Correct Option: C

Detailed Breakdown:

The given equation is \(x^2 - y^2 = 2024\).

We can factor the left side using the difference of squares identity: \((x-y)(x+y) = 2024\).

Let \(A = x-y\) and \(B = x+y\).

So, we have \(AB = 2024\).

Since \(x\) and \(y\) are positive integers, the following conditions must hold for \(A\) and \(B\):

1. \(x = \frac{A+B}{2}\) and \(y = \frac{B-A}{2}\). For \(x\) and \(y\) to be integers, \(A+B\) and \(B-A\) must both be even. This implies that \(A\) and \(B\) must have the same parity (both even or both odd).

2. Since their product \(AB = 2024\) (an even number), both \(A\) and \(B\) must be even.

3. Since \(y\) is a positive integer, \(y > 0\), which means \(B-A > 0\), so \(B > A\).

4. Since \(x\) is a positive integer, \(x > 0\). Also, since \(y\) is positive, \(x-y

Now, we need to find the pairs of even factors \((A, B)\) of 2024 such that \(A First, find the prime factorization of 2024:\(2024 = 2 \times 1012 = 2^2 \times 506 = 2^3 \times 253\)To check for factors of 253, we can try prime numbers. Not divisible by 3, 5, 7. \(253 \div 11 = 23\).So, \(2024 = 2^3 \times 11 \times 23\).Let's list all factors of 2024 and then identify pairs \((A,B)\) that satisfy the conditions:Factors of 2024: 1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024.We are looking for pairs \((A, B)\) where \(A \cdot B = 2024\), \(A 1. If \(A=1\), \(B=2024\). \(A\) is odd. (Invalid)2. If \(A=2\), \(B=1012\). Both are even. (Valid) \(x = \frac{2+1012}{2} = 507\), \(y = \frac{1012-2}{2} = 505\). This is a valid pair \((507, 505)\).3. If \(A=4\), \(B=506\). Both are even. (Valid) \(x = \frac{4+506}{2} = 255\), \(y = \frac{506-4}{2} = 251\). This is a valid pair \((255, 251)\).4. If \(A=8\), \(B=253\). \(B\) is odd. (Invalid)5. If \(A=11\), \(B=184\). \(A\) is odd. (Invalid)6. If \(A=22\), \(B=92\). Both are even. (Valid) \(x = \frac{22+92}{2} = 57\), \(y = \frac{92-22}{2} = 35\). This is a valid pair \((57, 35)\).7. If \(A=23\), \(B=88\). \(A\) is odd. (Invalid)8. If \(A=44\), \(B=46\). Both are even. (Valid) \(x = \frac{44+46}{2} = 45\), \(y = \frac{46-44}{2} = 1\). This is a valid pair \((45, 1)\).

Any further pairs where \(A\) is larger than 44 would result in \(A \ge B\) (e.g., \(A=46, B=44\) or greater). We only consider \(A We found 4 pairs of positive integers \((x, y)\) that satisfy the given condition.

The final answer is \(\boxed{C}\)