2014 中国数学奥林匹克（CMO）

首先理解集合 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的定义：它是 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 的所有因子对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}$（其中 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{b}$）之差 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}$ 构成的集合。我们需要找到 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 个数 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \dots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{n}_{\htmlData{tutor-start=17,tutor-end=18}{k}}$，使得它们的差集中至少有两个公共元素，设为 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{y}$。

若 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{)}$，则存在 $\htmlData{tutor-start=0,tutor-end=1}{b}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{d}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{n}$，即 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{b}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{d}$。这意味着 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 可以表示为关于 $\htmlData{tutor-start=0,tutor-end=1}{b}$ 的二次多项式值。 为了使多个 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 共享同一个差值 $\htmlData{tutor-start=0,tutor-end=1}{d}$，我们可以固定 $\htmlData{tutor-start=0,tutor-end=1}{d}$，并让每个 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 对应不同的起始因子 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$。具体地，若取 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{b}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{d}\htmlData{tutor-start=21,tutor-end=22}{)}$，则显然 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}$。

题目要求交集大小至少为 2，因此我们需要两个固定的差值 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{y}$。这提示我们构造形如 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{b}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{c}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{c}_{\htmlData{tutor-start=34,tutor-end=35}{i}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{y}\htmlData{tutor-start=38,tutor-end=39}{)}$ 的数。也就是说，对于每个 $\htmlData{tutor-start=0,tutor-end=1}{i}$，数 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 既是某个数与其加 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 后的乘积，也是某个数与其加 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 后的乘积。

我们需要找到无穷多组正整数 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{)}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{)}$。整理该方程： $\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{b} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{c}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{y}\htmlData{tutor-start=22,tutor-end=23}{c}$ 两边乘以 4 并配方： $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{b}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{b} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{x}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{4}\htmlData{tutor-start=32,tutor-end=33}{c}^{\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{+} \htmlData{tutor-start=40,tutor-end=41}{4}\htmlData{tutor-start=41,tutor-end=42}{y}\htmlData{tutor-start=42,tutor-end=43}{c} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{y}^{\htmlData{tutor-start=49,tutor-end=50}{2}} \htmlData{tutor-start=52,tutor-end=53}{-} \htmlData{tutor-start=54,tutor-end=55}{y}^{\htmlData{tutor-start=57,tutor-end=58}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{y}\htmlData{tutor-start=26,tutor-end=27}{)}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{y}^{\htmlData{tutor-start=37,tutor-end=38}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{c}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{y}^{\htmlData{tutor-start=37,tutor-end=38}{2}}$ 令 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{Y} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{c}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{K} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{-} \htmlData{tutor-start=32,tutor-end=33}{y}^{\htmlData{tutor-start=35,tutor-end=36}{2}}$。方程化为双曲线方程： $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{K}$ 只要 $\htmlData{tutor-start=0,tutor-end=1}{K} \neq \htmlData{tutor-start=7,tutor-end=8}{0}$（即 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{y}$），该方程就有无穷多组整数解。例如，取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}$，则 $\htmlData{tutor-start=0,tutor-end=1}{K} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{2}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}$。方程变为 $\htmlData{tutor-start=0,tutor-end=1}{Y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{X}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{3}$，即 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{Y}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{X}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{Y}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{X}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{3}$。虽然这个特例只有有限解，但我们可以选择其他 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 或者利用更一般的佩尔型方程理论。

更简单地，我们可以直接选取特殊的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 使得方程退化为线性或有明显参数解。或者，注意到我们只需要 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 个解，不需要无穷多。但为了严谨，我们说明对于一般的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$，只要 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}$ 不是某些特殊形式，或者通过调整 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 的取值（例如取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{m}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{m}$ 等技巧），总能找到足够多的解。

实际上，有一个更直接的构造法：取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。我们需要 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{)}$。这看起来像三角数问题。让我们换一个思路，利用恒等式。 考虑 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{t}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{t}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{t}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)}$？不，太复杂。 回到 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{y}^{\htmlData{tutor-start=27,tutor-end=28}{2}}$。如果我们取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{8}$。解为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{.}\htmlData{tutor-start=15,tutor-end=16}{.}$ 有限。 我们需要 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}$ 有足够多的因子分解，或者使用佩尔方程 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{d}\htmlData{tutor-start=9,tutor-end=10}{Y}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{N}$ 的形式。

