Lời giải:
\(4xy-x-y=y^2\)
\(\Leftrightarrow x(4y-1)=y^2+y\)
\(\Rightarrow x=\frac{y^2+y}{4y-1}\)
Để \(x\in\mathbb{Z}\Rightarrow \frac{y^2+y}{4y-1}\in\mathbb{Z}\)
\(\Rightarrow y^2+y\vdots 4y-1\)
\(\Rightarrow 4y^2+4y\vdots 4y-1\)
\(\Leftrightarrow 4y^2-y+4y-1+y+1\vdots 4y-1\)
\(\Leftrightarrow y(4y-1)+(4y-1)+y+1\vdots 4y-1\)
\(\Rightarrow y+1\vdots 4y+1\Rightarrow 4y+4\vdots 4y-1\)
\(\Rightarrow 4y-1+5\vdots 4y-1\Rightarrow 5\vdots 4y-1\)
Do đó: \(4y-1\left\{\pm 1; \pm 5\right\}\Rightarrow y\in\left\{0; -1\right\}\)
Với $y=0$ thì $x=0$
Với $y=-1$ thì $x=0$
Vậy \((x,y)=(0,-1); (0,0)\)
Lời giải:
\(4xy-x-y=y^2\)
\(\Leftrightarrow x(4y-1)=y^2+y\)
\(\Rightarrow x=\frac{y^2+y}{4y-1}\)
Để \(x\in\mathbb{Z}\Rightarrow \frac{y^2+y}{4y-1}\in\mathbb{Z}\)
\(\Rightarrow y^2+y\vdots 4y-1\)
\(\Rightarrow 4y^2+4y\vdots 4y-1\)
\(\Leftrightarrow 4y^2-y+4y-1+y+1\vdots 4y-1\)
\(\Leftrightarrow y(4y-1)+(4y-1)+y+1\vdots 4y-1\)
\(\Rightarrow y+1\vdots 4y+1\Rightarrow 4y+4\vdots 4y-1\)
\(\Rightarrow 4y-1+5\vdots 4y-1\Rightarrow 5\vdots 4y-1\)
Do đó: \(4y-1\left\{\pm 1; \pm 5\right\}\Rightarrow y\in\left\{0; -1\right\}\)
Với $y=0$ thì $x=0$
Với $y=-1$ thì $x=0$
Vậy \((x,y)=(0,-1); (0,0)\)
