1981 IMO Problems/Problem 3
Problem
Determine the maximum value of , where and are integers satisfying and .
Solution
We first observe that since , and are relatively prime. In addition, we note that , since if we had , then would be the sum of two negative integers and therefore less than . We now observe
,
i.e., is a solution iff. is also a solution. Therefore, for a solution , we can perform the Euclidean algorithm to reduce it eventually to a solution . It is easy to verify that if is a positive integer, it must be either 2 or 1. Thus by trivial induction, all the positive integer solutions are of the form , where the are the Fibonacci numbers. Simple calculation reveals 987 and 1597 to be the greatest Fibonacci numbers less than 1981, giving as the maximal value Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1981 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
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