If n is an even positive integer, then `a^(n)+b^(n)` is divisible by

Options

A. a+b
B. a-b
C. `a^(2)-b^(2)`
D. none of these

Answer

To solve the problem, we will use the properties of even integers and algebraic identities. We need to show that if \( n \) is an even positive integer, then \( a^n + b^n \) is divisible by \( a^2 + b^2 \) and not divisible by \( a + b \).

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression \( a^n + b^n \) where \( n \) is an even positive integer.

2. **Using the Binomial Theorem**:
Since \( n \) is even, we can express \( n \) as \( n = 2k \) for some integer \( k \). Thus, we can rewrite the expression as:
\[ a^{2k} + b^{2k} \]

3. **Factoring the Expression**:
We can use the identity for the sum of squares. The expression \( a^{2k} + b^{2k} \) can be factored as follows:
\[ a^{2k} + b^{2k} = (a^k + b^k)(a^k - b^k) \]
However, for \( n = 2 \), we have:
\[ a^2 + b^2 \]
which is not divisible by \( a + b \) in general.

4. **Testing with Specific Values**:
Let's test with \( n = 2 \):
\[ a^2 + b^2 \]
We can check if this is divisible by \( a + b \):
- For \( a = 1 \) and \( b = 1 \):
\[ 1^2 + 1^2 = 2 \quad \text{and} \quad 1 + 1 = 2 \quad \text{(divisible)} \]
- For \( a = 1 \) and \( b = -1 \):
\[ 1^2 + (-1)^2 = 2 \quad \text{and} \quad 1 + (-1) = 0 \quad \text{(not applicable)} \]
- For \( a = 2 \) and \( b = 2 \):
\[ 2^2 + 2^2 = 8 \quad \text{and} \quad 2 + 2 = 4 \quad \text{(divisible)} \]
- For \( a = 2 \) and \( b = 0 \):
\[ 2^2 + 0^2 = 4 \quad \text{and} \quad 2 + 0 = 2 \quad \text{(divisible)} \]

5. **Conclusion**:
We can conclude that \( a^n + b^n \) is divisible by \( a^2 + b^2 \) when \( n \) is even, but not necessarily by \( a + b \).

### Final Result:
Thus, if \( n \) is an even positive integer, \( a^n + b^n \) is divisible by \( a^2 + b^2 \) but not by \( a + b \).