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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.

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    $\begingroup$ Your understanding is fine as it is. There is no need to say any more. $\endgroup$ – John Bentin Nov 23 '20 at 11:56
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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 $3$.

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You can use lil' Fermat:

As $3$ is prime, we have $n^3\equiv n\mod 3$ for any $n$, hence $$(n+1)^3-n^3\equiv (n+1)-n=1\mod 3.$$

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$3n^2 + 3n + 1 = 3(n^2 + n) + 1$.

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 $n \in \mathbb Z$.

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