To prove by induction that $n^3 - n$ is divisible by 3 for all positive integers $n$, follow these steps:

Step 1: Base Case

We check for $n = 1$: $n^3 - n = 1^3 - 1 = 0.$ Since $0$ is divisible by 3, the base case holds.

Step 2: Inductive Hypothesis

Assume that $n^3 - n$ is divisible by 3 for some positive integer $n = k$. That is: $k^3 - k = 3m \quad \text{for some integer } m.$

Step 3: Inductive Step

We must prove that $(k+1)^3 - (k+1)$ is also divisible by 3.

Expand $(k+1)^3$: $(k+1)^3 = k^3 + 3k^2 + 3k + 1.$ Subtract $(k+1)$ from this: $(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 + 3k^2 + 3k - k.$ Simplify: $(k+1)^3 - (k+1) = k^3 - k + 3k^2 + 3k.$

By the inductive hypothesis, $k^3 - k = 3m$, so substitute: $(k+1)^3 - (k+1) = 3m + 3k^2 + 3k.$ Factor out 3: $(k+1)^3 - (k+1) = 3(m + k^2 + k).$

Since $m + k^2 + k$ is an integer, $(k+1)^3 - (k+1)$ is divisible by 3.

Step 4: Conclusion

By induction, $n^3 - n$ is divisible by 3 for all positive integers $n$.

Do you have any questions or want further details?
Here are 5 related questions to explore:

1. What are other common examples of divisibility proofs using induction?
2. How can we generalize the approach to prove divisibility by numbers other than 3?
3. Why is mathematical induction a valid proof method?
4. What is the importance of the base case in an induction proof?
5. How does this divisibility relate to modular arithmetic?

Tip: When using induction, always ensure the base case and inductive step are carefully proven to validate the result!