\(x^3+3x=x^2y+2y+5\left(1\right)\\ \Leftrightarrow x^2y+2y=x^3+3x-5\\ \Leftrightarrow\left(x^2+2\right)y=x^3+3x-5\\ \Leftrightarrow y=\dfrac{x^3+3x-5}{x^2+2}=\dfrac{x^3+2x+x-5}{x^2+2}\\ =\dfrac{x\left(x^2+2\right)+\left(x-5\right)}{x^2+2}=\dfrac{x\left(x^2+2\right)}{x^2+2}+\dfrac{x-5}{x^2+2}\\ =x+\dfrac{x-5}{x^2+2}\)
Mà \(x;y\in Z\)
\(\Rightarrow\dfrac{x-5}{x^2+2}\in Z\\ \Rightarrow x-5⋮x^2+2\\ \Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\\ \Rightarrow x^2-25⋮x^2+2\\ \Rightarrow x^2+2-27⋮x^2+2\\ \Rightarrow27⋮x^2+2\\ \Rightarrow x^2+2\inƯ_{\left(27\right)}\)
Mà \(Ư_{\left(27\right)}=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)
Nhưng: \(x^2+2\ge2\forall x\)
\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)
Lập bảng giá trị:
| \(x^2+2\) | \(3\) | \(9\) | \(27\) |
| \(x^2\) | \(1\) | \(7\) | \(25\) |
| \(x\) | \(\pm1\) | \(\sqrt{7}\) | \(\pm5\) |
Mà \(x\in Z\Rightarrow x\in\left\{\pm1;\pm5\right\}\left(2\right)\)
Thay \(\left(2\right)\) vào \(\left(1\right):\)
+) Với \(x=-1\Rightarrow y=-3\left(T/m\right)\)
+) Với \(x=1\Rightarrow y=-\dfrac{1}{3}\left(loại\right)\)
+) Với \(x=-5\Rightarrow-\dfrac{145}{27}\left(loại\right)\)
+) Với \(x=5\Rightarrow y=5\left(T/m\right)\)
Vậy các số nguyên \(\left\{x;y\right\}\) cần tìm là \(\left\{\left(-1;-3\right);\left(5;5\right)\right\}\)
