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

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(my own solution)
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== Solution ==
 
== Solution ==
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===Solution 1===
 
[[Completing the square]], <math>y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244</math>. Thus <math>244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)</math> by [[difference of squares]].  
 
[[Completing the square]], <math>y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244</math>. Thus <math>244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)</math> by [[difference of squares]].  
  
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Indeed, by solving, we find <math>(x,y) = (18,62)</math> is the unique solution.  
 
Indeed, by solving, we find <math>(x,y) = (18,62)</math> is the unique solution.  
 +
 +
===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:
 +
:<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>.
  
 
== See also ==
 
== See also ==

Revision as of 20:47, 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, $y=60$, and $x+y=80$.

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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