Lời giải:
PT $\Leftrightarrow x^3+3x-5=x^2y+2y=y(x^2+2)$
$\Rightarrow y=\frac{x^3+3x-5}{x^2+2}$
Để $y$ nguyên thì $x^3+3x-5\vdots x^2+2$
$\Leftrightarrow x(x^2+2)+x-5\vdots x^2+2$
$\Leftrightarrow x-5\vdots x^2+2(1)$
$\Rightarrow x^2-5x\vdots x^2+2$
$\Leftrightarrow x^2+2-(5x+2)\vdots x^2+2$
$\Leftrightarrow 5x+2\vdots x^2+2(2)$
Từ $(1);(2)\Rightarrow 5(x-5)-(5x+2)\vdots x^2+2$
$\Leftrightarrow 27\vdots x^2+2$. Do $x^2+2\geq 2$ nên:
$\Rightarrow x^2+2\in\left\{3;9;27\right\}$
$\Rightarrow x^2\in\left\{1;7;25\right\}$
Do $x$ nguyên nên $x\in\left\{\pm 1; \pm 5\right\}$
Thay vào $y$ ta tìm được:
$x=-1\Rightarrow y=-3$
$x=5\Rightarrow y=5$
