Show that the difference of two consecutive cubes is never divisible by $3$. Here is my proof:
Let $n \in \Bbb Z$.  Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.
Without loss of generality (not sure if I can use this), let $n = 2k$.
Then, $n + 1 = 2k + 1$.
$$\begin{align}
(n + 1)^3 - n^3 & = (2k + 1)^3 - (2k)^3 \\
& = 8k^3 + 12k^2 + 6k + 1 - 8k^3 \\
& = 3(4k^2 + 2k) + 1
\end{align}$$
Let $m = (n + 1)^3 - n^3$.
Then, $m = 1 + 3(4k^2 + 2k) \Rightarrow m \equiv 1 \pmod 3$.
Therefore, $\forall m$, when divided by $3$, there remains a remainder $1$.  So $(n + 1)^3 - n^3$ is not divisible by $3$.
Do I also have to prove this for the other case where $n$ is odd, or is this proof sufficient?

$(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3  = 3(n^2 + n) + 1 \equiv 1$ (mod $3$)
