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?