\(x\left(x+1\right)=y^2+1\)
\(\Leftrightarrow x^2+x=y^2+1\)
\(\Leftrightarrow4x^2+4x=4y^2+4\)
\(\Leftrightarrow4x^2+4x+1=4y^2+5\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2y\right)^2=5\)
\(\left(2x+1-2y\right)\left(2x+1+2y\right)=5=1.5=5.1=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1-2y=1\\2x+1+2y=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=5\\2x+1+2y=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=-1\\2x+1+2y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=-5\\2x+1+2y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left\{\left(x;y\right)\right\}=\left\{\left(1;1\right);\left(1;-1\right);\left(-2;-1\right);\left(-2;1\right)\right\}\)
