Difference between revisions of "1981 IMO Problems/Problem 3"
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== Solution == | == Solution == | ||
− | We first observe that since <math>\gcd(m,n) | + | We first observe that since <math>\gcd(m,n)=1 </math>, <math>m</math> and <math>n</math> are [[relatively prime]]. In addition, we note that <math>n \ge m</math>, since if we had <math>n < m</math>, then <math>n^2 -nm -m^2 = n(n-m) - m^2 </math> would be the sum of two negative integers and therefore less than <math>-1</math>. We now observe |
<cmath> | <cmath> |
Latest revision as of 10:23, 13 May 2019
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
and
to be the greatest Fibonacci numbers less than
, giving
as the maximal value.
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 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |