# Mathematical induction proof

1. Nov 25, 2011

### dpesios

Hello everybody,
I am doing my reading lately to prepare for some exams to join a mathematics department.

And I would very much like, if anyone could help, the solution (or a hint) to the following induction proof.

" Show that n^3 + (n+1)^3 + (n+2)^3 is a multiple of 9 "

I can deal with proofs that say " *this* is divisible with *that* " but in that case I have no clue how to start.

I have noone to ask for help ... but you in the forum.

2. Nov 25, 2011

### Mentallic

Show it's true for the base case n=1, then assume it's true for n=k, that is

$$k^3+(k+1)^3+(k+2)^3 = 9M$$

where M is an integer, now prove it's true for n=k+1,

$$(k+1)^3+(k+2)^3+(k+3)^3=9N$$

So what you need to do is show that N is an integer as well. You'll need to make a substitution from the equation you assumed to get anywhere.

3. Nov 26, 2011

### dpesios

Okay, here is my solution. I thought that at the problem defintion "is divisible by" and "is a multiple of" makes a huge difference.

The base case is obvious.

Let that the statement holds for n=k, that is,
$$k^3+(k+1)^3+(k+2)^3$$ is a multiple of 9.
We will then show that it holds for n=k+1, that is $$(k+1)^3+(k+2)^3+(k+3)^3$$ is a multiple of 9.

We have,
$$k^3+(k+1)^3+(k+2)^3+(k+3)^3-k^3=9M+(k+3)^3-k^3$$,
So what we want to show is that $$(k+3)^3-k^3$$ is also a multiple of 9.
By using the formula $$a^3-b^3=(a-b)(a^2+b^2+ab)$$ we can reach our goal.

4. Nov 26, 2011

### Mentallic

Nicely done. The only thing I would add is to specifically state that the equation you assumed is equal to 9M, rather than telling us in words that it's a multiple of 9. This way there is no disputing anything when you make the substitution.