修正策略：不要固定 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 为小整数。取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{u}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{u}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{u}$ 这种参数化可能太复杂。 让我们使用一个已知的恒等式构造。取 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{m}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{m}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{m}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{1}$？这是平方数。

最稳妥的路径：取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。我们要找 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{)}$。这等价于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{y}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{y}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \implies \htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{c}\htmlData{tutor-start=48,tutor-end=49}{+}\htmlData{tutor-start=49,tutor-end=50}{y}\htmlData{tutor-start=50,tutor-end=51}{)}^{\htmlData{tutor-start=53,tutor-end=54}{2}} \htmlData{tutor-start=56,tutor-end=57}{-} \htmlData{tutor-start=58,tutor-end=59}{(}\htmlData{tutor-start=59,tutor-end=60}{2}\htmlData{tutor-start=60,tutor-end=61}{b}\htmlData{tutor-start=61,tutor-end=62}{+}\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{)}^{\htmlData{tutor-start=66,tutor-end=67}{2}} \htmlData{tutor-start=69,tutor-end=70}{=} \htmlData{tutor-start=71,tutor-end=72}{y}^{\htmlData{tutor-start=74,tutor-end=75}{2}}\htmlData{tutor-start=76,tutor-end=77}{-}\htmlData{tutor-start=77,tutor-end=78}{1}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{Y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{X}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}$。$\htmlData{tutor-start=0,tutor-end=1}{Y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{X}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{y}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}$。 如果我们取 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 有很多因子，或者取 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 为变量？不，$\htmlData{tutor-start=0,tutor-end=1}{y}$ 必须是常数。

让我们重新审视 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{K}$。如果 $\htmlData{tutor-start=0,tutor-end=1}{K}$ 是奇数且非平方数，或者 $\htmlData{tutor-start=0,tutor-end=1}{K}$ 是 4 的倍数且非平方数... 等等，$\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{Y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{K}$ 的解数取决于 $\htmlData{tutor-start=0,tutor-end=1}{K}$ 的因子个数。这对于固定的 $\htmlData{tutor-start=0,tutor-end=1}{K}$ 是有限的！ **关键发现**：对于固定的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$，方程 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{)}$ 只有有限组解。这意味着我们不能仅靠固定一对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)}$ 来生成任意大的 $\htmlData{tutor-start=0,tutor-end=1}{k}$。

**修正构造**：我们需要 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{n}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=15}{\cap }\dots \htmlData{tutor-start=21,tutor-end=26}{\cap }\htmlData{tutor-start=26,tutor-end=27}{D}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{n}_{\htmlData{tutor-start=31,tutor-end=32}{k}}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{|} \htmlData{tutor-start=36,tutor-end=40}{\ge }\htmlData{tutor-start=40,tutor-end=41}{2}$。这两个公共元素不必对所有 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 都是同一对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)}$ 吗？是的，交集定义如此。 那么，是否存在一对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{)}$ 有无穷多解？ 方程 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{c}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{y}^{\htmlData{tutor-start=35,tutor-end=36}{2}}$。左边是平方差。右边是常数。平方差等于常数的解确实只有有限组。

**难道题目理解有误？** 再读题：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{n}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=15}{\cap }\dots \htmlData{tutor-start=21,tutor-end=26}{\cap }\htmlData{tutor-start=26,tutor-end=27}{D}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{n}_{\htmlData{tutor-start=31,tutor-end=32}{k}}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{|} \htmlData{tutor-start=36,tutor-end=40}{\ge }\htmlData{tutor-start=40,tutor-end=41}{2}$。是的，必须有两个公共元素。 这说明我的“固定 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 反推 $\htmlData{tutor-start=0,tutor-end=1}{n}$”的思路在“无穷多解”上卡住了。必须换一种构造 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 的方式，使得 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 天然拥有多个差值。

