If n is a natural number the 9^ 2n −4^ 2n is always divisible by :

Options

A. a. 13
B. b. Both 5 and 13
C. c. 5
D. d. None of these

Answer

To solve the problem, we need to determine if \( 9^{2n} - 4^{2n} \) is divisible by certain numbers when \( n \) is a natural number.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ 9^{2n} - 4^{2n} \]
We can rewrite \( 9^{2n} \) as \( (3^2)^{2n} = 3^{4n} \) and \( 4^{2n} \) as \( (2^2)^{2n} = 2^{4n} \). Thus, we have:
\[ 9^{2n} - 4^{2n} = 3^{4n} - 2^{4n} \]

2. **Use the Difference of Squares**:
We can apply the difference of squares formula, which states that \( a^2 - b^2 = (a - b)(a + b) \). Here, we can treat \( 3^{2n} \) and \( 2^{2n} \) as \( a \) and \( b \):
\[ 3^{4n} - 2^{4n} = (3^{2n} - 2^{2n})(3^{2n} + 2^{2n}) \]

3. **Analyze the Factors**:
Now we need to determine if \( 3^{2n} - 2^{2n} \) and \( 3^{2n} + 2^{2n} \) are divisible by any specific numbers.

4. **Check Divisibility by 5**:
We can check the expression \( 3^{2n} - 2^{2n} \) modulo 5:
- For \( n = 1 \): \( 3^2 - 2^2 = 9 - 4 = 5 \equiv 0 \mod 5 \)
- For \( n = 2 \): \( 3^4 - 2^4 = 81 - 16 = 65 \equiv 0 \mod 5 \)
- Thus, \( 3^{2n} - 2^{2n} \) is divisible by 5 for all natural numbers \( n \).

5. **Check Divisibility by 13**:
Now we check \( 3^{2n} - 2^{2n} \) modulo 13:
- For \( n = 1 \): \( 3^2 - 2^2 = 9 - 4 = 5 \equiv 5 \mod 13 \)
- For \( n = 2 \): \( 3^4 - 2^4 = 81 - 16 = 65 \equiv 0 \mod 13 \)
- Thus, \( 3^{2n} - 2^{2n} \) is divisible by 13 for \( n = 2 \) and higher.

6. **Conclusion**:
Since \( 9^{2n} - 4^{2n} \) is divisible by both 5 and 13, we conclude that it is always divisible by 65 (since \( 65 = 5 \times 13 \)).

### Final Answer:
The expression \( 9^{2n} - 4^{2n} \) is always divisible by 5 and 13.