2016 AIME I Problems/Problem 7
For integers and consider the complex number
Find the number of ordered pairs of integers such that this complex number is a real number.
We consider two cases:
Case 1: .
In this case, if then and . Thus so . Thus , yielding values. However since , we have . Thus there are allowed tuples in this case.
Case 2: .
In this case, we want Squaring, we have the equations (which always holds in this case) and Then if and , let . If , Note that for every one of these solutions. If , then Again, for every one of the above solutions. This yields solutions. Similarly, if and , there are solutions. Thus, there are a total of solutions in this case.
Thus, the answer is .
(Solution by gundraja)
Similar to Solution 1, but concise:
First, we set the imaginary expression to , so that or , of which there are possibilities. But because the denominator would be . So this gives solutions.
Then we try to cancel the imaginary part with the square root of the real part, which must be negative. So . by Simon's Factoring Trick.
We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely and .
The factors of are , so the and the sets flipped.
Similarly from the second case of we also have solutions.
Thus, . Surely, all of their products, .
So there are solutions.
|2016 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|