Tìm tất cả các cặp số nguyên `(x;y)` thỏa mãn `x^3 - x^2 y + 2x = 5x^2 - 2y - 1`

$\begin{array}{l} {x^3} - {x^2}y + 2x = 5{x^2} - 2y - 1\\ \Rightarrow {x^3} - {x^2}y - 5{x^2} + 2x + 2y + 1 = 0\\ \Rightarrow y\left( {2 - {x^2}} \right) + {x^3} - 2x - 5{x^2} + 4x + 1 = 0\\ \Rightarrow y\left( {2 - {x^2}} \right) - x\left( {2 - {x^2}} \right) - 5{x^2} + 10 + 4x - 9 = 0\\ \Rightarrow \left( {y - x + 5} \right)\left( {2 - {x^2}} \right) + 4x - 9 = 0\\ \Rightarrow y - x + 5 = \dfrac{{9 - 4x}}{{2 - {x^2}}}\left( 1 \right)\\ x,y \in \mathbb Z \to \dfrac{{9 - 4x}}{{2 - {x^2}}} \in \mathbb Z\\ \to 2 - {x^2}|9 - 4x\\ \Rightarrow \left( {9 - 4x} \right)\left( {9 + 4x} \right) \vdots 2 - {x^2}\\ \Rightarrow 81 - 16{x^2} \vdots 2 - {x^2}\\ \Rightarrow 49 + 32 - 16{x^2} \vdots 2 - {x^2}\\ \Rightarrow 49 + 16\left( {2 - {x^2}} \right) \vdots 2 - {x^2}\\ \Rightarrow 49 \vdots 2 - {x^2}\\ \Leftrightarrow 2 - {x^2} \in U\left( {49} \right) = \left\{ { \pm 1; \pm 7 \pm 49} \right\}\\ \bullet 2 - {x^2} = 1 \Rightarrow {x^2} = 1 \Rightarrow x = \pm 1\\ \bullet 2 - {x^2} = - 1 \Rightarrow {x^2} = 3(L,x\notin \mathbb Z)\\ \bullet 2 - {x^2} = 7 \Rightarrow {x^2} = - 5(L,x\notin \mathbb Z)\\ \bullet 2 - {x^2} = - 7 \Rightarrow {x^2} = 9 \Rightarrow x = \pm 3\\ \bullet 2 - {x^2} = 49 \Rightarrow {x^2} = - 47(L,x\notin \mathbb Z)\\ \bullet 2 - {x^2} = - 49 \Leftrightarrow {x^2} = 51(L, x\notin \mathbb Z)\\ \text{Thử lại} \Rightarrow \left( {x;y} \right) = \left( {1;1} \right),\left( { - 1;7} \right),\left( { - 3; - 11} \right)\\ \end{array}$