Giải thích:
`4n^3-4n^2-n+4`
`=4n^3+2n^2-6n^2-3n+2n+1+3`
`=2n^2(2n+1)-3n(2n+1)+(2n+1)+3`
`=(2n+1)(2n^2-3n+1)+3`
Để `4n^3-4n^2-n+4` $\vdots$ `2n+1`
thì `(2n+1)(2n^2-3n+1)+3` $\vdots$ `2n+1`
Mà `(2n+1)(2n^2-3n+1)` $\vdots$ `2n+1`
`=> 3` $\vdots$ `2n+1`
`=> 2n+1 ∈ Ư(3)={-3; -1; 1; 3}`
`=> 2n ∈ {-2; -1; 0; 1 }`
`=> n ∈ {-2; -1; 0; 1 }`