Ta có:
\(2y^2x+x+y+1=x^2+2y^2+xy\)
\(\Leftrightarrow2y^2x+x+y+1-x^2-2y^2-xy=0\)
\(\Leftrightarrow(2y^2x-2y^2)+(x-x^2)+(y-xy)+1=0\)
\(\Leftrightarrow2y^2(x-1)-x(x-1)-y(x-1)=-1\)
\(\Leftrightarrow(2y^2-x-y)(x-1)=-1\)
Vì \(x,y\in Z\) nên \(2y^2-x-y\in Z\), \(x-1\in Z\)
Mà \((2y^2-x-y)(x-1)=-1\)
\(\Rightarrow\)\(2y^2-x-y\inƯ\left(-1\right)\), \(x-1\inƯ\left(-1\right)\)
+/ \(\left\{{}\begin{matrix}x-1=1\\2y^2-x-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\2y^2-x-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-2-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left[{}\begin{matrix}y=1\\y=\dfrac{-1}{2}\end{matrix}\right.\end{matrix}\right.\)
Mà \(x,y\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
+/ \(\left\{{}\begin{matrix}x-1=-1\\2y^2-x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\2y^2-x-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-0-y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y+1=0\end{matrix}\right.\)
Ta thấy: \(2y^2+y+1=2\left(y+\dfrac{1}{4}\right)^2+\dfrac{7}{8}>0\forall y\)
\(\Rightarrow2y^2+y+1=0\) (vô nghiệm)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\2y^2-y+1=0\end{matrix}\right.\)(vô nghiệm)
Vậy \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
