Difference between revisions of "2008 AIME I Problems/Problem 4"

Revision as of 21:51, 24 March 2008

Problem

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

Solution

Solution 1

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.

Solution 2

We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$ . Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework starting from $y-(x+42)=2$ , using the fact that consecutive squares differ by the consecutive odd numbers. If the results come out as integers, the ordered pair is a solution:

$2(x+42)+1+2(x+42)+3=244$

$\Rightleftarrow x=18$ (Error compiling LaTeX. Unknown error_msg)

Thus, $y=62$ , and $x+y=80$ .

@@ Line 14: / Line 14: @@
 :<math>2(x+42)+1+2(x+42)+3=244</math>
 :<math>\Rightleftarrow x=18</math>
-Thus, <math>y=60</math>, and <math>x+y=80</math>.
+Thus, <math>y=62</math>, and <math>x+y=80</math>.
 == See also ==

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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