Prove or disprove that $n^3-n$ is divisible by $6$, without using induction

Question

Prove or disprove that $n^3-n$ is divisible by $6$, without using induction

I have no idea how to go about this.

I should add that n is an integer. I started by looking for some integer that was a single case disproving it, but I couldn't find one.

Any help/suggestions are welcome!

Answer 1

We have $$n^3 - n = n(n^2-1) = n(n+1)(n-1)$$

One of these $(n, n+1, n-1)$ must be even (why?) and one must be divisible by 3 (why?).

Answer 2

Hint:

Fermat's theorem and Chinese remainder theorem.

Answer 3

By the division theorem we have $n=6k+r$ for some $0 \le r < 6$. Then $$(n^3-n) = (6k+r)^3 - (6k+r) = 6(x-k) + (r^3-r)$$ where $x = \sum_{i=1}^3 \binom{3}{i} 6^{i-1}k^ir^{3-i} \in \mathbb{Z}$ by the binomial theorem.

But $r \in \{0,1,2,3,4,5\}$, so $r^3-r \in \{0,0,6,24,60,120\}$, each of which is a multiple of $6$.

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Disclaimer: I would argue that both this answer and twnly's answer 'use induction', because both depend on results that themselves require induction in some form to be proved.