Question

CMR: \(\left(n^6+n^4-2n^2\right)⋮72\) với mọi n la số nguyên.

Answer 1

\(n^6+n^4-2n^2\)

\(=n^2\left(n^4+n^2-2\right)\)

\(=n^2\left[\left(n^4-1\right)+n^2-1\right]\)

\(=n^2\left[\left(n^2-1\right)\left(n^2+1\right)+n^2-1\right]\)

\(=n^2\left(n^2-1\right)\left(n^2+1+1\right)\)

\(=n^2\left(n^2-1\right)\left(n^2+2\right)\)

\(=n\left(n-1\right)\left(n+1\right)n\left(n^2+2\right)\)

Answer 2

Xét \(n=2k\) , ta có :

\(\left(2k\right)^2\left[\left(2k\right)^2-1\right]\left[\left(2k\right)^2+2\right]=4k^2\left(2k-1\right)\left(2k+1\right)\left(4k^2+2\right)\)

\(=8k^2\left(2k-1\right)\left(2k+1\right)\left(2k^2+1\right)⋮8\left(1\right)\)

Xét \(n=2k+1\) , ta có :

\(\left(2k+1\right)^2\left[\left(2k+1\right)^2-1\right]\left[\left(2k+1\right)^2+2\right]=\left(2k+1\right)^2.2k\left(2k+2\right)\left(4k^2+4k+1+2\right)\)

\(=\left(2k+1\right)^2.4k\left(k+1\right)\left(4k^2+4k+3\right)⋮8\left(2\right)\)

( do \(k\left(k+1\right)⋮2\Rightarrow4k\left(k+1\right)⋮8\) )

Với n \(⋮3\Rightarrow n^2⋮9\) \(\Rightarrow n^2\left(n^2-1\right)\left(n^2+2\right)⋮9\left(3\right)\)

Với n \(⋮3̸\) \(\Rightarrow n^2:3\) ( dư 1 ) \(\Rightarrow n^2-1⋮3\Rightarrow n^2+2⋮3\)

Do \(n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n^2\left(n-1\right)\left(n+1\right)\left(n^2+2\right)⋮9\left(4\right)\)

Từ ( 1 ) ; ( 2 ) ; ( 3 ) ; ( 4 )

\(\Rightarrow n^6+n^4-2n^2⋮72\left(đpcm\right)\)