Question

For how many ordered pairs $(x, y)$ of integers is it true that $0 \lt x \lt y \lt 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y?$

答案

Solution:

The condition is $\frac{x + y}{2} = \sqrt{xy} + 2,$ that is, $x + y - 2\sqrt{xy} = 4,$ so $(\sqrt{y} - \sqrt{x})^2 = 4$ and (as $y \gt x$ ) $\sqrt{y} - \sqrt{x} = 2.$ Note $\sqrt{xy} = \frac{x + y - 4}{2}$ is rational, hence $\sqrt{y} + \sqrt{x} = \frac{y - x}{\sqrt{y} - \sqrt{x}} = \frac{y - x}{2}$ is rational too, so $\sqrt{x}$ and $\sqrt{y}$ are rational — and a rational square root of an integer is an integer.

Therefore $x = a^2$ and $y = (a + 2)^2$ for a positive integer $a.$ The constraint $y \lt 10^6$ means $a + 2 \le 999,$ so $a$ ranges over $1, 2, \ldots, 997,$ and each value gives a valid pair.

Hence there are $997$ ordered pairs.

2000 AIME I Problem 6

Solution:

Problem 6 in Other Years