Tìm nghiệm nguyên của phương trình \({(x + y)^2} + 2{y^2}\left( {x + 1} \right) + {(y + 2)^2} - 9 = 0\).

\({\left( {x + y} \right)^2} + 2{y^2}\left( {x + 1} \right) + {\left( {y + 2} \right)^2} - 9 = 0\left( {\rm{*}} \right)\)

\( \Leftrightarrow {x^2} + {y^2} + 2xy + 2x{y^2} + 2{y^2} + {y^2} + 4y + 4 - 9 = 0\)

\( \Leftrightarrow {x^2} - 4 + 2x\left( {{y^2} + y} \right) + 4\left( {{y^2} + y} \right) = 1\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 2} \right) + 2\left( {{y^2} + y} \right)\left( {x + 2} \right) = 1\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 2 + 2{y^2} + 2y} \right) = 1\)

TH1: \(\left\{ {\begin{array}{*{20}{c}}{x + 2 = 1}\\{x - 2 + 2{y^2} + 2y = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 1}\\{2{y^2} + 2y = 4}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 1}\\{\left[ {\begin{array}{*{20}{c}}{y = 1}\\{y =  - 2}\end{array}} \right.}\end{array}} \right. \Rightarrow \left( { - 1;1} \right),{\rm{\;}}\left( { - 1;{\rm{\;}} - 2} \right)\)

TH2: \(\left\{ {\begin{array}{*{20}{c}}{x + 2 = 1}\\{x - 2 + 2{y^2} + 2y = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 3}\\{2{y^2} + 2y = 4}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 3}\\{\left[ {\begin{array}{*{20}{c}}{y = 1}\\{y =  - 2}\end{array}} \right.}\end{array}} \right. \Rightarrow \left( { - 3;1} \right),{\rm{\;}}\left( { - 3;{\rm{\;}} - 2} \right)\)