\(x^2+xy+y^2=3x+y-1\)
\(\Leftrightarrow x^2+\left(y-3\right)x+y^2-y+1=0\left(1\right)\)
Theo định lý Vi -ét của \(\left(1\right):\left\{{}\begin{matrix}x_1+x_2=3-y\in Z\\x_1x_2=y^2-y+1\in Z\end{matrix}\right.\) \(\left(2\right)\)
Để \(\left(1\right)\) có nghiệm \(x\) nguyên khi và chỉ khi
\(\Leftrightarrow\Delta=\left(y-3\right)^2-4\left(y^2-y+1\right)\ge0;\forall y\in Z\) và là số chính phương
\(\Leftrightarrow-3y^2-2y+5\ge0\)
\(\Leftrightarrow-\dfrac{5}{3}\le y\le1\)
\(\Rightarrow y\in\left\{-1;0;1\right\}\)
\(TH_1:t=-1\Rightarrow\Delta=-3.\left(-1\right)^2-2.\left(-1\right)+4=0\left(tm\right)\)
\(\left(2\right)\Rightarrow\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\) \(\left(tm\right)\)
\(TH_2:t=0\Rightarrow\Delta=-3.0^2-2.0+5=5\left(ktm\right)\)
\(TH_3:t=1\Rightarrow\Delta=-3.1^2-2.1+5=0\left(tm\right)\)
\(\left(2\right)\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=1\end{matrix}\right.\) \(\Rightarrow x=1\left(tm\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(1;-1\right);\left(3;-1\right);\left(1;1\right)\right\}\)
