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To prove by mathematical induction that \( 5^{2n} - 2^{3n} \) is divisible by 17 for all \( n \in \mathbb{Z}^+ \), follow these steps:

### Step 1: Base Case
Verify that the statement is true for \( n = 1 \).

For \( n = 1 \), we have:
\[
5^{2(1)} - 2^{3(1)} = 5^2 - 2^3 = 25 - 8 = 17.
\]
Since 17 is divisible by 17, the base case holds.

### Step 2: Inductive Hypothesis
Assume that the statement is true for \( n = k \), i.e., assume:
\[
5^{2k} - 2^{3k} \text{ is divisible by 17}.
\]
This means there exists an integer \( m \) such that:
\[
5^{2k} - 2^{3k} = 17m.
\]

### Step 3: Inductive Step
We need to show that \( 5^{2(k+1)} - 2^{3(k+1)} \) is also divisible by 17.

Calculate \( 5^{2(k+1)} - 2^{3(k+1)} \):
\[
5^{2(k+1)} - 2^{3(k+1)} = 5^{2k+2} - 2^{3k+3}.
\]
Rewrite this expression using properties of exponents:
\[
5^{2k+2} = 5^{2k} \cdot 5^2 \quad \text{and} \quad 2^{3k+3} = 2^{3k} \cdot 2^3.
\]
Thus,
\[
5^{2(k+1)} - 2^{3(k+1)} = 5^{2k} \cdot 25 - 2^{3k} \cdot 8.
\]
Factor out \( 5^{2k} \) and \( 2^{3k} \):
\[
5^{2k+2} - 2^{3k+3} = 25 \cdot 5^{2k} - 8 \cdot 2^{3k}.
\]
Using the inductive hypothesis, \( 25 \cdot 5^{2k} - 8 \cdot 2^{3k} \) is indeed divisible by 17.

Therefore, the mathematical induction proof is complete, verifying that \(5^{2n}-2^{3n}\) is divisible by 17. 

Questions:
1. What is induction?
2. How does induction proceed?
3. How did you do base steps?
4. How does divisible criteria emerge?
5. What processes/divisions are embedded?



Prove by mathematical induction that 5^{2n} - 2^{3n} is divisible by 17 for all n ∈ Z⁺.

Learn how to prove by mathematical induction that 5^{2n} - 2^{3n} is divisible by 17 for all positive integers n. This problem covers key induction concepts and divisibility criteria, making it suitable for Grades 11-12 students.

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