Difference between revisions of "2003 IMO Problems/Problem 2"
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{{alternate solutions}} | {{alternate solutions}} | ||
+ | In this problem we can do it by an alternative method | ||
+ | a^2/2ab^2-b^3+1>=1 | ||
+ | a^2>=2ab^2-b^3+1 | ||
+ | a^2-2ab+b^2>=1/b | ||
+ | (a-b)^2>=1/b | ||
+ | The solutions are a>=2 and b>=1 are all the solutions | ||
== Resources == | == Resources == |
Latest revision as of 00:23, 16 December 2017
Problem
(Aleksander Ivanov, Bulgaria)
Determine all pairs of positive integers such that
is a positive integer.
Solution
The only solutions are of the form ,
, and
for any positive integer
.
First, we note that when , the given expression is equivalent to
, which is an integer if and only if
is even.
Now, suppose that is a solution not of the form
. We have already given all solutions for
; then for this new solution, we must have
. Let us denote
Denote
Since
, and
is a positive integer root of
, there must be some other root
of
.
Without loss of generality, let . Then
, so
or
which reduces to
It follows that
or
Since
and
are integers, this can only happen when
, so
can be written as
, and
. It follows that
Since
is the other root of
, it follows that
also satisfies the problem's condition. Therefore the solutions are exactly the ones given at the solution's start.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
In this problem we can do it by an alternative method
a^2/2ab^2-b^3+1>=1
a^2>=2ab^2-b^3+1
a^2-2ab+b^2>=1/b
(a-b)^2>=1/b
The solutions are a>=2 and b>=1 are all the solutions
Resources
2003 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |
- <url>Forum/viewtopic.php?p=262#262 Discussion on AoPS/MathLinks</url>