Ta có :

`a^5-a`

`=(a^4-1)a`

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

`=a(a^2-1)(a^2-4+5)`

`=a(a^2-1)(a^2-4)+5a(a^2-1)`

`=a(a-1)(a+1)(a-2)(a+2)+5a(a-1)(a+1)`

`=(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)`

Do `(a-2)(a-1)a(a+1)(a+2)\vdots5` và `5a(a-1)(a+1)\vdots5` nên `[(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)]\vdots5` hay `a^5-a\vdots5`

Vậy `a^5-a\vdots5=>đpcm`

`________`

HDT `A^2-B^2=(A-B)(A+B)`