1997 IMO Problems/Problem 5
Problem
Find all pairs of integers that satisfy the equation
Solution
Case 1:
Looking at this expression since then .
Here we look at subcase which gives for all . This contradicts condition , and thus can't be more than one giving the solution of with . So we substitute the value of into the original equation to get which solves to and our first pair
Case 2:
since , then and we multiply both sides of the equation by to get:
Since , then and . This gives
This implies that for
Let with . Since , then
, which gives
subcase :
and . which provides 2nd pair
subcase :
, thus and . which provides 3rd pair
subcase :
, thus which decreases with and as . From subcase , we know that , thus for subcase , . Therefore this subcase has no solution because it contradicts the condition for Case 2 of .
Final solution for all pairs is
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1997 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |