[tex](2x+1)\cdot (y-1) = 21 \\ \\ x\geq 0 \Big|\cdot 2 \Rightarrow 2x \geq 0 |+1 \Rightarrow 2x+1 \geq1 \\ \\ y\geq0\Big|-1 \Rightarrow y-1 \geq -1 \\ \\\bullet (2x+1) \in D_{21} \Rightarrow (2x+1) \in \Big\{\pm1,\pm3,\pm7,\pm21\Big\}, $ dar, 2x+1\geq1 \Rightarrow \\ \\ \Rightarrow (2x+1)\in\Big\{1,3,7,21\Big\}$ \\ \\ $\bullet$ $ (y-1)\in D_{21} \Rightarrow (y-1)\in \Big\{\pm1,\pm3,\pm7,\pm21\Big\}, $dar, 2x+1 pozitiv \Rightarrow \\ \\ [/tex]
[tex]\Rightarrow $ si y - 1 trebuie sa fie pozitiv ca produsul sa dea pozitiv \Rightarrow \\ \\ \Rightarrow (y-1) \in \Big\{1,3,7,21\Big\} \\ \\ \\ $Avem:
\left\{ \begin{array}{c} (2x+1)\in\Big\{1,3,7,21\Big\} \\ (y-1) \in \Big\{1,3,7,21\Big\} \end{array} \right| \Rightarrow ($formam perechile de numere$) \\ \\ \Rightarrow \Big((2x+1),(y-1)\Big) = \Big\{(1,21);(3,7);(7,3);(21,1)\Big\} \\ \\ $(-1 in stanga si +1 in dreapta)$ \\ \\ \Rightarrow \Big((2x),y\Big) = \Big\{(0,22);(2,8);(6,4);(20,2)\Big\} [/tex]
[tex]\\ $(:2 in stanga ) \\ \\ \Rightarrow\boxed{ \Big(x,y\Big) = \Big\{(0,22);(1,8);(3,4);(10,2)\Big\}} $ - $ $solutie finala[/tex]
