IMO 1984 SL 16

Origin: POL

Problem

Let a, b, c, d be odd positive integers such that a < b < c < d, ad = bc, and a + d = 2k, b + c = 2m for some integers k and m. Prove that a = 1.

Solution

From the first two conditions we can easily conclude that a + d > b + c (indeed, (d + a)2 −(d −a)2 = (c + b)2 −(c −b)2 = 4ad = 4bc and d −a > c −b > 0). Thus k > m. From d = 2k −a and c = 2m −b we get a(2k −a) = b(2m −b), or equivalently, (b + a)(b −a) = 2m(b −2k−ma). (1) Since 2k−ma is even and b is odd, the highest power of 2 that divides the right-hand side of (1) is m. Hence (b + a)(b −a) is divisible by 2m but not by 2m+1, which implies b+a = 2m1p and b−a = 2m2q, where m1, m2 \geq1, m1 + m2 = m, and p, q are odd. Furthermore, b = (2m1p + 2m2q)/2 and a = (2m1p −2m2q)/2 are odd, so either m1 = 1 or m2 = 1. Note that m1 = 1 is not possible, since it would imply that b −a = 2m−1q \geq2m−1, although b + c = 2m and b < c imply that b < 2m−1. Hence m2 = 1 and m1 = m −1. Now since a + b < b + c = 2m, we obtain a + b = 2m−1 and b −a = 2q, where q is an odd integer. Substituting these into (1) and dividing both sides by 2m we get q = 2m−2 + q −2k−ma =⇒ 2k−ma = 2m−2. Since a is odd and k > m, it follows that a = 1. Remark. Now it is not difficult to prove that all fourtuplets (a, b, c, d) that satisfy the given conditions are of the form (1, 2m−1−1, 2m−1+1, 22m−2− 1), where m \inN, m \geq3.