To prove that $5^n - 1$ is divisible by 4 for all natural numbers $n$ using mathematical induction, we proceed as follows:

Step 1: Base Case ($n = 1$)

For $n = 1$: $5^n - 1 = 5^1 - 1 = 5 - 1 = 4$ Since $4$ is divisible by $4$, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary $n = k$, i.e., assume: $5^k - 1 \text{ is divisible by } 4.$ This implies: $5^k - 1 = 4m \quad \text{for some integer } m.$

Step 3: Inductive Step

We need to prove that $5^{k+1} - 1$ is also divisible by 4.

Start with $5^{k+1} - 1$: $5^{k+1} - 1 = 5 \cdot 5^k - 1$ Rewriting it: $5^{k+1} - 1 = (5 \cdot 5^k - 5) + 4$ Factoring $5$: $5^{k+1} - 1 = 5(5^k - 1) + 4$

From the inductive hypothesis, $5^k - 1$ is divisible by 4. This means $5(5^k - 1)$ is also divisible by 4 because $5$ is an integer. Adding $4$ does not change divisibility by $4$.

Thus, $5^{k+1} - 1$ is divisible by 4.

Step 4: Conclusion

By mathematical induction, $5^n - 1$ is divisible by 4 for all natural numbers $n$.

Would you like a detailed example or explanation of any step? Here are 5 related questions:

1. Can this induction process be adapted for other bases besides $5$?
2. How does the inductive step work when proving divisibility properties?
3. Why does divisibility by $4$ hold consistently in this case?
4. Can we generalize $5^n - 1$ to a modular arithmetic proof?
5. What role does the base case play in mathematical induction?

Tip: Always verify divisibility in the base case to ensure the foundation of induction is valid.