Question

如果一个三项严格递增的整数等差数列的三项平方和等于中项与公差平方的乘积，就称它为特殊数列。求所有特殊数列第三项的和。

Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

答案

答案：31

知识点：等差数列丢番图方程整除性

难度评级：2390

解答：

将三项写成 $a - d$ 、 $a$ 、 $a + d$ ，其中整数 $d \ge 1$ 。条件为所以 $d^2(a - 2) = 3a^2$ ，即 $d^2 = \frac{3a^2}{a - 2}$ 。为了使 $d^2$ 为正，需要 $a \gt 2$ （若 $a = 0$ ，则 $d = 0$ ，不严格递增；而 $a \lt 2$ 或 $a = 2$ 会使右边在可检查的情形中为负或非整数）。 $\begin{aligned} &(a-d)^2 + a^2 \\ &\quad {}+ (a+d)^2 = ad^2 \\ &\quad\Longleftrightarrow\quad 3a^2 + 2d^2 = ad^2, \end{aligned}$

令 $t = a - 2 \ge 1$ ，得到所以 $t \mid 12$ 。检验 $t = 1, 2, 3, 4, 6, 12$ ，得到 $d^2 = 27, 24, 25, 27, 32, 49$ ：只有 $t = 3$ 与 $t = 12$ 给出完全平方数。 $d^2 = \frac{3(t+2)^2}{t} = 3t + 12 + \frac{12}{t},$

它们分别给出 $(a, d) = (5, 5)$ ，数列为 $0, 5, 10$ ，以及 $(a, d) = (14, 7)$ ，数列为 $7, 14, 21$ 。第三项之和为 $10 + 21 = 31$ 。

Write the terms as $a - d,$ $a,$ $a + d$ with integer $d \ge 1.$ The condition is $\begin{aligned} &(a-d)^2 + a^2 \\ &\quad {}+ (a+d)^2 = ad^2 \\ &\quad\Longleftrightarrow\quad 3a^2 + 2d^2 = ad^2, \end{aligned}$ so $d^2(a - 2) = 3a^2$ and $d^2 = \frac{3a^2}{a - 2}.$ For $d^2$ to be positive we need $a \gt 2.$ Indeed, $a = 0$ forces $d = 0,$ while for any nonzero integer $a \lt 2$ the right side is negative; $a = 2$ makes the original equation impossible.

Substituting $t = a - 2 \ge 1$ gives $d^2 = \frac{3(t+2)^2}{t} = 3t + 12 + \frac{12}{t},$ so $t \mid 12.$ Testing $t = 1, 2, 3, 4, 6, 12$ gives $d^2 = 27, 24, 25, 27, 32, 49:$ only $t = 3$ and $t = 12$ yield perfect squares.

These give $(a, d) = (5, 5)$ with sequence $0, 5, 10,$ and $(a, d) = (14, 7)$ with sequence $7, 14, 21.$ The sum of the third terms is $10 + 21 = 31.$

2021 AIME I 第 5 题

答案：31

难度评级：2390

解答：

其他年份的第 5 题