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 $x = \frac{a + b}{2}$ and $y = \frac{b - a}{2}$. For $x$ and $y$ to be integers, $a + b$ and $b - a$ must both be even, so $a$ and $b$ must have the same parity.

["Solving $x^2 - y^2 = 2025$: A Complete Guide to Integer Solutions", "The equation $x^2 - y^2 = 2025$ represents a classic difference of squares and offers a clear path to finding all integer solutions using factorization. By transforming the equation into a product of two integers, we unlock elegant and systematic methods for solving this Diophantine equation.", "---", "### Understanding the Equation", "Start with the identity:\n$$\nx^2 - y^2 = (x - y)(x + y)\n$$\nSo,\n$$\n(x - y)(x + y) = 2025\n$$", "Let:\n$$\na = x - y, \quad b = x + y\n$$\nThen $ab = 2025$, and solving for $x$ and $y$ gives:\n$$\nx = \frac{a + b}{2}, \quad y = \frac{b - a}{2}\n$$", "---", "### Ensuring Integer Solutions", "For $x$ and $y$ to be integers, both $a + b$ and $b - a$ must be even. This happens only when $a$ and $b$ share the same parity — that is, both are odd or both are even.", "Since $2025$ is an odd number, all its factor pairs $(a, b)$ must consist of odd integers only (because odd × odd = odd). Therefore, every factor pair of $2025$ consists of odd numbers, ensuring $a + b$ and $b - a$ are even — and thus $x$ and $y$ are integers.", "---", "### Finding All Factor Pairs of 2025", "First, factor $2025$:\n$$\n2025 = 3^4 \ imes 5^2\n$$", "The number of positive divisors is $(4+1)(2+1) = 15$, so there are 15 positive factor pairs $(a, b)$ such that $ab = 2025$. Each pair satisfies $a \leq b$ and both $a$, $b$ positive.", "Since $a$ and $b$ are both odd, all 15 positive pairs yield integer solutions.", "But we must also consider negative factor pairs, because $a$ and $b$ can both be negative (since $(-a)(-b) = 2025$). For each positive pair $(a, b)$, the negative counterpart $(-a, -b)$ also works.", "Thus, the full set of integer solutions includes:\n- 15 solutions from positive factor pairs $(a, b)$\n- 15 solutions from negative factor pairs $(-a, -b)$", "Total: $30$ integer solutions $(x, y)$", "---", "### Computing $x$ and $y$ from Factor Pairs", "For each factor pair $(a, b)$ of $2025$ (positive or negative), compute:\n$$\nx = \frac{a + b}{2}, \quad y = \frac{b - a}{2}\n$$", "Because $a$ and $b$ are both odd and of the same parity, $a + b$ and $b - a$ are even, so $x, y \in \mathbb{Z}$.", "Example: Take $a = 1$, $b = 2025$\n$$\nx = \frac{1 + 2025}{2} = 1013, \quad y = \frac{2025 - 1}{2} = 1012\n$$", "Check: $1013^2 - 1012^2 = (1013 - 1012)(1013 + 1012) = 1 \ imes 2025 = 2025$ ✓", "Similarly, the negative pair $(-1, -2025)$ gives $x = -1013$, $y = -1012$, which also satisfies the equation.", "---", "### Summary", "Solving $x^2 - y^2 = 2025$ reduces to iterating over all factor pairs $(a, b)$ of $2025$, computing integer solutions via:\n$$\nx = \frac{a + b}{2}, \quad y = \frac{b - a}{2}\n$$\nAll 30 integer solutions come from the 15 positive and 15 negative factor pairs, all valid due to the oddness of $a$ and $b$.", "This method combines number theory with algebraic manipulation to efficiently find all integer solutions — a powerful technique for Diophantine equations of this form.", "---", "Key takeaways:\n- Factor $2025$ into all pairs $(a, b)$ such that $ab = 2025$\n- Use $x = \frac{a + b}{2}$, $y = \frac{b - a}{2}$\n- All 30 sign combinations (positive/negative factor pairs) yield valid integer solutions\n- Result: $x^2 - y^2 = 2025$ has exactly 30 integer solutions", "---", "### Further Reading\n- The method of factorization in Diophantine equations\n- Pythagorean and hyperbolic integer solutions\n- Exploring symmetric forms of integer equations", "---", "Optimize your problem-solving with factorization techniques — unlock infinite possibilities in number theory!"]