To prove that $n^3 - n$ is divisible by 6, we must demonstrate that $n^3 - n$ is divisible by both 2 and 3 (since 6 = 2 × 3).

1. Divisibility by 2

A number is divisible by 2 if it is even. For $n^3 - n$:

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

Here, $n, n-1,$ and $n+1$ are three consecutive integers. Among any three consecutive integers, at least one is even. Therefore, the product $n(n-1)(n+1)$ is always divisible by 2.

2. Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3, or if it is of the form $3k$. Again, considering $n(n-1)(n+1)$, among any three consecutive integers, at least one is divisible by 3. Therefore, $n(n-1)(n+1)$ is divisible by 3.

3. Divisibility by 6

Since $n(n-1)(n+1)$ is divisible by both 2 and 3, it is divisible by their least common multiple, 6. Thus, $n^3 - n$ is divisible by 6 for all integers $n$.

Proof Example for Clarity

Let $n = 4$: $n^3 - n = 4^3 - 4 = 64 - 4 = 60$ $60 \div 6 = 10$, which is an integer.

This completes the proof.

---

Do you want a deeper explanation or examples?
Here are some related questions to explore:

1. Can you generalize this divisibility result for other polynomials?
2. What is the significance of consecutive integers in divisibility problems?
3. Can $n^3 - n$ be divisible by 12? Why or why not?
4. How does this proof work if $n$ is negative?
5. Can modular arithmetic simplify the proof further?

Tip: Understanding how divisibility rules work with consecutive integers is crucial for many number theory proofs!