IMO 1993 SL 7

Origin: GEO

Problem

Let a, b, c be given integers a > 0, ac −b2 = P = P1 \cdot \cdot \cdot Pm where P1, . . . , Pm are (distinct) prime numbers. Let M(n) denote the num- ber of pairs of integers (x, y) for which ax2 + 2bxy + cy2 = n. Prove that M(n) is finite and M(n) = M(P k \cdot n) for every integer k \geq0.

Solution

Multiplying by a and c the equation ax2 + 2bxy + cy2 = P kn, (1) gives (ax + by)2 + Py2 = aP kn and (bx + cy)2 + Px2 = cP kn. It follows immediately that M(n) is finite; moreover, (ax+by)2 and (bx+ cy)2 are divisible by P, and consequently ax + by, bx + cy are divisible by P because P is not divisible by a square greater than 1. Thus there exist integers X, Y such that bx + cy = PX, ax + by = −PY . Then x = −bX −cY and y = aX + bY . Introducing these values into (1) and simplifying the expression obtained we get aX2 + 2bXY + cY 2 = P k−1n. (2) Hence (x, y) "\to(X, Y ) is a bijective correspondence between integral so- lutions of (1) and (2), so that M(P kn) = M(P k−1n) = \cdot \cdot \cdot = M(n).