vi n-2 ∈ Ư(n-3)
⇒ n-3 ⋮ n-2
⇒ n-2-1 ⋮ n-2
vi n-2 ⋮ n-2
⇒ -1 ⋮ n-2 hay n-2 ∈ Ư(-1) = { 1;-1}
⇒ n ∈ { 3; 1}
Vậy n = 3 hoac n=1
Vì (n-2) ∈ Ư (n-3)
=> (n-3) ⋮ (n-2)
=> (n-2) -1 ⋮ (n-2)
=> -1 ⋮ (n-2)
hay (n-2) ∈ Ư ( -1)
=> \(\left\{{}\begin{matrix}n-2=1\\n-2=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=3\\n=1\end{matrix}\right.\)
Vậy n = 3 hay n = 1 .