Let a, b, c, d be odd integers such that 0 < a < b < c < d and ad = bc. Prove that if a + d = 2k and b + c = 2m for some integers k and m, then a = 1.
Solution
a < c, so a(d - c) < c (d - c) and hence bc - ac < c(d - c). So b - a < d - c, or a + d > b + c, so k > m.
bc = ad, so b(2m - b) = a(2k - a). Hence b2 - a2 = 2m(b - 2k-ma). But b2 - a2 = (b + a)(b - a), and (b + a) and (b - a) cannot both be divisible by 4 (since a and b are odd), so 2m-1 must divide b + a or b - a. But if it divides b - a, then b - a ≥ 2m-1, so b and c > 2m-1 and b + c > 2m. Contradiction. Hence 2m-1 divides b + a. If b + a ≥ 2m = b + c, then a ≥ c. Contradiction. Hence b + a = 2m-1.
So we have b = 2m-1 - a, c = 2m-1 + a, d = 2k - a. Now using bc = ad gives: 2ka = 22m-2. But a is odd, so a = 1.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 1998