2008 AIME I Problems/Problem 4

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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 $x+y=80$.

See also

2008 AIME I (ProblemsAnswer KeyResources)
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