**新策略**：利用倍率关系。若 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{)}$，则对于任意 $\htmlData{tutor-start=0,tutor-end=1}{m}$，是否有 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{d} \htmlData{tutor-start=3,tutor-end=7}{\in }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{m}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{)}$？ 设 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{d}$。则 $\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{n} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{b}\htmlData{tutor-start=17,tutor-end=18}{)}$，且 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{m}\htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{m}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{m}\htmlData{tutor-start=18,tutor-end=19}{d}$。所以 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{d} \htmlData{tutor-start=3,tutor-end=7}{\in }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{m}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{)}$。 这告诉我们，如果有一组数 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \dots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{n}_{\htmlData{tutor-start=17,tutor-end=18}{k}}$ 共享差值集合 $\htmlData{tutor-start=0,tutor-end=1}{S}$，那么 $\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{n}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{,} \dots\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{m}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{n}_{\htmlData{tutor-start=29,tutor-end=30}{k}}$ 共享差值集合 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{S} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=7}{\{}\htmlData{tutor-start=7,tutor-end=8}{m}\htmlData{tutor-start=8,tutor-end=9}{s} \htmlData{tutor-start=10,tutor-end=15}{\mid }\htmlData{tutor-start=15,tutor-end=16}{s} \htmlData{tutor-start=17,tutor-end=21}{\in }\htmlData{tutor-start=21,tutor-end=22}{S}\htmlData{tutor-start=22,tutor-end=24}{\}}$。 但这并没有增加交集的大小，只是放大了数值。

**再次修正**：我们需要找到一组数，它们本身就有很多共同的差。考虑 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 为完全平方数的情况？若 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{s}^{\htmlData{tutor-start=5,tutor-end=6}{2}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{s}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0} \notin \htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{)}$（因为 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{b}$）。但 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 接近平方数时差值很小。 考虑 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{j}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)}$ 这类连续整数乘积？ $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{j}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{j}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{j}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{)}$。令 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{j}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{j}$，则 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}$。所以 $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{)}$。 同时 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{j}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{j}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{1}$。这不是因子差的形式。 但注意 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{j}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=20}{\times }\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{j}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{j}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{)}$？不，$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{n}$。 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{j}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{j}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{]} \htmlData{tutor-start=13,tutor-end=19}{\cdot }\htmlData{tutor-start=19,tutor-end=20}{[}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{j}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{j}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{]}$。令 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{j}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{j}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}$, $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{j}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{j}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{j}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{j}$。则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$。所以对任意 $\htmlData{tutor-start=0,tutor-end=1}{j}$， $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}_{\htmlData{tutor-start=11,tutor-end=12}{j}}\htmlData{tutor-start=13,tutor-end=14}{)}$。 现在我们需要第二个公共差值。观察 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{j}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)}$。 还可以写成 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{[}\htmlData{tutor-start=9,tutor-end=10}{j}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{j}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{]} \htmlData{tutor-start=17,tutor-end=23}{\cdot }\htmlData{tutor-start=23,tutor-end=24}{[}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{j}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{j}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{]}$？不，$\htmlData{tutor-start=0,tutor-end=1}{j}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{j}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{j}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{)}$。差值为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{j}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{j}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{4}\htmlData{tutor-start=26,tutor-end=27}{j}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{6}$。随 $\htmlData{tutor-start=0,tutor-end=1}{j}$ 变化，不是公共的。 还可以写成 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{[}\htmlData{tutor-start=9,tutor-end=10}{j}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{j}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{]} \htmlData{tutor-start=17,tutor-end=23}{\cdot }\htmlData{tutor-start=23,tutor-end=24}{[}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{j}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{j}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{]}$？$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{j}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{j}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{j}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{j}$。差值 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{j}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}$。随 $\htmlData{tutor-start=0,tutor-end=1}{j}$ 变化。

看来 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{j}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)}$ 只能保证 $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}_{\htmlData{tutor-start=11,tutor-end=12}{j}}\htmlData{tutor-start=13,tutor-end=14}{)}$。我们需要另一个固定的差。 让我们尝试 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{j}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{j}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{)}$？ $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{j}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{j}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{j}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{j}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{)} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{j}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{j}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{2}$。差值 $\htmlData{tutor-start=0,tutor-end=1}{2}$。依然只有 2。

