Cmr: với mọi số nguyên lẻ thì n^6-n^4-n^2+1 chia hết cho 128

`\color{#ef7e80}{\text{B}} \color{#f28e90}{\text{t}} \color{#f49d9f}{\text{r}} \color{#f5acad}{\text{a}} \color{#f7bbbc}{\text{m}} \color{#f8cacb}{\text{m}} \color{#fad9da}{\text{m}} \color{#fce8e9}{\text{m}}`

Ta có:

`A=n^6-n^4-n^2+1`

`=n^4(n^2-1)-(n^2-1)`

`=(n^4-1)(n^2-1)`

`=(n^2-1)(n^2+1)(n-1)(n+1)`

`=(n-1)(n+1)(n-1)(n+1)(n^2+1)`

`=(n-1)^2.(n+1)^2.(n^2+1)`

Vì `n` là số nguyên lẻ nên có dạng `n=2k+1(k\in ZZ)`

`=>` `A=(2k+1-1)^2. (2k+1+1)^2. [(2k+1)^2+1]`

`=(2k)^2.(2k+2)^2.(4k^2+4k+1+1)`

`=4k^2.4(k+1)^2.(4k^2+4k+2)`

`=4k^2.4(k+1)2.(2k^2+2k+1)`

`=32[k.(k+1)]^2.(2k^2+2k+1)`

Vì `k(k+1)\vdots2`

nên `[k(k+1)]^2\vdots4`

`=>` `32[k.(k+1)]^2.(2k^2+2k+1)\vdots128`

Vậy `n^6-n^4-n^2+1\vdots128 AA n` lẻ `\inZZ`