How many ordered pairs $(a,b)$ of positive integers satify the following equation $1/a+1/b=1/120$? So far I've only agebra bashed it, as there seems to be no other way of solving it. Am I missing some key insight into this equation so I can solve it, preferably not using algebra?

\begin{align} \frac 1a + \frac 1b & = \frac1{120}\qquad \implies a,b >120\\\ \frac {b+a}{ab} & = \frac1{120}\\\ \frac {ab}{b+a} & = 120\\\ ab & = 120(a+b)\\\ ab - 120 a & = 120b\\\ a(b-120) & = 120b\\\ (b-120) & = \frac{120b}{a}\\\ \text{Similarly } \quad(a-120) & = \frac{120a}{b} = \frac {120^2}{b-120}\\\ (a-120)(b-120) &= 120^2\\\ \end{align}

Now find the factors of $120^2$...

For example $(9,1600)$ is a factor pair of $120^2$ so $(a,b)=(129,1720)$ is a solution. Or $60 \times 240 = 120^2$ gives $(a,b)=(180,360)$ etc.