TITLE: Solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$?
QUESTION [7 upvotes]: I encountered this problem yesterday and successfully solved it. I'm interested in seeing other people's approach to solving this problem. 
Problem: How many ordered pairs $(a, b)$ are solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$ where $a, b \in \mathbb{Z}$?
Edit: The solution is:

 $49$

REPLY [2 votes]: Taking seriously the request for alternative approaches, here's one that'll work if you don't mind doing a large (but finite) amount of trial and error:
After noting that $a=b=200$ is one solution, we can count the rest by doubling the number of solutions with $0\lt|a|\lt |b|$.  These we can find after observing that $|b|\gt|a|\ge200$ implies
$$\left|{1\over a}+{1\over b}\right|\le{1\over|a|}+{1\over|b|}\lt{1\over200}+{1\over200}={1\over100}$$
so it suffices to look in the range $-199\le a\le 199$ (skipping $a=0$), to see how often $b=100a/(a-100)$ is an integer whose absolute value is greater than $|a|$.
Please note, I am not saying this is a smart way to solve the problem, just that it's an approach that is guaranteed to work.