Prove that n3 - n is divisible by 6 for all n >= 1.

4. Prove that: $n^3 - n$ is divisible by 6 for all $n \geq 1$.

• Step 1: Factor the expression. We can factor out $n$ from the expression $n^3 - n$:

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

  Recognize that $n^2 - 1$ is a difference of squares, which can be factored as $(n-1)(n+1)$:

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

  Rearranging the terms, we get the product of three consecutive integers:

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

• Step 2: Show divisibility by 2. In any set of two consecutive integers, one must be even. Therefore, in any set of three consecutive integers, at least one integer must be even. Since $(n-1)n(n+1)$ is a product of three consecutive integers, it must contain at least one even number. Thus, $(n-1)n(n+1)$ is always divisible by 2.

• Step 3: Show divisibility by 3. In any set of three consecutive integers, one integer must be a multiple of 3. Since $(n-1)n(n+1)$ is a product of three consecutive integers, it must contain exactly one multiple of 3. Thus, $(n-1)n(n+1)$ is always divisible by 3.

• Step 4: Conclude divisibility by 6. Since $n^3 - n$ is divisible by both 2 and 3, and 2 and 3 are coprime (their greatest common divisor is 1), $n^3 - n$ must be divisible by their product, $2 \times 3 = 6$.

Therefore, $n^3 - n$ is divisible by 6 for all $n \geq 1$.
$\boxed{\text{Proven}}$