Question

How to prove a^3 -a divisible by 3 for all a?
Please, help

Answer 1

Factors into:
*a(a-1)(a+1)*

Answer 2

I know a^3 -a = a(a+1)(a-1) are 3 consecutive numbers, hence it divisible by3

Answer 3

so what's the prob/

Answer 4

All I want is to go this way
a is odd, a = 2k +1
a is even, a = 2k

Answer 5

but I got stuck at a is odd.

Answer 6

It is back to 3 consecutive numbers.

Answer 7

whether you start from odd or even integer, the product of 3 consequtive integers will be divisible by 3....

Answer 8

:) Hence, no way to avoid "3 consecutive numbers" method?? right?OOOOOOk, that's all I want to know. Thank you.

Answer 9

Yes, 3 conseq.... (or not another way that I aware of using my small knowledge)

Answer 10

you could do induction if you don't like that method above...
\[\text{ assume } 3i=a^3-a \text{ for some integer } a \\ ... \\ (a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1 \\ (a+1)^3-(a+1)=a^3-a+3a^2+3a \\ (a+1)^3-(a+1)=3i+3a^2+3a \\ (a+1)^3-(a+1)=3(i+a^2+a) \\ \text{ so } 3 \text{ is a factor of } (a+1)^3-(a+1)\]

Answer 11

and i is an integer of course

Answer 12

Got you. Thanks @freckles