<s>[INST] Using the Well Ordering Principle to Show that n^3 - n is divisible by 6. A Guide On Well-Ordering Principle
MIT OCW Guide on WOP
I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the  following problem: 


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*Use the well-ordering principle to show that n^3 - n is divisible by 6. 


Any help would be greatly appreciated! 
Note: After factoring out (n-1)^3 - (n-1) it took the form of n^3 - 3n^2 + 2n and the only difference between that and n^3 - n is "3n^2 - 3n" which I proved was divisible by 6 via induction. Is there a better way to solve this problem?
Edit: Show that $n^3-n$ is divisible by $6$ using induction, post is useful however I not supposed to use induction for the main part of the proof.
 [/INST] Define $f(n) = n^3-n$.
Since $f(-n) = -f(n)$, we only have to prove that $f(n)$ is divisible by $6$ for $n \ge 0$. So our domain is the well-ordered set of natural numbers and we always have $f(n) \ge 0$.
We note that both $f(0)$ and $f(1)$ are divisible by $6$.
Let $k \gt 1$ be the least number such that $f(k)$ is not divisible by $6$, so we employ Euclidean division and write
$\tag 1 f(k) = k^3 - k = 6q + r \quad \text{ with } q \ge 0 \text{ and } 1 \le r \le 5$
Plugging $k-1$ into $f$ we get
$\tag 2 f(k-1) = (k-1)[(k-1)^2-1]$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (k-1) \,(k^2 -2k)$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = k^3 -3k^2 + 2k$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (k^3 - k) + k + 2k - 3k^2$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = 6q + r + 3k\,(1+k)$
Now both $2$ and $3$ must divide $3k\,(1+k)$, so we write that
$\tag 3 f(k-1) =  6q + r + 6q^{'} = 6(q+q^{'}) + r$
So $f(k-1)$ is also not divisible by $6$, a contradiction.
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