<s>[INST] Elementary Algebra The exercise is to solve this equation: $$ \frac{2}{x} + \frac{3}{y} = 1,\quad x, y \in  \mathbb{Z}^{*}$$
I tried as follows: 
$$ \frac{2}{x} = 1 - \frac{3}{y} \implies  \frac{2}{x} = \frac{y-3}{y} \implies x = \frac{2y}{y-3} \in \mathbb{Z}^{*}, \quad y \neq 3 $$
$$ \frac{3}{y} = 1 - \frac{2}{x} \implies \frac{3}{y} = \frac{x-2}{x} \implies y = \frac{3x}{x-2} \in \mathbb{Z}^{*}, \quad x \neq 2 $$
But how further I have no idea.
 [/INST] Note that if $x,y\in \mathbb{Z}$ and $x\not=0$, $x\not=2$,
$$y = \frac{3x}{x-2} =3+\frac{6}{x-2}\Leftrightarrow \mbox{ $(x-2)$ divides 6}\Leftrightarrow  (x-2)\in\{\pm 1, 2,\pm 3,\pm 6\}\\
\Leftrightarrow x\in\{-4,-1,1,3,4,5,8\}.$$
Hence the equation has $7$ solutions:
$$(-4, 2),\;
(-1, 1),\;
(1, -3),\;
(3, 9),\;
(4, 6),\;
(5, 5),\;
(8, 4).$$
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