Đặt \(Q=n^6+n^4-2n^2\)
\(\Rightarrow Q=n^2\left(n^4+n^2-2\right)\)
\(=n^2\left[\left(n^4-1\right)+\left(n^2-1\right)\right]\)
\(=n^2\left[\left(n^2-1\right)\left(n^2+1\right)+\left(n^2-1\right)\right]\)
\(=n^2\left(n^2-1\right)\left(n^2+2\right)\)
\(=n\cdot n\left(n+1\right)\left(n-1\right)\left(n^2+2\right)\)
* Nếu n chẵn. Đặt n = 2k (với k thuộc Z)
\(\Rightarrow Q=4k^2\left(2k+1\right)\left(2k-1\right)\left(4k^2+2\right)\)
\(=4k^2\left(2k-1\right)\left(2k+1\right)\cdot2\left(2k^2+1\right)\)
\(=8k^2\left(2k^2+1\right)\left(2k+1\right)\left(2k-1\right)⋮8\)
* Nếu n lẻ. Đặt n = 2k+1 (với k thuộc Z)
\(\Rightarrow\)\(Q = (2k + 1)^2 .2k (2k + 2)(4k^2 + 4k + 1 + 2) \)
\(= 4k(k + 1)(2k + 1)^2 (4k^2 + 4k + 3) \)
Vì \(k\left(k+1\right)⋮2\) \(\Rightarrow Q⋮8\)
Vậy \(Q⋮8\)
** Nếu \(n⋮3\)
\(\Rightarrow n^2⋮9\Rightarrow Q⋮9\)
** Nếu \(n⋮̸3\)
Vì \(\left(n-1\right)n\left(n+1\right)⋮3\)
Mà \(n⋮̸3\Rightarrow n^2+2⋮3\)
\(\Rightarrow Q⋮9\)
Có \(\left(8;9\right)=1\Rightarrow Q⋮72\)