**突破点**：考虑 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 具有多种对称分解。例如 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{y}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{m}\htmlData{tutor-start=18,tutor-end=19}{)}$。前面证明了固定 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{m}$ 解有限。但如果我们不固定 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{m}$，而是让 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 的结构本身蕴含多个差呢？ 回顾 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}$ 给出差 2。如果 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 还能写成 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{L}\htmlData{tutor-start=5,tutor-end=6}{)}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{L}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{j}$ 无关？ 刚才试了 $\htmlData{tutor-start=0,tutor-end=1}{j}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{j}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{j}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{j}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{)}$，只找到了 $\htmlData{tutor-start=0,tutor-end=1}{L}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$。 试试 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{j}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{j}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{j}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}^{\htmlData{tutor-start=29,tutor-end=30}{2}}$？太大。

让我们回到 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{K}$ 的局限性。是否有可能 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 不是常数，而是依赖于 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 的某种结构，但在交集中表现为常数？不，交集元素必须是具体的整数。

**正确构造方向**：利用 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{m}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{d}^{\htmlData{tutor-start=15,tutor-end=16}{2}}$ 的形式？不，$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{d} \implies \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{b}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{d}\htmlData{tutor-start=25,tutor-end=26}{/}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{)}^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{d}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{)}^{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{n}$（若 $\htmlData{tutor-start=0,tutor-end=1}{d}$ 偶）。即 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{d}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{d}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}}$ 是完全平方数。 所以 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{)} \iff \htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{d}^{\htmlData{tutor-start=22,tutor-end=23}{2}}$ 是完全平方数。 我们要找 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \dots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{n}_{\htmlData{tutor-start=17,tutor-end=18}{k}}$ 和两个整数 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$，使得 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}_{\htmlData{tutor-start=4,tutor-end=5}{i}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{u}^{\htmlData{tutor-start=12,tutor-end=13}{2}}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}_{\htmlData{tutor-start=4,tutor-end=5}{i}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{v}^{\htmlData{tutor-start=12,tutor-end=13}{2}}$ 都是完全平方数。 设 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}_{\htmlData{tutor-start=4,tutor-end=5}{i}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{u}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{X}_{\htmlData{tutor-start=20,tutor-end=21}{i}}^{\htmlData{tutor-start=24,tutor-end=25}{2}}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}_{\htmlData{tutor-start=4,tutor-end=5}{i}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{v}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{Y}_{\htmlData{tutor-start=20,tutor-end=21}{i}}^{\htmlData{tutor-start=24,tutor-end=25}{2}}$。 相减得 $\htmlData{tutor-start=0,tutor-end=1}{X}_{\htmlData{tutor-start=3,tutor-end=4}{i}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{Y}_{\htmlData{tutor-start=15,tutor-end=16}{i}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{u}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{-} \htmlData{tutor-start=32,tutor-end=33}{v}^{\htmlData{tutor-start=35,tutor-end=36}{2}}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{u}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{v}^{\htmlData{tutor-start=15,tutor-end=16}{2}}$（常数）。我们需要 $\htmlData{tutor-start=0,tutor-end=1}{X}_{\htmlData{tutor-start=3,tutor-end=4}{i}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{Y}_{\htmlData{tutor-start=15,tutor-end=16}{i}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{C}$ 有 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 组解 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{X}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{Y}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}$，且对应的 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{X}_{\htmlData{tutor-start=12,tutor-end=13}{i}}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{u}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{4}$ 是正整数且互不相同。 前面说过 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}$ 只有有限解。**这似乎是个死胡同。**

