Đáp án:
$\left( {x;y} \right) = \left\{ {\left( {5;9} \right);\left( { - 5; - 3} \right);\left( { - 5;9} \right);\left( {5; - 3} \right)} \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{x^2} + 2 = y\left( {y - 6} \right)\\
\Leftrightarrow {x^2} + 2 = {y^2} - 6y\\
\Leftrightarrow {x^2} + 11 = {y^2} - 6y + 9\\
\Leftrightarrow {x^2} + 11 = {\left( {y - 3} \right)^2}\\
\Leftrightarrow {\left( {y - 3} \right)^2} - {x^2} = 11\\
\Leftrightarrow \left( {y - 3 - x} \right)\left( {y - 3 + x} \right) = 11\\
\Leftrightarrow \left( {y - x - 3} \right)\left( {y + x - 3} \right) = 11\left( 1 \right)
\end{array}$
Mà $x,y \in Z$
$ \Rightarrow y - x - 3;y + x - 3$ là cặp ước của $11$
Khi đó:
$\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y - x - 3 = 1\\
y + x - 3 = 11
\end{array} \right.\\
\left\{ \begin{array}{l}
y - x - 3 = - 1\\
y + x - 3 = - 11
\end{array} \right.\\
\left\{ \begin{array}{l}
y - x - 3 = 11\\
y + x - 3 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y - x - 3 = - 11\\
y + x - 3 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2y - 6 = 12\\
2x = 10
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 6 = - 12\\
2x = - 10
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 6 = 12\\
2x = - 10
\end{array} \right.\\
\left\{ \begin{array}{l}
2y - 6 = - 12\\
2x = 10
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 9\\
x = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - 3\\
x = - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 9\\
x = - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - 3\\
x = 5
\end{array} \right.
\end{array} \right.$
Vậy $\left( {x;y} \right) = \left\{ {\left( {5;9} \right);\left( { - 5; - 3} \right);\left( { - 5;9} \right);\left( {5; - 3} \right)} \right\}$
