\(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\)
\(\Rightarrow xy=\dfrac{4-x^2-y^2}{2}\)
Thay vào \(xy-z^2=1\Rightarrow\dfrac{4-x^2-y^2}{2}-z^2=1\)
\(\Leftrightarrow x^2+y^2+2z^2=2\Rightarrow x^2+y^2=2-2z^2\le2\)
\(\Rightarrow x^2+y^2=\left\{0;1;2\right\}\) (do đề ko có dữ kiện \(z\) nguyên nên vẫn tồn tại \(z\) để \(2-2z^2=1\), nếu có \(z\) nguyên thì loại trường hợp \(x^2+y^2=1\) ko cần xét)
TH1: \(x^2+y^2=0\Rightarrow x=y=0\)
TH2: \(x^2+y^2=1\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=1\\y^2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=0\\y^2=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\pm1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\end{matrix}\right.\)
TH3: \(x^2+y^2=2\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\end{matrix}\right.\)
