Question

[Anh Đỗ Ngọc]

Tìm nghiệm nguyên của phương trình

Answer 1

[Tạ Đức Hoàng Anh]

Ta có: \(x^2+5y^2+2y-4xy-3=0\)

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

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

Vì \(\hept{\begin{cases}\left(x-4y\right)^2\ge0\forall x,y\\\left(y+1\right)^2\ge0\forall y\end{cases}}\)\(\Rightarrow\)\(\left(x-4y\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

mà \(\left(x-4y\right)^2+\left(y+1\right)^2=4\)\(\Rightarrow\)\(0\le\left(x-4y\right)^2+\left(y+1\right)^2\le4\forall x,y\)

Vì \(x,y\in Z\)\(\Rightarrow\)\(\hept{\begin{cases}\left(x-4y\right)^2\inℤ\\\left(y+1\right)^2\inℤ\end{cases}}\)

+) \(\hept{\begin{cases}\left(x-4y\right)^2=0\\\left(y+1\right)^2=4\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=0\\y+1=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=4\\y=1\end{cases}}\)( TM )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=1\\\left(y+1\right)^2=3\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=1\\y+1=\sqrt{3}\end{cases}}\)( loại )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=2\\\left(y+1\right)^2=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=\sqrt{2}\\y+1=\sqrt{2}\end{cases}}\)( loại )

+) \(\hept{\begin{cases}\left(x-4y\right)^2=4\\\left(y+1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-4y=2\\y+1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=-2\\y=-1\end{cases}}\)( TM )

Vậy \(\left(x,y\right)\in\left\{\left(-2,-1\right);\left(4,1\right)\right\}\)

Answer 2

[Trần Thu Hà]

\(x^2+5y^2+2y-4xy-3=0\)

\(\Leftrightarrow x^2-4xy+4y^2+y^2+2y+1-4=0\)

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

Vì \(\hept{\begin{cases}\left(x-2y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}\Leftrightarrow0\le\left(x-2y\right)^2+\left(y+1\right)^2\le4}\)

Ta xét lần lượt là ra nha