Question

[Thơ Nụ =))]

Tìm x,y nguyên: \(2xy^2+x+y-1=x^2+2y^2+xy\)

Answer 1

[Toru]

\(2xy^2+x+y-1=x^2+2y^2+xy\\\Leftrightarrow 2xy^2+x+y-1-x^2-2y^2-xy=0\\\Leftrightarrow(2xy^2-2y^2)-(xy-y)-(x^2-x)=1\\\Leftrightarrow2y^2(x-1)-y(x-1)-x(x-1)=1\\\Leftrightarrow(x-1)(2y^2-y-x)=1\)

Vì \(x,y\) nguyên \(\Rightarrow x-1;2y^2-y-x\) có giá trị nguyên

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

Do đó ta có các trường hợp xảy ra là:

\(+,\left\{{}\begin{matrix}x-1=1\\2y^2-y-x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left(2y-3\right)\left(y+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y\in\left\{\dfrac{3}{2};-1\right\}\end{matrix}\right.\)

Mà \(x,y\) nguyên nên: \(x=2;y=-1\)

\(+,\left\{{}\begin{matrix}x-1=-1\\2y^2-y-x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2\left(y-\dfrac{1}{4}\right)^2+\dfrac{7}{8}=0\left(\text{vô lí}\right)\end{matrix}\right.\)

Vậy \(x=2;y=-1\) là các giá trị cần tìm.

\(\text{#}Toru\)