\(x^2-16=y^2+6y\Rightarrow x^2-7=\left(y+3\right)^2\Rightarrow x^2-\left(y+3\right)^2=7\)
\(\Rightarrow\left(x-y-3\right)\left(x+y+3\right)=7\)
\(\Rightarrow x-y-3\) và \(x+y+3\) là các ước của 7, \(Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
TH1: \(\left\{{}\begin{matrix}x-y-3=-7\\x+y+3=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=-4\\x+y=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x-y-3=-1\\x+y+3=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=2\\x+y=-10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x-y-3=1\\x+y+3=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=4\\x+y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x-y-3=7\\x+y+3=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=10\\x+y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)
Vậy có 4 cặp số nguyên thỏa mãn \(\left(x;y\right)=\left(-4;0\right);\left(-4;-6\right);\left(4;0\right);\left(4;-6\right)\)
