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Home » Discover the ‘Ampharos’ Secrets That Are Changing Archaeology Forever! - Baxtercollege

Discover the ‘Ampharos’ Secrets That Are Changing Archaeology Forever! - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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.

📌 We factor $2025 = 3^4 \cdot 5^2$, so the number of positive divisors is $(4+1)(2+1) = 15$. Each divisor pair $(a, b)$ corresponds to a solution, and since $a$ and $b$ must have the same parity, we count such pairs.

📌 Since $2025$ is odd, all its divisors are odd, so all pairs $(a, b)$ have the same parity. Thus, each of the 15 positive divisor pairs gives a solution. Including negative divisors (since $(-a)(-b) = 2025$), we double this count:

📌 \times 2 = 30

📌 Hence, the total number of lattice points is $\boxed{30}$.

📌 Question: What is the smallest three-digit number that is divisible by both $12$ and $15$ and leaves a remainder of $1$ when divided by $7$?

📌 Solution: A number divisible by both $12$ and $15$ must be divisible by their least common multiple:

📌 \text{lcm}(12, 15) = 60

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