Cách này không được hay cho lắm. Thông cảm nhé!
$\begin{array}{l} {y^2} = 1 + \sqrt {9 - 4x + {x^2}} \in \mathbb{Z}\\ \Rightarrow \sqrt {9 - 4x + {x^2}} = {y^2} - 1 \in \mathbb{Z}\\ \Rightarrow \sqrt {9 - 4x + {x^2}} = k\left( {k \ge 0,k \in \mathbb{Z}} \right)\\ \Rightarrow 9 - 4x + {x^2} = {k^2}\\ \Leftrightarrow 13 = {k^2} + {\left( {x + 2} \right)^2}\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} {k^2} = 4\\ {\left( {x + 2} \right)^2} = 9 \end{array} \right.\\ \left\{ \begin{array}{l} {k^2} = 9\\ {\left( {x + 2} \right)^2} = 4 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} k = 2\\ \left[ \begin{array}{l} x = 1\\ x = - 5 \end{array} \right. \end{array} \right.\\ \left\{ \begin{array}{l} k = 3\\ \left[ \begin{array}{l} x = 0\\ x = - 4 \end{array} \right. \end{array} \right. \end{array} \right.\\ \left\{ \begin{array}{l} k = 2\\ x = 1 \end{array} \right.. \Rightarrow 9 - 4.1 + {1^2} = 6 \ne {k^2} = 4(L)\\ \left\{ \begin{array}{l} k = 2\\ x = - 5 \end{array} \right. \Rightarrow 9 - 4.\left( { - 5} \right) + {\left( { - 5} \right)^2} = 54 \ne {k^2} = 4(L)\\ \left\{ \begin{array}{l} k = 3\\ x = 0 \end{array} \right. \Rightarrow 9 - 4.0 + {0^2} = 9 = {k^2} = 9(TM) \Rightarrow y = \pm 2\\ \left\{ \begin{array}{l} k = 3\\ x = - 4 \end{array} \right. \Rightarrow 9 - 4.\left( { - 4} \right) + {\left( { - 4} \right)^2} = 9 = {k^2} = 9\left( {TM} \right) \Rightarrow y = \pm 2\\ \Rightarrow \left( {x;y} \right) = \left( {0;2} \right),\left( {0; - 2} \right),\left( { - 4;2} \right),\left( { - 4; - 2} \right) \end{array}$
