# Difference between revisions of "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

### Solution 1

$\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}$

Because $y > x$, we only consider $+2$.

For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.

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 $\boxed{997}$.

### Solution 2

Let $a^2$ = $x$ and $b^2$ = $y$

Then $$\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2$$ $$a^2 + b^2 = 2ab + 4$$ $$(a-b)^2 = 4$$ $$(a-b) = \pm 2$$

This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.

Because $\sqrt{10^6} = 10^3$, then we can use all positive integers less than 1000 for $a$ and $b$.

Without loss of generality, let's say $a < b$.

We can count even and odd pairs separately to make things easier*:

Odd: $$(1,3) , (3,5) , (5,7) . . . (997,999)$$

Even: $$(2,4) , (4,6) , (6,8) . . . (996,998)$$

This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.

$*$Note: We are counting the pairs for the values of $a$ and $b$, which, when squared, translate to the pairs of $(x,y)$ we are trying to find.

### Solution 3(2-Liner)

Our equation is $x+y-4=2\sqrt{xy} \implies \sqrt{y}-\sqrt{x}=2$ since $y>x$. As a result $y$ must be a perfect square and cannot be $10^6, 4, 1$ so the answer is $\boxed{997}$.