<s>[INST] Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$
 [/INST] Gibbs has already presented a perfect proof. However, there is another one, that does not involve decomposition of $n^3-n$ and is relatively straightforward.
To say $n^3-n$ is divisible by $6$ is the same as saying $n^3-n \equiv 0 \pmod 6$. Thus, once we compute $n^3-n \pmod 6$ for all possible $n$, we are done. By general algebra, this is the same as computing $n^3-n$ for all elements of the ring $\mathbb Z / 6 \mathbb Z$, and there are exactly $6$ elements in this ring, namely the equivalence classes of $0,1,\dots,5$ modulo $6$.
So, let's compute this $6$ values $\mod 6$:
$$
\begin{array}{ l|c }
n & n^3-n \\ \hline
0 & 0-0 \equiv 0 \pmod 6\\ \hline
1 & 1-1 \equiv 0 \pmod 6 \\ \hline
2 & 8-2 \equiv 6 \equiv 0 \pmod 6 \\ \hline
3 & 27-3 \equiv 24 \equiv 0 \pmod 6 \\ \hline
4 & 64 - 4 \equiv 60 \equiv 0 \pmod 6 \\ \hline
5 & 125 - 5 \equiv 120 \equiv 0 \pmod 6
\end{array}
$$
As you can see, $n^3-n\equiv 0 \pmod 6$ holds for any $n$.
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