Prove by induction that for all positive integers n, 5^n − 1 is divisible by 4. Do not assume
the laws of modular arithmetic in your proof; instead, use the definition of divisibility.

We will prove by **mathematical induction** that for all positive integers \( n \), \( 5^n - 1 \) is divisible by 4. In other words, we want to show that for every positive integer \( n \), there exists an integer \( k \) such that:

\[
5^n - 1 = 4k
\]

### **Step 1: Base Case**
We begin with the base case \( n = 1 \).

\[
5^1 - 1 = 5 - 1 = 4
\]

Since 4 is divisible by 4, the base case holds.

---

### **Step 2: Inductive Hypothesis**
Assume that the statement is true for some 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 show that the statement holds for \( n = k + 1 \). That is, we need to prove:

\[
5^{k+1} - 1 \text{ is divisible by 4}.
\]

We start by manipulating \( 5^{k+1} - 1 \) using the properties of exponents:

\[
5^{k+1} - 1 = 5 \cdot 5^k - 1.
\]

Now, add and subtract 5 in the expression:

\[
5 \cdot 5^k - 1 = (5 \cdot 5^k - 5) + 4.
\]

Factor out 5 from the first part:

\[
(5(5^k - 1)) + 4.
\]

By the **inductive hypothesis**, we know that \( 5^k - 1 = 4m \) for some integer \( m \). Substituting this into the expression:

\[
5(4m) + 4 = 20m + 4 = 4(5m + 1).
\]

Since \( 5m + 1 \) is an integer, the expression \( 4(5m + 1) \) is divisible by 4.

---

### **Step 4: Conclusion**
By mathematical induction, we have shown that \( 5^n - 1 \) is divisible by 4 for all positive integers \( n \).

---

Do you have any questions or need further clarifications?  

Here are 5 related questions for further exploration:  
1. How would the proof change if the statement involved divisibility by 8 instead of 4?  
2. Can you prove that \( 5^n + 1 \) is divisible by 3 for odd \( n \)?  
3. How is mathematical induction applied differently in strong induction?  
4. What is the relationship between divisibility proofs and modular arithmetic?  
5. Can we generalize the result for powers of other numbers, like \( 6^n - 1 \)?  

**Tip:** When using mathematical induction, always carefully link the inductive step to the hypothesis to avoid logical gaps.

Prove by induction that for all positive integers n, 5^n − 1 is divisible by 4. Do not assume the laws of modular arithmetic in your proof; instead, use the definition of divisibility.

Learn how to prove by mathematical induction that 5^n - 1 is divisible by 4 for all positive integers. This proof uses the definition of divisibility and the properties of exponents. Suitable for high school students in Grades 11-12.

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