Đặt A = n⁵ - n = n.(n⁴ - 1)
= n.(n² + 1)(n² - 1)
= n.(n² + 1)(n - 1)(n + 1) (\(⋮6\), vì \(⋮2,3\)) (1)
= n.(n² - 4 + 5)(n - 1)(n + 1)
= n[(n-2)(n+2)+5](n - 1)(n + 1)
= [n(n-2)(n+2)+5n](n - 1)(n + 1)
= n(n-2)(n+2)(n - 1)(n + 1) + 5n(n - 1)(n + 1)
Do \(\left\{{}\begin{matrix}\text{n(n-2)(n+2)(n - 1)(n + 1) ⋮ 5 }\\\text{5n(n - 1)(n + 1) ⋮ 5 }\end{matrix}\right.\)
\(\Rightarrow\text{ n(n-2)(n+2)(n - 1)(n + 1) + 5n(n - 1)(n + 1) }⋮5\)
\(\Rightarrow A⋮5\) (2)
Từ (1)(2)=> \(A⋮30\) do (5,6)=1
