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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.