Solution: The equation $ x^2 - y^2 = 2025 $ factors as $ (x - y)(x + y) = 2025 $. Since $ x $ and $ y $ are integers, both $ x - y $ and $ x + y $ must be integers. Let $ a = x - y $ and $ b = x + y $, so $ ab = 2025 $. Then $ x = raca + b2 $ and $ y = racb - a2 $. For $ x $ and $ y $ to be integers, $ a + b $ and $ b - a $ must both be even, meaning $ a $ and $ b $ must have the same parity. Since $ 2025 = 3^4 \cdot 5^2 $, it has $ (4+1)(2+1) = 15 $ positive divisors. Each pair $ (a, b) $ such that $ ab = 2025 $ gives a solution, but only those with $ a $ and $ b $ of the same parity are valid. Since 2025 is odd, all its divisors are odd, so $ a $ and $ b $ are both odd, ensuring $ x $ and $ y $ are integers. Each positive divisor pair $ (a, b) $ with $ a \leq b $ gives a unique solution, and since 2025 is a perfect square, there is one square root pair. There are 15 positive divisors, so 15 such factorizations, but only those with $ a \leq b $ are distinct under sign and order. Considering both positive and negative factor pairs, each valid $ (a,b) $ with $ a - Nelissen Grade advocaten