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

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Step 1: Base Case

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

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Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer $k$, i.e., $k^3 - k \text{ is divisible by 3.}$ This means: $k^3 - k = 3m \quad \text{for some integer } m.$

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Step 3: Inductive Step

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

Expand $(k+1)^3 - (k+1)$: $(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 + 3k^2 + 3k - k = k^3 + 3k^2 + 3k.$

Group terms: $k^3 + 3k^2 + 3k = (k^3 - k) + 3k^2 + 3k.$

From the inductive hypothesis, $k^3 - k = 3m$ for some integer $m$. Thus: $k^3 + 3k^2 + 3k = 3m + 3k^2 + 3k = 3(m + k^2 + k).$

Since $m + k^2 + k$ is an integer, $k^3 + 3k^2 + 3k$ is divisible by 3.

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Step 4: Conclusion

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

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Would you like additional details or clarifications? 😊

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Related Questions:

1. Can you generalize this proof to show divisibility by other numbers (e.g., $n^3 - n$ divisible by 6)?
2. What does the divisibility imply about the factorization of $n^3 - n$?
3. Can you visualize this result with modular arithmetic?
4. How does this relate to the roots of the equation $n(n^2 - 1) = 0$?
5. What are some examples of polynomial divisibility tests in number theory?

Tip:

When proving divisibility, modular arithmetic often simplifies calculations and reveals patterns quickly.