IMO 2001 SL N5

Origin: BUL | Category: Number Theory

Problem

Let a > b > c > d be positive integers and suppose ac + bd = (b + d + a −c)(b + d −a + c). Prove that ab + cd is not prime.

Solution

The given equality is equivalent to a2 −ac + c2 = b2 + bd + d2. Hence (ab + cd)(ad + bc) = ac(b2 + bd + d2) + bd(a2 −ac + c2), or equivalently, (ab + cd)(ad + bc) = (ac + bd)(a2 −ac + c2). (1) Now suppose that ab + cd is prime. It follows from a > b > c > d that ab + cd > ac + bd > ad + bc; (2) hence ac + bd is relatively prime with ab + cd. But then (1) implies that ac + bd divides ad + bc, which is impossible by (2). Remark. Alternatively, (1) could be obtained by applying the law of cosines and Ptolemy’s theorem on a quadrilateral XY ZT with XY = a, Y Z = c, ZT = b, TX = d and \angleY = 60◦, \angleT = 120◦.