For how many natural number values of `N,N^(4)+4` will be a prime number?

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To determine for how many natural number values of \( N \) the expression \( N^4 + 4 \) will be a prime number, we can follow these steps:

### Step 1: Factor the Expression
We start with the expression \( N^4 + 4 \). We can rewrite this as follows:
\[ N^4 + 4 = N^4 + 4N^2 + 4 - 4N^2 = (N^2 + 2)^2 - (2N)^2 \]
This is a difference of squares, which can be factored using the identity \( a^2 - b^2 = (a + b)(a - b) \).

### Step 2: Apply the Difference of Squares Formula
Applying the difference of squares, we have:
\[ (N^2 + 2)^2 - (2N)^2 = (N^2 + 2 + 2N)(N^2 + 2 - 2N) \]
Thus, we can express \( N^4 + 4 \) as:
\[ (N^2 + 2 + 2N)(N^2 + 2 - 2N) \]

### Step 3: Analyze the Factors
Now, we have two factors:
1. \( N^2 + 2 + 2N \)
2. \( N^2 + 2 - 2N \)

### Step 4: Check for Prime Conditions
For \( N^4 + 4 \) to be a prime number, one of these factors must equal 1 (since a prime number has exactly two positive divisors: 1 and itself).

#### Factor 1: \( N^2 + 2 - 2N = 1 \)
Setting the first factor equal to 1:
\[ N^2 - 2N + 2 = 1 \implies N^2 - 2N + 1 = 0 \implies (N-1)^2 = 0 \implies N = 1 \]

#### Factor 2: \( N^2 + 2 + 2N = 1 \)
Setting the second factor equal to 1:
\[ N^2 + 2 + 2N = 1 \implies N^2 + 2N + 1 = 0 \implies (N+1)^2 = 0 \]
This gives \( N = -1 \), which is not a natural number.

### Conclusion
The only natural number value of \( N \) that makes \( N^4 + 4 \) a prime number is \( N = 1 \).

Thus, there is only **one** natural number value of \( N \) for which \( N^4 + 4 \) is prime.

### Final Answer
The answer is \( \boxed{1} \).

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