= 3^4 \cdot 5^2

February 22, 2026

$= 3^4 \cdot 5^2$

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Solution: We are tasked with finding the number of integer solutions $(x, y)$ to the equation $ x^2 - y^2 = 2025 $. This can be factored as:
Let $ a = x - y $ and $ b = x + y $. Then $ ab = 2025 $, and $ x = \frac{a + b}{2} $, $ y = \frac{b - a}{2} $. For $ x $ and $ y $ to be integers, $ a + b $ and $ b - a $ must both be even, which implies $ a $ and $ b $ must be both even or both odd.
Now, factor 2025:
The number of positive divisors is $ (4+1)(2+1) = 15 $, so there are 15 positive divisor pairs $ (a, b) $ with $ a > 0, b > 0 $, and another 15 with $ a < 0, b < 0 $, giving 30 total integer divisor pairs (since $ a $ and $ b $ can both be negative).
But we only count pairs where $ a $ and $ b $ have the same parity. Since 2025 is odd, all its divisors are odd. Therefore, $ a $ and $ b $ are both odd in every factor pair, and the sum and difference are even. So all 30 pairs yield integer $ (x, y) $.
However, each solution $ (x, y) $ corresponds to a unique pair $ (a, b) $, and since $ x $ and $ y $ are determined uniquely, and the hyperbola is symmetric, we must check for duplicates. But since $ (a, b) $ and $ (b, a) $ would give different $ x $ and $ y $, and both are included, all 30 pairs produce valid, distinct lattice points.

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