n3 + 2n is divisible by 3 for all n≥ 1 by induction

Question

n3 + 2n is divisible by 3 for all  n≥ 1 by induction

myininaya · Answer

So we want to show 3|(n^3+2n) for all n>=1.

You basically have two steps, but I will write 3.
1) Show it is true for n=1.
2) Assume it is true for some integer k greater than 1. (no work to be done on step 2)
3) Show it is true for k+1 using step 2.

anonymous · Answer

thank u

myininaya · Answer

Let me know if you run into trouble with the steps.

anonymous · Answer

i need the answer for my assignment so i can use what u wrote ?

myininaya · Answer

For 1 and 3 you have to do some work.

anonymous · Answer

ok

anonymous · Answer

thank u once again

myininaya · Answer

So is that n^3+2n?
Did I translate that right?

anonymous · Answer

yes

myininaya · Answer

I find it helpful to recall that 3|(n^3+2n) just means we have 3m=n^3+2n for some integer m. 
Whenever one of my proofs that call for induction has that little divides symbol I use that definition. 
If a|b then ak=b for integer k.

zarkon · Answer

I like this ...

$$n^3+2n=n^3-n+3n=n(n^2-1)+3n=n(n-1)(n+1)+3n$$

$$=(n-1)n(n+1)+3n$$

zarkon · Answer

no induction required ;)

zarkon · Answer

since $3|(n-1)n(n+1)$ and $3|3n$

myininaya · Answer

I'm trying to think why (n-1)n(n+1) would be divisible by 3. I mean of course I could go through numbers to see it is divisible by 3 but that would be just induction again. 
What elementary thingy am I missing?

zarkon · Answer

it is the product of 3 consecutive numbers  (mod 3 one of them must be 0)

myininaya · Answer

also it is every 3 integer numbers.
1 2 3
2 3 4
4 5 6
one of those numbers has to be divsible by 3
Okay.

myininaya · Answer

Yea, consecutive. Whatever. Nice Zarkon.

zarkon · Answer

though I'm sure the OP is learning induction and this approach would get them 0 points  :)