## anonymous one year ago mathematical induction to prove that n 3 − n is divisible by 3, for every positive integer n

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1. SolomonZelman

$$n^3-n=n(n^2-1)=n(n-1)(n+1)$$ that should help.

2. anonymous

thank you

3. SolomonZelman

No further questions?

4. anonymous

not yet... im working it thru

5. anonymous

ok this is where im at P(x+1)=(x+1)((x+1)−1)((x+1)+1) = 3m do i divide

6. SolomonZelman

you don't need anything, except for a little logic.

7. SolomonZelman

here, tell me what happens if you have a product that consists of integers and one of these integers are divisible by 3? Do you agree that the result is divisible by 3?

8. anonymous

yes

9. SolomonZelman

Ok, good.

10. SolomonZelman

Now, lets come back to the fact that you want to prove that: *n(n-1)(n+1)* is divislbe by 3, $$\forall {\bf n \in \mathbb Z}$$

11. SolomonZelman

Ok, lets consider 3 possible cases (for possible integer k) 3k, 3k+1, and 3k+2

12. SolomonZelman

If your number n falls under the category 3k (Such that n --> 3k) then the *n* component of *n(n-1)(n+1)*, is divisible by 3, and thus the entire product *n(n-1)(n+1)* is divislbe by 3.

13. SolomonZelman

If your number n falls under the category 3k+1 (Such that n --> 3k+1) then the *n-1* component of *n(n-1)(n+1)*, is divisible by 3, and thus the entire product *n(n-1)(n+1)* is divislbe by 3.

14. SolomonZelman

And then if: n --> 3k+2 then the *n+1* makes it divisible by 3.

15. SolomonZelman

Do I sound rediculous, or is it understandable.

16. anonymous

no u dont.. im just trying to absorb the concept

17. anonymous

my brain is a bit mathed out

18. SolomonZelman

$$\displaystyle\int$$$$\theta$$ $$f(u)+nn=y$$

19. SolomonZelman

If you have more questions to ask, I wil be back if I am online.

20. anonymous

thank you