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:

 $49$  


Here's my solution. 
So we start with $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$.
We can rewrite this as 
$$\frac{100}{a} + \frac{100}{b} = 1$$
$$100a + 100b = ab$$
$$ab - 100a - 100b = 0$$
$$ab - 100a - 100b + 100^2 = 10000$$
$$(a - 100)(b - 100) = 10000$$
So now I realize my solutions $(a, b)$ correspond to the negative and positive   factors $(a - 100) = f$, $(b-100) = g$ where $fg = 10000$.   
Then I find the prime factorization of $10000 = 2^45^4$. So we have $5\cdot5 = 25$ possible positive factors of $10000$. Corresponding to $50$ positive and negative solutions. Except I realized we have to exclude the pair $(-100, -100)$
factors of $10000$ since these correspond to the pair $(a, b) = (0, 0)$ and we can't divide by zero.   

 Therefore there are $49$ solutions.    

