Đặt :
\(\left\{{}\begin{matrix}n+2019=a^2\\n+1960=b^2\end{matrix}\right.\) \(\left(a>b\right)\)
\(\Leftrightarrow a^2-b^2=59\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=59\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=1.59\)
Lại có : \(a>b\Leftrightarrow a+b>a-b\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\a+b=59\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=30\\b=29\end{matrix}\right.\)
Thay vào rồi tìm n nhóe <3