1994 IMO Problems/Problem 4
Problem
Find all ordered pairs where and are positive integers such that is an integer.
Solution
Suppose where is a positive integer. Then and so it is clear that . So, let where is a positive integer. Then we have which by cancelling out the s and dividing by yields . The equation is a quadratic. We are given that is one of the roots. Let be the other root. Notice that since we have that is an integer, and so from we have that is positive.
It is obvious that is a solution. Now, if not, and are all greater than , we have the inequalities and which contradicts the equations . Thus, at least one of is equal to .
If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .
If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .
From these, we have all solutions .
See also
1994 IMO (Problems) • Resources) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |