Find all pairs of integers $(x,y)$ for
$$\frac{1}{x} + \frac{1}{y} = \frac{1}{2020}$$
Attempt:
Notice $x,y \ne 0$.
$$ \frac{1}{x} + \frac{1}{y} = \frac{1}{2020} $$
$$ \frac{x+y}{xy} = \frac{1}{2020} $$
$$ 2020(x+y) = xy $$
$$2020x + 2020y = xy$$
$$x(2020-y) = -2020y $$
$$x = \frac{2020y}{y-2020}$$
Notice $2020 = 101(5)2^{2}$, so the divisors are $1,2,4,5,10,20,101,202,404,505,1010,2020$.
If we check where $y-2020 | 2020$, then the solutions for $y$ are: $$y = 2020+1, 2020+2, 2020+4, 2020+5, 2020+10, 2020+20, ..., 2020+2020$$ along with the corresponding $x$.
If we check $y-2020 | y$, then notice that $\frac{y}{y-2020} = 1 + \frac{2020}{y-2020}$, so it will be integer only if $\frac{2020}{y-2020}$ is integer, which means the $y$ solution is the same as before.
So far I have only found $12$ solutions.
