Giải thích các bước giải:
Ta có:
$\begin{array}{l}
4{x^2} - {y^2} = 27\\
\Leftrightarrow 4{x^2} - 2xy + 2xy - {y^2} = 27\\
\Leftrightarrow 2x\left( {2x - y} \right) + y\left( {2x - y} \right) = 27\\
\Leftrightarrow \left( {2x - y} \right)\left( {2x + y} \right) = 27\left( 1 \right)
\end{array}$
Mà $x,y \in Z$ nên
$\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - y = 1\\
2x + y = 27
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = 3\\
2x + y = 9
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = 9\\
2x + y = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = 27\\
2x + y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = - 1\\
2x + y = - 27
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = - 3\\
2x + y = - 9
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = - 9\\
2x + y = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - y = - 27\\
2x + y = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4x = 28\\
2y = 26
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = 12\\
2y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = 12\\
2y = - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = 28\\
2y = - 26
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = - 28\\
2y = - 26
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = - 12\\
2y = - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = - 12\\
2y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
4x = - 28\\
2y = 26
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7;y = 13\\
x = 3;y = 3\\
x = 3;y = - 3\\
x = 7;y = - 13\\
x = - 7;y = - 13\\
x = - 3;y = - 3\\
x = - 3;y = 3\\
x = - 7;y = 13
\end{array} \right.$
Vậy $\left( {x;y} \right) \in \left\{ {\left( {7;13} \right);\left( {3;3} \right);\left( {3; - 3} \right);\left( {7; - 13} \right);\left( { - 7; - 13} \right);\left( { - 3; - 3} \right);\left( { - 3;3} \right);\left( { - 7;13} \right)} \right\}$ thỏa mãn.
