Question

Tìm các số nguyên x,y thỏa mãn :

x² + 8y² + 4xy - 2x - 4y = 4

Answer 1

\(x^2+4y^2+1+4xy-2x-4y+4y^2=5\)

\(\Leftrightarrow\left(x+2y-1\right)^2=5-4y^2\)

Do \(VT\ge0\Rightarrow VP\ge0\Rightarrow5-4y^2\ge0\)

\(\Rightarrow y^2\le\frac{5}{4}\Rightarrow y^2=\left\{0;1\right\}\Rightarrow y=\left\{-1;0;1\right\}\)

Thay lần lượt \(y\) vào ta thấy \(y=\left\{-1;1\right\}\) thỏa mãn, khi đó \(\left[{}\begin{matrix}\left(x-3\right)^2=1\\\left(x+1\right)^2=1\end{matrix}\right.\) \(\Rightarrow x=...\)