**重新思考**：是不是 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$ 可以不同？不，交集要求公共元素固定。 难道 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{d}^{\htmlData{tutor-start=6,tutor-end=7}{2}}$ 是完全平方数这个等价条件用错了？ $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{d}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{n} \implies \htmlData{tutor-start=21,tutor-end=22}{a}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{b}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{d} \implies \htmlData{tutor-start=36,tutor-end=37}{b}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{b}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{d}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{n} \implies \htmlData{tutor-start=54,tutor-end=55}{4}\htmlData{tutor-start=55,tutor-end=56}{b}^{\htmlData{tutor-start=58,tutor-end=59}{2}}\htmlData{tutor-start=60,tutor-end=61}{+}\htmlData{tutor-start=61,tutor-end=62}{4}\htmlData{tutor-start=62,tutor-end=63}{b}\htmlData{tutor-start=63,tutor-end=64}{d}\htmlData{tutor-start=64,tutor-end=65}{=}\htmlData{tutor-start=65,tutor-end=66}{4}\htmlData{tutor-start=66,tutor-end=67}{n} \implies \htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{2}\htmlData{tutor-start=79,tutor-end=80}{b}\htmlData{tutor-start=80,tutor-end=81}{+}\htmlData{tutor-start=81,tutor-end=82}{d}\htmlData{tutor-start=82,tutor-end=83}{)}^{\htmlData{tutor-start=85,tutor-end=86}{2}} \htmlData{tutor-start=88,tutor-end=89}{-} \htmlData{tutor-start=90,tutor-end=91}{d}^{\htmlData{tutor-start=93,tutor-end=94}{2}} \htmlData{tutor-start=96,tutor-end=97}{=} \htmlData{tutor-start=98,tutor-end=99}{4}\htmlData{tutor-start=99,tutor-end=100}{n}$。 没错。所以 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{)} \iff \htmlData{tutor-start=16,tutor-end=24}{\exists }\htmlData{tutor-start=24,tutor-end=25}{b} \htmlData{tutor-start=26,tutor-end=30}{\in }\mathbb{\htmlData{tutor-start=38,tutor-end=39}{N}}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{b}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{d}\htmlData{tutor-start=47,tutor-end=48}{)}^{\htmlData{tutor-start=50,tutor-end=51}{2}} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{4}\htmlData{tutor-start=56,tutor-end=57}{n}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{d}^{\htmlData{tutor-start=61,tutor-end=62}{2}}$。 这意味着 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{d}^{\htmlData{tutor-start=6,tutor-end=7}{2}}$ 必须是一个与 $\htmlData{tutor-start=0,tutor-end=1}{d}$ 同奇偶的完全平方数（因为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{d}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{d}$ 同奇偶）。

如果 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}$ 只有有限解，那怎么可能有任意大的 $\htmlData{tutor-start=0,tutor-end=1}{k}$？ 除非... $\htmlData{tutor-start=0,tutor-end=1}{C}$ 可以是 0？即 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{v}$，但这只能提供一个公共元素。 或者，我的“有限解”结论在什么情况下不成立？ $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}$。若 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$，则 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{Y}$，无穷多解。但对应 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{v}$，只有一个差值。 若 $\htmlData{tutor-start=0,tutor-end=1}{C} \neq \htmlData{tutor-start=7,tutor-end=8}{0}$，因子分解 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{Y}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{X}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{Y}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{C}$。因子对有限，解有限。

