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

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===Solution 2===
 
===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 with the even numbers, starting with <math>y-(x+42)=2</math>.
 
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>.
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<center><math>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\
 
<center><math>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\
 
\Rightleftarrow x&=18\end{align*}</math></center>
 
\Rightleftarrow x&=18\end{align*}</math></center>
  
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Thus, <math>y=62</math>, which works, so <math>x+y=\boxed{80}</math>.
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Thus, <math>y=62</math> and the answer is <math>\boxed{080}</math>.
  
 
===Solution 3===
 
===Solution 3===
 
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>y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}</math>, so <math>x \equiv 0, 2 \mod{4}</math>. Combining, we see that <math>x \equiv 0 \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>y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}</math>, so <math>x \equiv 0, 2 \mod{4}</math>. Combining, we see that <math>x \equiv 0 \mod{6}</math>.  
  
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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.
+
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. <math>18+62=\boxed{080}</math>
  
 
===Solution 4===
 
===Solution 4===

Revision as of 14:29, 19 April 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 the answer is $\boxed{080}$.

Solution 3

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, $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$, 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. $18+62=\boxed{080}$

Solution 4

We solve for x: $x^2 + 84x + 2008-y^2 = 0$

$x=\dfrac{-84+\sqrt{84^2-4\cdot 2008+4y^2}}{2}=-42+\sqrt{y^2-244}$

So $y^2-244$ is a perfect square. Since 244 is even, the difference $\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$: $-244=-4y+4$, $y=62$.

Plugging into our equation, we find that $x=18$, and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=\boxed{080}$

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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