Ta có: \(n^2-15⋮n+3\)
\(\Leftrightarrow n^2-9-6⋮n+3\)
mà \(n^2-9=\left(n+3\right)\left(n-3\right)⋮n+3\)
nên \(-6⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(6\right)\)
\(\Leftrightarrow n+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(n\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)(tm)
Vậy: \(n\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)