An Equation in Positive Integers

Number Theory Level pending

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$ .

The answer is 80.

1 solution

Alan Yan
Sep 2, 2015

$x^2 + 84x + 2008 = y^2 \implies (y-x-42)(y+x+42) = 244$

The only case that works is

$y-x-42 = 2$

$y+x+42 = 122$

$\implies x+y = \boxed{80}$