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

Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives $((x+y)-4)^2 = 4xy$ $((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy$ $x^2 - 2xy + y^2 + 16 - 8x - 8y = 0$

Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$ , so the problem asks for solutions of $(x-y-4)^2 = 16y$ Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$ , $y$ must be from $1^2$ to $999^2$ , giving at most 999 options for $y$ .

However if $y = 1^2$ , you get $(x-5)^2 = 16$ , which has solutions $x = 9$ and $x = 1$ . Both of those solutions are not less than $y$ , so $y$ cannot be equal to 1. If $y = 2^2 = 4$ , you get $(x - 8)^2 = 64$ , which has 2 solutions, $x = 16$ , and $x = 0$ . 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$ , you get exactly 1 solution for $x$ , and that gives a total of $999 - 2 = \boxed{997}$ pairs.

- asbodke

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