There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$. Find $x + y$.
[Completing the square](https://artofproblemsolving.com/wiki/index.php/Completing_the_square), $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$. Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by [difference of squares](https://artofproblemsolving.com/wiki/index.php/Difference_of_squares). 
Since $244$ is even, one of the factors is even. A [parity](https://artofproblemsolving.com/wiki/index.php/Parity) check shows that if one of them is even, then both must be even. Since $244 = 2^2 \cdot 61$, the factors must be $2$ and $122$. Since $x,y > 0$, we have $y - x - 42 = 2$ and $y + x + 42 = 122$; the latter equation implies that $x + y = \boxed{080}$. 
Indeed, by solving, we find $(x,y) = (18,62)$ is the unique solution.