# Simple Proof

1. Feb 26, 2008

### sapiental

1. The problem statement, all variables and given/known data

Prove the following: If the integer n is divisible by 3 then n^3 is divisible by 3.

2. Relevant equations

Direct Proof

3. The attempt at a solution

n = 3m

n^2 = 9m^2

n^2 = 3(3m^2)

I think the proof is done at this point because the 3 factors out but I also did this:

n^2 = 9m^2

(n^2)/9 = m^2

(n/3)(n/3) = (m)(m)

which also implies n is divisible by 3 since integer x integer = integer

My professor is kind of harsh on proofs so Im not sure if there are intermediate steps I'm missing. Thanks!

2. Feb 26, 2008

### Gib Z

If 3 is a factor of n, it must also be a factor of any natural power of n. Do you mean n^3 or n^2? Your working looks like n^2. Anyway, the first way you did it is correct and satisfactory though if your teacher is really harsh, you may want to go like:

$$n= 3m$$ for some natural value of m, because n is divisible by 3 (Data).
Squaring both sides, $$n^2 = 9m^2 = 3 ( 3m^2)$$. Since m is natural, 3m^2 must also be natural, and hence 3 is a factor of n^2 as well.