2003 IMO Problems/Problem 2
(Aleksander Ivanov, Bulgaria) Determine all pairs of positive integers such that is a positive integer.
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
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