1988 IMO Problems/Problem 6
Let and be positive integers such that divides . Show that is the square of an integer.
Choose integers such that Now, for fixed , out of all pairs choose the one with the lowest value of . Label . Thus, is a quadratic in . Should there be another root, , the root would satisfy: Thus, isn't a positive integer (if it were, it would contradict the minimality condition). But , so is an integer; hence, . In addition, so that . We conclude that so that .
This construction works whenever there exists a solution for a fixed , hence is always a perfect square.
|1988 IMO (Problems) • Resources|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|