# 2016 AIME I Problems/Problem 7

## Problem

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

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 $$0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}$$ 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: $ab < -2016$.

In this case, we want $$0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \frac{\sqrt{-ab-2016} - \sqrt{|a+b|}}{ab+100}$$ 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 $87 + 16 = \boxed{103}$.