2008 Indonesia MO Problems/Problem 5
Let are integers which satisfy and . Is it a must that ?
Solution 1 (credit to mehmetcantu)
Since and , we must have . Expanding results in . Note that both and are multiples of , and so is also a multiple of . Therefore, but we must also have because . As a result, and so .
Since , we must have for a positive integer . Then by solving for and doing some substitution, we must have be an integer. Now we need to show that if is an integer, then .
Assume that there is such that is an integer. Consider the fraction . Since and , we must have and so can not be an integer. Therefore, is not an integer. Since is an integer and the product of integers result in another integer, we must have not be an integer. This contradicts being an integer, so by proof by contradiction, and so .
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