2000 AIME II 第 2 题

Factor $x^2 - y^2 = (x - y)(x + y)$ and note the two factors have the same parity

Write $x - y = 2a$ and $x + y = 2b$ with $ab = 10^6;$ count ordered pairs, and remember negative values of $x$

Factor $(x - y)(x + y) = 2000^2$ $= 2^8 \cdot 5^6.$ The factors $x - y$ and $x + y$ have the same parity, and their product is even, so both must be even. Writing $x - y = 2a$ and $x + y = 2b$ gives $ab = 2^6 \cdot 5^6 = 10^6.$

Each ordered pair of positive integers $(a, b)$ with $ab = 10^6$ yields exactly one solution $x = a + b,$ $y = b - a$ with $x \gt 0,$ and $10^6$ has $7 \cdot 7 = 49$ divisors, hence $49$ such pairs. Replacing $(a, b)$ by $(-a, -b)$ gives the $49$ solutions with $x \lt 0,$ and $x = 0$ is impossible since $-y^2 \lt 2000^2.$

In total there are $49 + 49 = 98$ lattice points.