31st BMO 1995 Problem Find all positive integers a ≥ b ≥ c such that (1 + 1/a)(1 + 1/b)(1 + 1/c) = 2. Solution
: (a, b, c) = (5, 4, 3), (8, 3, 3), (7, 6, 2), (9, 5, 2), (15, 4, 2). We have 1 + 1/a ≤ 1 + 1/b ≤ 1 + 1/c. So 2 ≤ (1 + 1/c) 3 . But (1 + 1/c) 3 is a decreasing function of c and c = 4 gives 125/64  3b + 3. Contradiction. (b - 3)a = 3b + 3, so b - 3 must be positive. Thus we need only consider b = 4, 5 and 6. These give the three solutions with c = 2 shown above. Suppose c = 3. Then 2ab = 4(a + b + 1), so ab = 2a + 2b + 2. So (b - 2)a = 2b + 2. If b ≥ 5, then (b - 2)a ≥ 3a ≥ 3b ≥ 2b + 5 > 2b + 2. Contradiction. But b - 2 must be positive, so we need only consider b = 3 and 4. That gives the two solutions shown above for c = 3. 31st BMO 1995 © John Scholes jscholes@kalva.demon.co.uk 14 Aug 2002