IMO 1992 SL 13

Origin: NZL

Problem

Find all integer triples (p, q, r) such that 1 < p < q < r and (p −1)(q −1)(r −1) is a divisor of (pqr −1).

Solution

Let us define F(p, q, r) = (pqr −1) (p −1)(q −1)(r −1) = 1 + p −1 + q −1 + r −1 + (p −1)(q −1) + (q −1)(r −1) + (r −1)(p −1). Obviously F is a decreasing function of p, q, r. Suppose that 1 < p < q < r are integers for which F(p, q, r) is an integer. Observe that p, q, r are either all even or all odd. Indeed, if for example p is odd and q is even, then pqr−1 is odd while (p−1)(q−1)(r−1) is even, which is impossible. Also, if p, q, r are even then F(p, q, r) is odd. If p \geq4, then 1 < F(p, q, r) \leqF(4, 6, 8) = 191/105 < 2, which is impossi- ble. Hence p \leq3. Let p = 2. Then q, r are even and 1 < F(2, q, r) \leqF(2, 4, 6) = 47/15 < 4. Therefore F(2, q, r) = 3. This equality reduces to (q −3)(r −3) = 5, with the unique solution q = 4, r = 8. Let p = 3. Then q, r are odd and 1 < F(3, q, r) \leqF(3, 5, 7) = 104/48 < 3. Therefore F(3, q, r) = 2. This equality reduces to (q −4)(r −4) = 11, which leads to q = 5, r = 15. Hence the only solutions (p, q, r) of the problem are (2, 4, 8) and (3, 5, 15).