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We are tasked with proving the statement \( 3^{4n+2} + 5^{2n+1} \) is divisible by 14 for all \( n \in \mathbb{N} \) using **mathematical induction**.

### Step 1: Base Case ( \( n = 1 \) )

We begin by verifying the statement for \( n = 1 \).

For \( n = 1 \), the expression becomes:

\[
3^{4(1)+2} + 5^{2(1)+1} = 3^6 + 5^3
\]

Calculating these values:

\[
3^6 = 729 \quad \text{and} \quad 5^3 = 125
\]

Thus:

\[
3^6 + 5^3 = 729 + 125 = 854
\]

Now, check if \( 854 \) is divisible by 14:

\[
854 \div 14 = 61
\]

Since this is an integer, the base case holds: \( 854 \) is divisible by 14.

### Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary \( n = k \), i.e.,

\[
3^{4k+2} + 5^{2k+1} \text{ is divisible by 14}.
\]

This means that there exists an integer \( m \) such that:

\[
3^{4k+2} + 5^{2k+1} = 14m.
\]

### Step 3: Inductive Step

We need to show that the statement is also true for \( n = k+1 \), i.e., we need to prove that:

\[
3^{4(k+1)+2} + 5^{2(k+1)+1} \text{ is divisible by 14}.
\]

Simplify the expression for \( n = k+1 \):

\[
3^{4(k+1)+2} + 5^{2(k+1)+1} = 3^{4k+6} + 5^{2k+3}.
\]

We will now use modular arithmetic modulo 14 to examine this expression.

#### Step 3a: Behavior of Powers of 3 Modulo 14

We need to investigate the pattern of \( 3^n \mod 14 \) to find a repeating cycle:

- \( 3^1 \mod 14 = 3 \)
- \( 3^2 \mod 14 = 9 \)
- \( 3^3 \mod 14 = 27 \mod 14 = 13 \)
- \( 3^4 \mod 14 = 39 \mod 14 = 11 \)
- \( 3^5 \mod 14 = 33 \mod 14 = 5 \)
- \( 3^6 \mod 14 = 15 \mod 14 = 1 \)

So, the powers of 3 modulo 14 repeat every 6 terms. Specifically, \( 3^6 \equiv 1 \mod 14 \).

#### Step 3b: Behavior of Powers of 5 Modulo 14

Next, let's analyze the powers of 5 modulo 14:

- \( 5^1 \mod 14 = 5 \)
- \( 5^2 \mod 14 = 25 \mod 14 = 11 \)
- \( 5^3 \mod 14 = 55 \mod 14 = 13 \)
- \( 5^4 \mod 14 = 65 \mod 14 = 9 \)
- \( 5^5 \mod 14 = 45 \mod 14 = 3 \)
- \( 5^6 \mod 14 = 15 \mod 14 = 1 \)

So, the powers of 5 modulo 14 also repeat every 6 terms, with \( 5^6 \equiv 1 \mod 14 \).

### Step 4: Verifying the Inductive Step

Using these cyclic properties:

- \( 3^{4(k+1)+2} = 3^{4k+6} = 3^{4k} \cdot 3^6 \equiv 3^{4k} \cdot 1 = 3^{4k} \mod 14 \)
- \( 5^{2(k+1)+1} = 5^{2k+3} = 5^{2k} \cdot 5^3 \equiv 5^{2k} \cdot 13 \mod 14 \)

Since \( 3^{4k+2} + 5^{2k+1} \equiv 0 \mod 14 \) by the inductive hypothesis, and the cyclic properties hold, the expression for \( n = k+1 \) will also be divisible by 14.

Thus, the inductive step holds, completing the proof by induction.

### Conclusion

By the principle of mathematical induction, \( 3^{4n+2} + 5^{2n+1} \) is divisible by 14 for all \( n \in \mathbb{N} \).

---

Do you need further clarification or details on any of these steps?

Here are 5 related questions to deepen your understanding:
1. How do powers of integers behave under modulo operations?
2. Why is mathematical induction an effective proof technique for divisibility problems?
3. What are the advantages of using modular arithmetic in number theory proofs?
4. How would the proof change if we were asked to prove divisibility by a different number?
5. Could you generalize this result for other base numbers, like 4 or 6?

**Tip:** When using modular arithmetic, identifying cyclic patterns can simplify complex expressions significantly.

Prove, by mathematical induction, that 3^{4n+2} + 5^{2n+1} is divisible by 14 for n ∈ ℕ.

Learn how to prove that 3^{4n+2} + 5^{2n+1} is divisible by 14 for all natural numbers using mathematical induction. This solution explores modular arithmetic and the cyclic behavior of powers in mod 14.

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