Problem:
There exist unique positive integers x and y that satisfy the equation x2+84x+2008=y2. Find x+y.
Solution:
Observe that if x is a positive integer, then
(x+42)2=x2+84x+1764<x2+84x+2008<x2+90x+2025=(x+45)2.
Because x2+84x+2008 is a square, it is either (x+43)2 or (x+44)2. In the first case, 2x=159, which is impossible, and in the second case, 4x=72, which implies x=18. In that case, y=x+44=62, and x+y=18+62=80.
OR
Complete the square in x to find that x2+84x+1764=(x+42)2= y2−244. Letting v=x+42, the condition y2−v2=244 is equivalent to (y−v)(y+v)=2⋅2⋅61, whose only positive integer solutions are given by y−v=2 and y+v=2⋅61. Thus the only two perfect squares that differ by 244 are 602 and 622. Hence x=60−42=18, and y=62.
The problems on this page are the property of the MAA's American Mathematics Competitions