Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

## Arithmetic and geometric mean with Algebra – AIME 2000

Find the number of ordered pairs (x,y) of integers is it true that \(0 \lt y \lt 10^{6}\) and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

- is 107
- is 997
- is 840
- cannot be determined from the given information

**Key Concepts**

Algebra

Equations

Ordered pair

## Check the Answer

But try the problem first…

Answer: is 997.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

given that \(\frac{x+y}{2}=2+({xy})^\frac{1}{2}\) then solving we have \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and-2

Second Hint

given that \(y \gt x\) then \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and here maximum integer value of \(y^\frac{1}{2}\)=\(10^{3}-1\)=999 whose corresponding \(x^\frac{1}{2}\)=997 and decreases upto \(y^\frac{1}{2}\)=3 whose corresponding \(x^\frac{1}{2}\)=1

Final Step

then number of pairs (\(x^\frac{1}{2}\),\(y^\frac{1}{2}\))=number of pairs of (x,y)=997.

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## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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