Define a function `phi:NtoN` as follows `phi(1)=1,phi(P^(n))=P^(n-1)(P-1)` is prime and `n epsilonN` and `phi(mn)=phi(m)phi(n)` if m & n are relatively prime natural numbers.
`phi(8n+4)` when `n epsilonN` is equal to

Options

A. `2phi(4n+2)`
B. `phi(2n+1)`
C. `2pi(2n+1)`
D. `4pi(2n+1)`

Answer

To solve the problem, we need to evaluate the function \( \phi(8n + 4) \) based on the properties of the Euler's totient function \( \phi \).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression \( 8n + 4 \). We can factor out a 4:
\[ 8n + 4 = 4(2n + 1) \]

2. **Identify the Prime Factorization**:
The number \( 4 \) can be expressed as \( 2^2 \). Therefore, we can rewrite \( 8n + 4 \) as:
\[ 8n + 4 = 2^2(2n + 1) \]

3. **Check if \( 2n + 1 \) is Prime**:
For \( n \in \mathbb{N} \), \( 2n + 1 \) is always an odd number. We need to check if it is prime. Since \( n \) is a natural number, \( 2n + 1 \) can be prime for certain values of \( n \).

4. **Apply the Property of \( \phi \)**:
Since \( 2^2 \) and \( 2n + 1 \) are relatively prime (as \( 2n + 1 \) is odd), we can use the property:
\[ \phi(mn) = \phi(m) \cdot \phi(n) \]
Thus,
\[ \phi(8n + 4) = \phi(2^2) \cdot \phi(2n + 1) \]

5. **Calculate \( \phi(2^2) \)**:
Using the formula \( \phi(p^n) = p^{n-1}(p - 1) \):
\[ \phi(2^2) = 2^{2-1}(2 - 1) = 2^1 \cdot 1 = 2 \]

6. **Calculate \( \phi(2n + 1) \)**:
Since \( 2n + 1 \) is prime, we can use the formula:
\[ \phi(2n + 1) = 2n + 1 - 1 = 2n \]

7. **Combine the Results**:
Now, substituting back into our expression, we have:
\[ \phi(8n + 4) = \phi(2^2) \cdot \phi(2n + 1) = 2 \cdot 2n = 4n \]

### Final Answer:
Thus, the value of \( \phi(8n + 4) \) is:
\[ \phi(8n + 4) = 4n \]