If n is a natural number, then $(9^{2n} – 4^{2n})$ is always divisible by :

Q: If n is a natural number, then $(9^{2n} – 4^{2n})$ is always divisible by :

To determine why the expression $9^{2n} - 4^{2n}$ is always divisible by both 5 and 13, we can analyze the expression using algebraic identities and modular arithmetic. ### Explanation: 1. **Expression Analysis:** The expression $9^{2n} - 4^{2n}$ can be rewritten using the difference of squares: \[ 9^{2n} - 4^{2n} = (9^n - 4^n)(9^n + 4^n) \] 2. **Divisibility by 5:** - Consider $9 \equiv 4 \pmod{5}$. Therefore, $9^n \equiv 4^n \pmod{5}$. - This implies $9^n - 4^n \equiv 0 \pmod{5}$, meaning $9^n - 4^n$ is divisible by 5. - Similarly, $9^n + 4^n \equiv 4^n + 4^n \equiv 2 \times 4^n \equiv 0 \pmod{5}$ when $n$ is such that $4^n \equiv 0 \pmod{5}$. 3. **Divisibility by 13:** - Consider $9 \equiv -4 \pmod{13}$. Therefore, $9^n \equiv (-4)^n \equiv (-1)^n \times 4^n \pmod{13}$. - For even $n$, $9^n \equiv 4^n \pmod{13}$, so $9^n - 4^n \equiv 0 \pmod{13}$. - For odd $n$, $9^n \equiv -4^n \pmod{13}$, so $9^n + 4^n \equiv 0 \pmod{13}$. 4. **Conclusion:** Since $9^{2n} - 4^{2n}$ is divisible by both 5 and 13 for any natural number $n$, it is divisible by $5 \times 13 = 65$. ### Why Other Options are Incorrect: - **Option 1 (Only 5):** This option is incorrect because, as shown, the expression is divisible by both 5 and 13, not just 5. - **Option 2 (Only 13):** This option is incorrect because the expression is also divisible by 5, not just 13. - **Option 4 (None of these):** This option is incorrect because the expression is indeed divisible by both 5 and 13, as demonstrated. Thus, the correct answer is Option 3, as $9^{2n} - 4^{2n}$ is always divisible by both 5 and 13.

Question

Accepted Answer

To determine why the expression $9^{2n} - 4^{2n}$ is always divisible by both 5 and 13, we can analyze the expression using algebraic identities and modular arithmetic.

### Explanation:

1. **Expression Analysis:**
   The expression $9^{2n} - 4^{2n}$ can be rewritten using the difference of squares:
   $$
   9^{2n} - 4^{2n} = (9^n - 4^n)(9^n + 4^n)
   $$

2. **Divisibility by 5:**
   - Consider $9 \equiv 4 \pmod{5}$. Therefore, $9^n \equiv 4^n \pmod{5}$.
   - This implies $9^n - 4^n \equiv 0 \pmod{5}$, meaning $9^n - 4^n$ is divisible by 5.
   - Similarly, $9^n + 4^n \equiv 4^n + 4^n \equiv 2 \times 4^n \equiv 0 \pmod{5}$ when $n$ is such that $4^n \equiv 0 \pmod{5}$.

3. **Divisibility by 13:**
   - Consider $9 \equiv -4 \pmod{13}$. Therefore, $9^n \equiv (-4)^n \equiv (-1)^n \times 4^n \pmod{13}$.
   - For even $n$, $9^n \equiv 4^n \pmod{13}$, so $9^n - 4^n \equiv 0 \pmod{13}$.
   - For odd $n$, $9^n \equiv -4^n \pmod{13}$, so $9^n + 4^n \equiv 0 \pmod{13}$.

4. **Conclusion:**
   Since $9^{2n} - 4^{2n}$ is divisible by both 5 and 13 for any natural number $n$, it is divisible by $5 \times 13 = 65$.

### Why Other Options are Incorrect:

- **Option 1 (Only 5):** This option is incorrect because, as shown, the expression is divisible by both 5 and 13, not just 5.
  
- **Option 2 (Only 13):** This option is incorrect because the expression is also divisible by 5, not just 13.

- **Option 4 (None of these):** This option is incorrect because the expression is indeed divisible by both 5 and 13, as demonstrated.

Thus, the correct answer is Option 3, as $9^{2n} - 4^{2n}$ is always divisible by both 5 and 13.