\(\left(y-2\right)x^2+1=y^2\Leftrightarrow\left(y-2\right)x^2=\left(y-1\right)\left(y+1\right)\)
- \(y=2\)không thỏa.
- \(y\ne2\): \(x^2=\frac{\left(y-1\right)\left(y+1\right)}{y-2}\)
Nếu \(y=1\Rightarrow x=0\).
Nếu \(y\ne1\)suy ra \(\left(y-1,y-2\right)=1\Rightarrow\left(y+1\right)⋮\left(y-2\right)\)
\(\Rightarrow3⋮\left(y-2\right)\Rightarrow y-2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\)
\(\Rightarrow y\in\left\{-1,3,5\right\}\)(do \(y\ne1\))
Ta chỉ có cặp \(\left(x,y\right)\in\left\{\left(0,-1\right)\right\}\)thỏa.
