Categories

# Equations and Integers | AIME I, 2008 | Question 4

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

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

But try the problem first…

Source

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.