Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and integers.

## Equations and integers – AIME I, 2008

There exists unique positive integers x and y that satisfy the equation \(x^{2}+84x+2008=y^{2}\)

- is 107
- is 80
- is 840
- cannot be determined from the given information

**Key Concepts**

Algebra

Equations

Integers

## Check the Answer

Answer: is 80.

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

## Try with Hints

\(y^{2}=x^{2}+84x+2008=(x+42)^{2}+244\) then 244=\(y^{2}-(x+42)^{2}=(y-x-42)(y+x+42)\)

here 244 is even and 244=\(2^{2}(61)\)=\( 2 \times 122\) for \(x,y \gt 0\)

(y-x-42)=2 and (y+x+42)=122 then y+x=80 and x=18 y=62.

## Other useful links

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