`A=k^(2)-1andB=(k+1)^(2)-1`, where k is a natural number greater than 1. How many prime number are there by which both A and B are divisible for at least one value of k?

Options

A. 0
B. 1
C. 2
D. More than 2

Answer

To solve the problem, we need to analyze the expressions for A and B given by:

1. \( A = k^2 - 1 \)
2. \( B = (k + 1)^2 - 1 \)

where \( k \) is a natural number greater than 1.

### Step 1: Simplify the expressions for A and B

First, we can simplify \( A \) and \( B \):

- For \( A \):
\[ A = k^2 - 1 = (k - 1)(k + 1) \]

- For \( B \):
\[ B = (k + 1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k = k(k + 2) \]

### Step 2: Analyze the factors of A and B

Now we have:
- \( A = (k - 1)(k + 1) \)
- \( B = k(k + 2) \)

### Step 3: Find the common divisors

Next, we need to determine if there are any prime numbers that divide both \( A \) and \( B \) for at least one value of \( k \).

1. **When \( k \) is odd**:
- \( k - 1 \) is even, \( k + 1 \) is even, hence \( A \) is even.
- \( k \) is odd, \( k + 2 \) is odd, hence \( B \) is odd.
- Therefore, when \( k \) is odd, \( A \) is even and \( B \) is odd. They have no common prime factors.

2. **When \( k \) is even**:
- \( k - 1 \) is odd, \( k + 1 \) is odd, hence \( A \) is odd.
- \( k \) is even, \( k + 2 \) is even, hence \( B \) is even.
- Therefore, when \( k \) is even, \( A \) is odd and \( B \) is even. They have no common prime factors.

### Step 4: Conclusion

Since in both cases (whether \( k \) is odd or even), \( A \) and \( B \) do not share any common prime factors, we conclude that:

**The number of prime numbers that divide both \( A \) and \( B \) for at least one value of \( k \) is zero.**

### Final Answer:
0