BMO 1967 Problem 7 Find all reals a, b, c, d such that abc + d = bcd + a = cda + b = dab + c = 2. Solution
: 1, 1, 1, 1 (one solution); 3, -1, -1, -1 (four solutions). Either all a, b, c, d are equal, which gives the first solution, or there is an unequal pair. Suppose a ≠ b. But bcd + a = cda + b, so (a - b)(cd - 1) = 0, and hence cd = 1. Hence a + b = 2. It follows that ab ≠ 1 (because the only solution to a + b = 2, ab = 1 is a = b = 1). Hence c = d (since (c - d)(ab - 1) = 0). So either c = d = 1, or c = d = -1. But the first case gives ab = 1, a + b = 2 (contradiction, since the only possibility then is a = b = 1, whereas we are assuming a ≠ b). So c = d = -1. Hence ab = -3, a + b = 2, so a = 3, b = -1 (or vice versa). 3rd BMO 1967 © John Scholes jscholes@kalva.demon.co.uk 30 Aug 2000