Question

Tìm \(x,y\in Z\) : \(1+x+y+2xy^2=xy+x^2+2y^2\)

Answer 1

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

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

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

Giải hệ nghiệm nguyên

\(\left(I\right)\left\{\begin{matrix}x-1=1\\x+y-2y^2=1\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\left\{\begin{matrix}x=2\\y=\left\{1\right\}\end{matrix}\right.\)

\(\left(II\right)\left\{\begin{matrix}x-1=-1\\x+y-2y^2=-1\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=0\\2y^2-y-1=0\end{matrix}\right.\Rightarrow}\left\{\begin{matrix}x=0\\y=1\end{matrix}\right.\)Kết luận

(x,y)=(2,1); (0,1)