2016 AIME I Problems/Problem 7

Problem

For integers $a$ and $b$ consider the complex number $\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i$

Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number.

Solution

We consider two cases:

Case 1: $ab \ge -2016$

In this case, if $0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = -\frac{\sqrt{|a+b|}}{ab+100}$ then $ab \ne -100$ and $|a + b| = 0 = a + b$ . Thus $ab = -a^2$ so $a^2 < 2016$ . Thus $a = -44,-43, ... , -1, 0, 1, ..., 43, 44$ , yielding $89$ values. However since $ab = -a^2 \ne -100$ , we have $a \ne \pm 10$ . Thus there are $87$ allowed tuples $(a,b)$ in this case.

Case 2: $ab \le -2016$ .

In this case, we want $0 = \text{Im}({\frac{\sqrt{ab+2016}}{ab+100}-({\frac{\sqrt{|a+b|}}{ab+100}})i}) = \frac{\sqrt{ab+2016} - \sqrt{|a+b|}}{ab+100}$ Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and $-(ab + 2016) = |a + b|.$ Then if $a > 0$ and $b < 0$ , let $c = -b$ . If $c > a$ , $ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).$ Note that $ab < -2016$ for every one of these solutions. If $c < a$ , then $ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).$ Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions. Similarly, if $a < 0$ and $b > 0$ , there are $8$ solutions. Thus, there are a total of $16$ solutions in this case.

Thus, the answer is $87 + 16 = \boxed{103}$ .

Solution by gundraja

2016 AIME I (Problems • Answer Key • Resources)
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2016 AIME I Problems/Problem 7

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