\(x^3+3x-5=\left(x^2+2\right)y\)
\(\Rightarrow y=\frac{x^3+3x-5}{x^2+2}=x+\frac{x-5}{x^2+2}\)
Để y nguyên \(\Rightarrow\frac{x-5}{x^2+2}\) nguyên \(\Rightarrow\frac{\left(x-5\right)\left(x+5\right)}{x^2+2}\) nguyên \(\Rightarrow\frac{x^2-25}{x^2+2}\) nguyên
\(\Rightarrow1-\frac{27}{x^2+2}\) nguyên \(\Rightarrow x^2+2=Ư\left(27\right)=\left\{27;9;3\right\}\) (chỉ cần quan tâm các ước lớn hơn 2 của 27)
\(x^2+2=27\Rightarrow\left[{}\begin{matrix}x=5\Rightarrow y=5\\x=-5\Rightarrow y=-\frac{145}{27}\left(l\right)\end{matrix}\right.\)
\(x^2+2=9\Rightarrow x^2=7\left(l\right)\)
\(x^2+2=3\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=-\frac{1}{3}\left(l\right)\\x=-1\Rightarrow y=-3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;-3\right);\left(5;5\right)\)
