[tex]\dfrac{12}{(x-1)(y-1)} \rightarrow $ fractie echiunitara \\ \\ \Rightarrow (x-1)(y-1) = 12 \\ \\ \bullet$ $x-1 \in D_{12}^+ \Rightarrow x-1\in \Big\{1,2,3,4,6,12\Big\} \\ \\ \bullet $ $ y-1 \in D_{12}^+ \Rightarrow y-1\in \Big\{1,2,3,4,6,12\Big\} \\ \\ $Dar, $ (x-1)(y-1) = 12 \\ \\x-1\in \Big\{1,2,3,4,6,12\Big\} $ si $ y-1 \in \Big\{12,6,4,3,2,1\Big\} \\ $(in aceasta ordine) \\ Adunam cu 1 in ambele relatii \Rightarrow \\ \\ [/tex]
[tex]\Rightarrow x\in \Big\{2,3,4,5,7,13\Big\} $ si $ y \in\Big\{13,7,5,4,3,2\Big\} \\ \\ $Le grupam in ordinea pozitiilor:\\ \\ \Rightarrow (x,y) = \Big\{(2, 13); (3,7); (4,5); (5,4); (7,3);(13,2)\Big\}[/tex]
