Solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$? 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:

 $49$  


This reminded me of problem 454 on project Euler which I wrote an algorithm for a year ago. (Here is the algorithm if anyone is interested)
So here is how I would solve this question.
Observe that
$$\frac{1}{2} = \frac{1}{3} + \frac{1}{6} \\ \frac{1}{3} = \frac{1}{4} + \frac{1}{12} \\ \frac{1}{4} = \frac{1}{5} + \frac{1}{20}$$
which can be generalized into 
$$\frac{1}{a} = \frac{1}{a+1} + \frac{1}{a^2+a}$$
So immediately we get our first solution:
$$\frac{1}{100} = \frac{1}{101} + \frac{1}{10100}$$
Note the equation above can be further generalized into 
$$\frac{1}{a} = \frac{1}{(a+n)} + \frac{n}{a^2+an}$$
in which case $n$ can be converted into 1 as long as $n$ is divisible by $a^2+an$. Thus to find all solutions to the equation is basically to find how many distinct $n$'s are there that are divisible by $a^2$ - how many factors do $a^2$ have. 
Note: the question is asking for the number of ordered pairs, in which case the summation of 2 distinct fractions count as 2.  
