Proof by Induction: $5^n - 1$ is divisible by 4 for all $n \in \mathbb{Z}^+$

We aim to prove, using mathematical induction, that for all positive integers $n$, $5^n - 1$ is divisible by 4. That is, $5^n - 1$ can be written as $4k$ for some integer $k$.

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

We check the statement for $n = 1$: $5^1 - 1 = 5 - 1 = 4$ Clearly, 4 is divisible by 4. Hence, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for some arbitrary positive integer $n = k$. That is, assume: $5^k - 1 = 4m$ for some integer $m$. This is our inductive hypothesis.

Step 3: Inductive Step

We need to prove that the statement holds for $n = k+1$, i.e., we need to show that: $5^{k+1} - 1 = 4q$ for some integer $q$.

Start with the expression for $5^{k+1}$: $5^{k+1} = 5 \cdot 5^k$ Now consider $5^{k+1} - 1$: $5^{k+1} - 1 = 5 \cdot 5^k - 1$ Using the inductive hypothesis, we know that $5^k - 1 = 4m$. So, substitute this into the equation: $5^{k+1} - 1 = 5 \cdot 5^k - 1 = 5 \cdot (4m + 1) - 1$ Distribute the $5$: $5^{k+1} - 1 = 5 \cdot 4m + 5 \cdot 1 - 1 = 20m + 5 - 1$ Simplify: $5^{k+1} - 1 = 20m + 4$ Factor out the 4: $5^{k+1} - 1 = 4(5m + 1)$ Thus, $5^{k+1} - 1$ is divisible by 4 because it is of the form $4q$, where $q = 5m + 1$ is an integer.

Step 4: Conclusion

By the principle of mathematical induction, $5^n - 1$ is divisible by 4 for all positive integers $n$.

Would you like to ask any further details or clarifications?

Relative Questions:

1. How can induction be used in other divisibility proofs?
2. Can this proof be adapted for different bases, like $3^n - 1$ being divisible by a specific number?
3. How does the principle of mathematical induction work in other branches of mathematics?
4. What would happen if we used modular arithmetic for this proof?
5. What is the difference between divisibility proofs and congruence proofs?

Tip:

Always ensure that the base case and the inductive step are clearly validated in any induction proof to avoid logical errors.