Bs: \(x,y\in \mathbb{Z}\)
Ta có \(36-y^2=8\left(x-2021\right)^2\ge0\Leftrightarrow y^2\le36\)
Mà \(8\left(x-2021\right)^2\) và 36 chẵn nên y chẵn
Do đó \(y^2\in\left\{4;16;36\right\}\)
Với \(y^2=4\Leftrightarrow8\left(x-2021\right)^2=32\Leftrightarrow\left(x-2021\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x=2025\\x=2017\end{matrix}\right.\)
Với \(y^2=16\Leftrightarrow8\left(x-2021\right)^2=20\Leftrightarrow\left(x-2021\right)^2=\dfrac{5}{2}\left(loại\right)\)
Với \(y^2=36\Leftrightarrow8\left(x-2021\right)^2=0\Leftrightarrow x=2021\)
Vậy \(\left(x;y\right)=\left(2025;2\right);\left(2025;-2\right);\left(2017;2\right);\left(2017;-2\right);\left(2021;6\right);\left(2021;-6\right)\)
