2008 AIME I Problems/Problem 4

Problem

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

Solution

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.

Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Sine $244 = 2^2 \cdot 61$ , 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.

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2008 AIME I Problems/Problem 4

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