸= 0, then p is a root of P(x). If p = 0, the above relation gives P(x) = −P(−x). This forces b = d = 0, so P(x) = x(ax2 + c). Thus 0 is a root of P(x).