2000 AIME I Problems/Problem 6

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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,

$\begin{align*} \frac{x + y}{2} - 2 &= \sqrt{xy}\\ x + y - 2\sqrt{xy} &= 4\\ (\sqrt{y} - \sqrt{x})^2 &= 4\\ \sqrt{y} - \sqrt{x} &= 2 \end{align*}$ (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, $999 - 3 + 1 = \boxed{997}$.

See also

2000 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions