The image contains the following statement:

For all $n \in \mathbb{Z}^+$, 5 divides $7^n - 2^n$.

This is a mathematical statement claiming that for all positive integers $n$, the expression $7^n - 2^n$ is divisible by 5, or equivalently:

$7^n - 2^n \equiv 0 \pmod{5}$

Proof by Induction:

Let's prove this statement using mathematical induction.

Base Case: For $n = 1$:

$7^1 - 2^1 = 7 - 2 = 5$

Clearly, $5$ is divisible by 5, so the base case holds.

Inductive Step: Assume the statement holds for some $n = k$, i.e.,

$7^k - 2^k \equiv 0 \pmod{5}$

This means:

$7^k \equiv 2^k \pmod{5}$

Now we need to prove the statement for $n = k+1$, i.e.,

$7^{k+1} - 2^{k+1} \equiv 0 \pmod{5}$

Using the property of exponents, we can express $7^{k+1}$ and $2^{k+1}$ as:

$7^{k+1} = 7^k \cdot 7$ $2^{k+1} = 2^k \cdot 2$

Thus, we need to show:

$7^k \cdot 7 - 2^k \cdot 2 \equiv 0 \pmod{5}$

From the inductive hypothesis, we know $7^k \equiv 2^k \pmod{5}$, so we replace:

$7^k \cdot 7 - 2^k \cdot 2 \equiv 2^k \cdot 7 - 2^k \cdot 2 \pmod{5}$ $2^k \cdot (7 - 2) \equiv 2^k \cdot 5 \equiv 0 \pmod{5}$

Thus, the statement holds for $n = k+1$.

By the principle of mathematical induction, the statement is true for all $n \in \mathbb{Z}^+$.

Would you like more details or have any questions about this proof?

Here are 5 related questions for you to consider:

1. What is modular arithmetic and how is it used in divisibility proofs?
2. Can this statement be proven using another method besides induction?
3. What other numbers besides 5 can be used to generalize this result?
4. How does the property of exponents help in simplifying complex expressions?
5. What are common pitfalls in applying mathematical induction?

Tip: When proving divisibility statements, modular arithmetic simplifies handling large powers and remainders, making the logic clearer.