2006 Indonesia MO Problems/Problem 7
Let be real numbers such that are rational numbers. Prove that there are integers , not all of them are , such that .
We divide the problem into three cases.
Case 1: At least one of equal zero
WLOG, let . Then , no matter what integer is. We can set and have a set . This solution set works because not all of are equal to zero.
Similar reasoning can be used if all three variables are zero and if two of the three variables are zero.
Case 2: None of equal zero
Note that non-zero rational numbers are closed under multiplication or division. In other words, multiplying or dividing two rational numbers (unless when dividing by 0) will always result in a rational number.
Therefore, and are also rational numbers. In the original equation, dividing both sides by results in .
Since and are rational, let and , where are integers. Therefore, we can let and to have be an integer. Since addition and subtraction are closed under integers, would also be an integer, so there is an integral solution that satisfies the equation
Thus, in all two cases, there are solutions (where all of are integers) that satisfy the equation.
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