# A more elegant manner of writing a proof.

By doing $$(n+1)^3 - n^3$$, I obtain $$3n^2+3n+1$$. Since the $$+1$$ prevents a multiple of 3 from forming, I understand the answer to this problem. However, what would be a more elegant manner to address why the +1 results in two consecutive terms never being a multiple of 3.

• Your understanding is fine as it is. There is no need to say any more. – John Bentin Nov 23 '20 at 11:56

By absurd, suppose that $$3(n^2+n)+1=3k$$ for some $$k \in \mathbb{Z}$$. Then, $$1=3(k-n^2-n)$$, that is, $$1$$ is multiple of $$3$$.
As $$3$$ is prime, we have $$n^3\equiv n\mod 3$$ for any $$n$$, hence $$(n+1)^3-n^3\equiv (n+1)-n=1\mod 3.$$
$$3n^2 + 3n + 1 = 3(n^2 + n) + 1$$.
Explanation: If $$n$$ is an integer, this will never be divisible by $$3$$ as $$3(n^2 + n)$$ is always divisible by $$3$$, but $$1$$ is not divisible by $$3$$. Thus the sum of these two is also never divisible by $$3$$.
In your proof, you can just mention the first line, but you have to add that this is never divisible by $$3$$ for all $$n \in \mathbb Z$$.