2008 AIME I 第 7 题

Consecutive squares differ by $2a + 1,$ which is less than $100$ for $a \le 49,$ so every set up through the one containing $50^2 = 2500$ has a square

Past $2500$ the gaps exceed $100,$ so each remaining set holds at most one square; count the squares up to $99999$ and subtract

Consecutive squares $a^2$ and $(a + 1)^2$ differ by $2a + 1 \le 99$ for $a \le 49,$ so the squares from $1^2$ to $50^2 = 2500$ never skip a hundred-block: every set $S_0, S_1, \ldots, S_{25}$ contains a perfect square. For $a \ge 50$ the gap $2a + 1 \ge 101$ exceeds $100,$ so each of the sets $S_{26}, \ldots, S_{999}$ contains at most one square.

The largest number involved is $99999,$ and $316^2 = 99856 \le 99999 \lt 317^2.$ So the squares landing in $S_{26}, \ldots, S_{999}$ are $51^2, 52^2, \ldots, 316^2$ — that is, $266$ squares occupying $266$ distinct sets out of those $974.$

Therefore $974 - 266 = 708$ sets contain no perfect square.