2008 AIME I Problem 4

Attempt Problem 4 of the 2008 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AIME I solutions, or check the answer key.

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4.

There exist unique positive integers xx and yy that satisfy the equation x2+84x+2008=y2.x^2 + 84x + 2008 = y^2. Find x+y.x + y.

Answer: 80
Concepts:Diophantine Equationcompleting the squaredifference of squares
Difficulty rating: 2230
Solution:

Completing the square, x2+84x+2008x^2 + 84x + 2008 =(x+42)2+244,= (x + 42)^2 + 244, so y2(x+42)2=244,y^2 - (x + 42)^2 = 244, which factors as (yx42)(y+x+42)=244=2261. \begin{aligned} &(y - x - 42) \\ &\quad {}\cdot (y + x + 42) \\ &= 244 \\ &= 2^2 \cdot 61. \end{aligned} The two factors have the same parity, and their product is even, so both are even: yx42=2y - x - 42 = 2 and y+x+42=122.y + x + 42 = 122.

Adding gives y=62,y = 62, and then x=18;x = 18; indeed 182+8418+200818^2 + 84 \cdot 18 + 2008 =3844= 3844 =622.= 62^2. Therefore x+y=18+62=80.x + y = 18 + 62 = 80.

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