Equations and Integers | AIME I, 2008 | Question 4

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

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

AMC - AIME Boot Camp... for brilliant students.

Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

Learn More

Subscribe to Cheenta at Youtube