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:

 $49$  

 A: 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.    

A: 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.  
A: We have
$$\dfrac1a+\dfrac1b = \dfrac1{n}$$
This means we have
$$n(a+b) = ab \implies ab - n a -n b = 0 \implies (a-n)(b-n) = n^2$$
Hence, each divisor pair of $n^2$ gives a possible value for $a$ and $b$, i.e., if $d_1d_2 = n^2$, we can set $a=n+d_1$ and $b=n+d_2$ to obtain all solutions. Hence, the number of pairs is the number of divisors (both positive and negative) of $n^2$.
A: 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.
A: $$ab=100(a+b)$$
$$(a-100)(b-100)=10^4$$
A: If you multiply through by $100ab,$ you get
\begin{equation*}
100b+100a=ab\\
\Rightarrow 0=ab-100a-100b \\
\Rightarrow (a-100)(b-100)=10000
\end{equation*} 
(the $10000$ is there to get rid of the $-100\times -100$).
Equate the Right side with the brackets on the left to get the pairs.