**这说明：不可能存在两个固定的整数 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$ ($\htmlData{tutor-start=0,tutor-end=1}{u} \neq \htmlData{tutor-start=7,tutor-end=8}{v}$) 使得有无穷多个 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 同时满足 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v} \htmlData{tutor-start=4,tutor-end=8}{\in }\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{)}$。** **等等，题目说的是“对于任意 $\htmlData{tutor-start=0,tutor-end=1}{k}$，存在 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{.}\htmlData{tutor-start=8,tutor-end=9}{n}_{\htmlData{tutor-start=11,tutor-end=12}{k}}$”。并没有说要有无穷多个这样的 $\htmlData{tutor-start=0,tutor-end=1}{n}$。只要有 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 个就够了。** 但是 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 是任意的。如果对于固定的 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$，最多只有 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 个 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 满足条件，那么当 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{M}$ 时就失败了。 因此，必须存在某对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}$ 使得解的数量无上界？或者，随着 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 增大，我们可以更换 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}$？ 不，题目是“Prove that for any integer $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$, there exists...”。量词顺序是 $\htmlData{tutor-start=0,tutor-end=8}{\forall }\htmlData{tutor-start=8,tutor-end=9}{k} \htmlData{tutor-start=10,tutor-end=18}{\exists }\htmlData{tutor-start=18,tutor-end=19}{n}_{\htmlData{tutor-start=21,tutor-end=22}{1}}\htmlData{tutor-start=23,tutor-end=24}{.}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{n}_{\htmlData{tutor-start=28,tutor-end=29}{k}}$。这意味着对于每个 $\htmlData{tutor-start=0,tutor-end=1}{k}$，我可以找不同的 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$。只要对于每个 $\htmlData{tutor-start=0,tutor-end=1}{k}$，我能找到一对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}_{\htmlData{tutor-start=4,tutor-end=5}{k}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{v}_{\htmlData{tutor-start=11,tutor-end=12}{k}}\htmlData{tutor-start=13,tutor-end=14}{)}$ 使得方程 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{Y}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{u}_{\htmlData{tutor-start=17,tutor-end=18}{k}}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{v}_{\htmlData{tutor-start=27,tutor-end=28}{k}}^{\htmlData{tutor-start=31,tutor-end=32}{2}}$ 至少有 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 组解即可。

**问题转化为**：对于任意 $\htmlData{tutor-start=0,tutor-end=1}{k}$，是否存在常数 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{k}}$，使得方程 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}_{\htmlData{tutor-start=19,tutor-end=20}{k}}$ 至少有 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 组正整数解 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{Y}\htmlData{tutor-start=4,tutor-end=5}{)}$，且这些解能还原出合法的 $\htmlData{tutor-start=0,tutor-end=1}{n}$？ 方程 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}$ 的解数等于 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的因子对数（满足同奇偶条件）。 要使解数任意大，只需取 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 有足够多的因子。例如取 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 为高合成数，或者 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}^{\htmlData{tutor-start=7,tutor-end=8}{m}} \htmlData{tutor-start=10,tutor-end=16}{\cdot }\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{m}} \dots$。 具体来说，若 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 有 $\htmlData{tutor-start=0,tutor-end=1}{T}$ 个满足条件的因子对，则有 $\htmlData{tutor-start=0,tutor-end=1}{T}$ 组解 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{Y}\htmlData{tutor-start=4,tutor-end=5}{)}$。 每组解 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{Y}\htmlData{tutor-start=4,tutor-end=5}{)}$ 对应一个 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{X}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{u}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{4}$。我们需要验证： 1. $\htmlData{tutor-start=0,tutor-end=1}{n}$ 是正整数。 2. 对应的 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{X}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}$ 是正整数。 3. $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 互不相同。

**构造细节**： 给定 $\htmlData{tutor-start=0,tutor-end=1}{k}$，选取整数 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{v}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{u}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{v}^{\htmlData{tutor-start=15,tutor-end=16}{2}}$ 拥有至少 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 组因子分解 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{d}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{d}_{\htmlData{tutor-start=13,tutor-end=14}{2}}$（其中 $d_{1} < d_{2}, d_{1} \equiv d_{2} \pmod 2$）。 由于 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{u}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{v}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{u}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{v}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{u}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{v}\htmlData{tutor-start=29,tutor-end=30}{)}$，我们可以先选定 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v}$ 都有大量因子，或者直接选 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 为高度合数，再反推 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{v}$。 更简单：取 $\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。则 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{u}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{1}$。我们需要 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 有至少 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 个因子对。 取 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 因子极多。例如 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}$ 都是高合成数附近的数。或者直接取 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{N}}$ 的倍数？ 注意 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{C}$ 要求 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{Y}$ 同奇偶，故 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 必须是奇数或 4 的倍数。 若取 $\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$（奇），则 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 必须为偶（使 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 为奇？不，$\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 若 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 偶则为奇；若 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 奇则为偶但不是4倍数？$\htmlData{tutor-start=0,tutor-end=1}{o}\htmlData{tutor-start=1,tutor-end=2}{d}\htmlData{tutor-start=2,tutor-end=3}{d}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{m}$。哦，$\htmlData{tutor-start=0,tutor-end=1}{u}$ 奇时 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 是 8 的倍数，符合条件）。 所以取 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 为奇数，$\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。$\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{u}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}$ 是 8 的倍数，必有解。 解的数量由 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 的因子数决定。因为因子数函数无界，所以对于任意 $\htmlData{tutor-start=0,tutor-end=1}{k}$，存在 $\htmlData{tutor-start=0,tutor-end=1}{u}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 的因子对数 $\htmlData{tutor-start=0,tutor-end=4}{\ge }\htmlData{tutor-start=4,tutor-end=5}{k}$。

