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


Solution 4 (similar to solution 3)

Rearranging our conditions to

\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]

Thus, $4|y-x.$

Now, let $y = 4k+x.$ Plugging this back into our expression, we get

\[(k-1)^2=x-1.\]

There, a unique value of $x, y$ is formed for every value of $k$. However, we must have

\[y<10^6 \implies (k+1)^2< 10^6-1\]

and

\[x=(k-1)^2+1>0.\]

Therefore, there are only $997$ pairs of $(x,y).$

Solution by Williamgolly

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

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