2000 AIME I Problems/Problem 6

Problem

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

Solution

From the condition given,

$ $\begin{aligned} \frac{x + y}{2} - 2 & = \sqrt{x y} \\ x + y - 2 \sqrt{x y} & = 4 \\ (\sqrt{y} - \sqrt{x})^{2} & = 4 \\ \sqrt{y} - \sqrt{x} & = 2 \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

The last equation is true because $y > x$ .

Here, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ satisfy our equation, rather than $(x,y)$ directly, because $(x,y)$ can get messy.

The maximum that $\sqrt{y}$ can be is $10^3 - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$ , an integer). Then $\sqrt{x} = 997$ , and we continue this downward until $\sqrt{y} = 3$ , in which case $\sqrt{x} = 1$ . The number of pairs of $(\sqrt{x},\sqrt{y})$ , and so $(x,y)$ is, then, $999 - 3 + 1 = \boxed{997}$ .

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2000 AIME I Problems/Problem 6

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