**验证合法性**： 对于每一组因子对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{d}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{d}_{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}$ 满足 $d_{1} d_{2} = u^{2}-1, d_{1} < d_{2}, d_{1} \equiv d_{2} \pmod 2$： 令 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{d}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{d}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{Y} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{d}_{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{d}_{\htmlData{tutor-start=35,tutor-end=36}{1}}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{/}\htmlData{tutor-start=39,tutor-end=40}{2}$。 则 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{u}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{1}$。 我们需要 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{X}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{u}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{4}$ 为正整数，且 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{X}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}$ 为正整数。 由 $\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{Y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{u}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{1} \implies \htmlData{tutor-start=35,tutor-end=36}{X}^{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{-} \htmlData{tutor-start=43,tutor-end=44}{u}^{\htmlData{tutor-start=46,tutor-end=47}{2}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{Y}^{\htmlData{tutor-start=54,tutor-end=55}{2}} \htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=60}{1}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{Y}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{1} \implies \htmlData{tutor-start=24,tutor-end=25}{n} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{Y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{4}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{Y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{X}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{u}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}$。又 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{d}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{d}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2}$。由于 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{d}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}$，且 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{d}_{\htmlData{tutor-start=11,tutor-end=12}{2}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}} \htmlData{tutor-start=23,tutor-end=24}{<} \htmlData{tutor-start=25,tutor-end=26}{u} \htmlData{tutor-start=27,tutor-end=28}{<} \htmlData{tutor-start=29,tutor-end=30}{d}_{\htmlData{tutor-start=32,tutor-end=33}{2}}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{d}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{d}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{0} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{u}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{2}$? 不一定大于 $\htmlData{tutor-start=0,tutor-end=1}{u}$。 我们需要 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{u}$ 才能保证 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$。 $X > u \iff (d_{1}+d_{2})/2 > u \iff d_{1}+d_{2} > 2u$。 已知 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{d}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}$。考虑函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{t} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{u}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{t}$。在 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{<} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{u}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}}$ 时单调递减。 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 的因子且 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}}$。最小的因子是 1。 若 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{u}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}$。$\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{d}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{u}$（当 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2}$）。成立。 若 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 较大，接近 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{u}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{d}_{\htmlData{tutor-start=9,tutor-end=10}{2}}$ 接近 $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{u}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{u}$。此时 $\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{u}$，导致 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}$，不合题意。 **危机**：并非所有因子对都产生合法的 $\htmlData{tutor-start=0,tutor-end=1}{n}$。只有那些满足 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{d}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{u}$ 的因子对才有效。 我们需要确保有效的因子对数量仍然可以任意大。 条件 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{u}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{d}_{\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{>} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{u} \iff \htmlData{tutor-start=34,tutor-end=35}{d}_{\htmlData{tutor-start=37,tutor-end=38}{1}}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{-} \htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{u} \htmlData{tutor-start=49,tutor-end=50}{d}_{\htmlData{tutor-start=52,tutor-end=53}{1}} \htmlData{tutor-start=55,tutor-end=56}{+} \htmlData{tutor-start=57,tutor-end=58}{u}^{\htmlData{tutor-start=60,tutor-end=61}{2}} \htmlData{tutor-start=63,tutor-end=64}{-} \htmlData{tutor-start=65,tutor-end=66}{1} \htmlData{tutor-start=67,tutor-end=68}{>} \htmlData{tutor-start=69,tutor-end=70}{0} \iff \htmlData{tutor-start=76,tutor-end=77}{(}\htmlData{tutor-start=77,tutor-end=78}{d}_{\htmlData{tutor-start=80,tutor-end=81}{1}}\htmlData{tutor-start=82,tutor-end=83}{-}\htmlData{tutor-start=83,tutor-end=84}{u}\htmlData{tutor-start=84,tutor-end=85}{)}^{\htmlData{tutor-start=87,tutor-end=88}{2}} \htmlData{tutor-start=90,tutor-end=91}{>} \htmlData{tutor-start=92,tutor-end=93}{1} \iff \htmlData{tutor-start=99,tutor-end=100}{|}\htmlData{tutor-start=100,tutor-end=101}{d}_{\htmlData{tutor-start=103,tutor-end=104}{1}}\htmlData{tutor-start=105,tutor-end=106}{-}\htmlData{tutor-start=106,tutor-end=107}{u}\htmlData{tutor-start=107,tutor-end=108}{|} \htmlData{tutor-start=109,tutor-end=110}{>} \htmlData{tutor-start=111,tutor-end=112}{1}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}} \htmlData{tutor-start=23,tutor-end=24}{<} \htmlData{tutor-start=25,tutor-end=26}{u}$，所以需 $\htmlData{tutor-start=0,tutor-end=1}{u} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{d}_{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=27}{} > 1 \iff d_{1} < }\htmlData{tutor-start=27,tutor-end=28}{u}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}$。 也就是说，只要因子 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 不等于 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}$（以及可能的其他边界情况），就是合法的。 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}$ 确实是 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 的因子。它对应 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{u}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{d}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{u}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1}$。此时 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{u}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{Y}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=23}{ \implies n=0, }\htmlData{tutor-start=23,tutor-end=24}{b}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{0}$。非法。 除此之外，其他因子对是否都合法？ 若 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{u}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{d}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{)}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{>} \htmlData{tutor-start=16,tutor-end=17}{1}$ 成立。所以除了 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{u}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}$ 这一组外，其余所有满足 $\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{u}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}}$ 的因子对都产生合法的 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{b} \htmlData{tutor-start=9,tutor-end=10}{>} \htmlData{tutor-start=11,tutor-end=12}{0}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 的因子总数可以任意大，去掉一个非法解后，剩余合法解的数量依然可以任意大。 且不同的因子对 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{d}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{d}_{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}$ 产生不同的 $\htmlData{tutor-start=0,tutor-end=1}{Y}$，从而产生不同的 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{Y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{4}$。所以 $\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 互不相同。

