2008 AIME I 第 4 题

Compare $x^2 + 84x + 2008$ with $(x + 42)^2 = x^2 + 84x + 1764$

Write $y^2 - (x + 42)^2 = 244$ and factor: $244 = 2 \cdot 2 \cdot 61,$ and both factors of the difference of squares must be even

Completing the square, $x^2 + 84x + 2008$ $= (x + 42)^2 + 244,$ so $y^2 - (x + 42)^2 = 244,$ which factors as $$\begin{aligned} &(y - x - 42) \\ &\quad {}\cdot (y + x + 42) \\ &= 244 \\ &= 2^2 \cdot 61. \end{aligned}$$ The two factors have the same parity, and their product is even, so both are even: $y - x - 42 = 2$ and $y + x + 42 = 122.$

Adding gives $y = 62,$ and then $x = 18;$ indeed $18^2 + 84 \cdot 18 + 2008$ $= 3844$ $= 62^2.$ Therefore $x + y = 18 + 62 = 80.$