IMO 1992 Problem A1 Find all integers a, b, c satisfying 1 < a < b < c such that (a - 1)(b -1)(c - 1) is a divisor of abc - 1. Solution
: a = 2, b = 4, c = 8; or a = 3, b = 5, c = 15. Let k = 2 1/3 . If a ≥ 5, then k(a - 1) > a. [Check: (k(a - 1) 3 - a 3 = a 3 - 6a 2 + 6a - 2. For a ≥ 6, a 3 ≥ 6a 2 and 6a > 2, so we only need to check a = 5: 125 - 150 + 30 - 2 = 3.] We know that c > b > a, so if a ≥ 5, then 2(a - 1)(b - 1)(c - 1) > abc > abc - 1. So we must have a = 2, 3 or 4. Suppose abc - 1 = n(a - 1)(b - 1)(c - 1). We consider separately the cases n = 1, n = 2 and n ≥ 3. If n = 1, then a + b + c = ab + bc + ca. But that is impossible, because a, b, c are all greater than 1 and so a = 9, then bc >= 9c > 4c + 4b. So we must have b = 5 or 7. If b = 5, then we find c = 15, which gives a solution. If b = 7, then we find c = 23/3 which is not a solution. The remaining case is n >= 3. If a = 2, we have n(bc - b - c + 1) = 2bc - 1, or (n - 2)bc + (n + 1) = nb + nc. But b ≥ 3, so (n - 2)bc ≥ (3n - 6)c ≥ 2nc for n ≥ 6, so we must have n = 3, 4 or 5. If n = 3, then bc + 4 = 3b + 3c. If b >= 6, then bc ≥ 6c > 3b + 3c, so b = 3, 4 or 5. Checking we find only b = 4 gives a solution: a = 2, b = 4, c = 8. If n = 4, then (n - 2)bc, nb and nc are all even, but (n + 1) is odd, so there is no solution. If n = 5, then 3bc + 6 = 5b + 5c. b = 3 gives c = 9/4, which is not a solution. b >= 4 gives 3bc > 10c > 5b + 5c, so there are no solutions. If a = 3, we have 2n(bc - b - c + 1) = 3bc - 1, or (2n - 3)bc + (2n + 1) = 2nb + 2nc. But b ≥ 4, so (2n - 3)bc ≥ (8n - 12)c ≥ 4nc > 2nc + 2nb. So there are no solutions. Similarly, if a = 4, we have (3n - 4)bc + (3n + 1) = 3nb + 3nc. But b ≥ 4, so (3n - 4)bc ≥ (12n - 16)c > 6nc > 3nb + 3nc, so there are no solutions. Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 33rd IMO 1992 © John Scholes jscholes@kalva.demon.co.uk 12 Nov 1998