综上，构造完成。

首先，由条件 i) 知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}$ 是唯一确定的。 对于 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$，代入条件 ii)： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{f}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{3} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{f}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{)} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{f}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{)} \htmlData{tutor-start=36,tutor-end=37}{+} \htmlData{tutor-start=38,tutor-end=39}{f}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{)} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{1} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{1} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{2}$ 此处 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}$，故右边两项的自变量分别为 $\htmlData{tutor-start=0,tutor-end=1}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{2}$，均小于 $\htmlData{tutor-start=0,tutor-end=1}{3}$ 且在定义域内，值唯一确定。 一般地，要使用数学归纳法证明唯一性，必须确保在计算 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{)}$ 时，等式右边出现的所有 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的下标 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 都严格小于 $\htmlData{tutor-start=0,tutor-end=1}{n}$。 观察递推式 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{f}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{n} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{f}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{)}$： 第一项自变量为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$。由于 $f: \mathbb{N}^* \to \mathbb{N}^*$，故 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=11}{\ge }\htmlData{tutor-start=11,tutor-end=12}{1}$。若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{n}$，则该项由归纳假设确定。 第二项自变量为 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{n}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}$。显然 $\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{n}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{n}$。又因 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=11}{\le }\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}$（这一点可由后续构造或简单归纳验证：若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{k}$ 对 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{n}$ 成立，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=11}{\le }\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}$，从而 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=13}{\ge }\htmlData{tutor-start=13,tutor-end=14}{2}$），故该下标也在 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{]}$ 范围内。 因此，每一步 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的值完全由前 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}$ 项唯一决定，不存在多解或循环依赖。