2000 AIME I Problem 6

Question

For how many ordered pairs

(x, y)

of integers is it true that

0 \lt x \lt y \lt 10^6

and that the arithmetic mean of

x

and

y

is exactly

2

more than the geometric mean of

x

and

y?

Answer

The condition rearranges to

(\sqrt{y} - \sqrt{x})^2 = 4,

so

\sqrt{y} = \sqrt{x} + 2

Show

\sqrt{x}

must be an integer, so

x = a^2

and

y = (a + 2)^2;

count the

a

with

y \lt 10^6

The condition is

\frac{x + y}{2} = \sqrt{xy} + 2,

that is,

x + y - 2\sqrt{xy} = 4,

so

(\sqrt{y} - \sqrt{x})^2 = 4

and (as

y \gt x

)

\sqrt{y} - \sqrt{x} = 2.

Note

\sqrt{xy} = \frac{x + y - 4}{2}

is rational, hence

\sqrt{y} + \sqrt{x} = \frac{y - x}{\sqrt{y} - \sqrt{x}} = \frac{y - x}{2}

is rational too, so

\sqrt{x}

and

\sqrt{y}

are rational — and a rational square root of an integer is an integer.

Therefore

x = a^2

and

y = (a + 2)^2

for a positive integer

a.

The constraint

y \lt 10^6

means

a + 2 \le 999,

so

a

ranges over

1, 2, \ldots, 997,

and each value gives a valid pair.

Hence there are

997

ordered pairs.

Problem 6 in other years