\(y^2+2y=x^2+11x+29\)
\(\Leftrightarrow x^2+11x+29-\left(y^2+2y\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{11}{2}\right)^2-\left(y+1\right)^2+29-\left(\dfrac{11}{2}\right)^2+1=0\)
\(\Leftrightarrow\dfrac{\left(2x+11\right)^2}{4}-\left(y+1\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left(2x+11\right)^2-4\left(y+1\right)^2=1\)
\(\Leftrightarrow\left[2x+11+2\left(y+1\right)\right]\left[2x+11-2\left(y+1\right)\right]=1\)
\(\Leftrightarrow\left(2x+2y+13\right)\left(2x-2y+9\right)=1\)
Do x, y nguyên nên : \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+2y+13=1\\2x-2y+9=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+2y+13=-1\\2x-2y+9=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-5\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy : Phương trình có cặp nghiệm : \(\left(x;y\right)\in\left\{\left(-5;-1\right);\left(-6;-1\right)\right\}\)