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

Revision as of 21:43, 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 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, $y=62$ , and $x+y=80$ .

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.

@@ Line 2: / Line 2: @@
 There exist unique positive integers <math>x</math> and <math>y</math> that satisfy the equation <math>x^2 + 84x + 2008 = y^2</math>. Find <math>x + y</math>.
+__TOC__
 == Solution ==
 ===Solution 1===
@@ Line 11: / Line 12: @@
 ===Solution 2===
-We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework starting from <math>y-(x+42)=2</math>, 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:
+We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework with the even numbers, starting with <math>y-(x+42)=2</math>.
-:<math>2(x+42)+1+2(x+42)+3=244</math>
-:<math>\Rightleftarrow x=18</math>
+<cmath>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\
+\Rightleftarrow x&=18\end{align*}</cmath>
 Thus, <math>y=62</math>, and <math>x+y=80</math>.
+===Solution 3===
+<math>\mod{6}</math>, we see that <math>y^2 \equiv x^2 + 4 \pmod{6}</math>; by [[quadratic residue]]s, we find that either <math>x \equiv 0, 3 \pmod{6}</math>. Also, <math>\mod{4}</math>, <math>y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}</math>, and so <math>x \equiv 0, 2 \mod{4}</math>. Combining, we see that <math>x \equiv 0 \mod{6}</math>.
+Testing <math>x = 6</math> and other multiples of <math>6</math>, we quickly find that <math>x = 18, y = 62</math> is the solution.
 == 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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