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

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 with the even numbers, starting with $y-(x+42)=2$ .

$\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ \Rightleftarrow x&=18\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Thus, $y=62$ , which works, so $x+y=80$ .

Solution 3

$\mod{6}$ , we see that $y^2 \equiv x^2 + 4 \pmod{6}$ ; by quadratic residues, we find that either $x \equiv 0, 3 \pmod{6}$ . Also, $\mod{4}$ , $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$ , and so $x \equiv 0, 2 \mod{4}$ . Combining, we see that $x \equiv 0 \mod{6}$ .

Testing $x = 6$ and other multiples of $6$ , we quickly find that $x = 18, y = 62$ is the solution.

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

Page

Toolbox

Search