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

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[[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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Since <math>244</math> 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 <math>244 = 2^2 \cdot 61</math>, the factors must be <math>2</math> and <math>122</math>. Since <math>x,y > 0</math>, we have <math>y - x - 42 = 2</math> and <math>y + x + 42 = 122</math>; the latter equation implies that <math>x + y = \boxed{080}</math>.  
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Since <math>244</math> is even, one of the factors is even. A [[parity]] check shows that if one of them is even, then both must be even. Since <math>244 = 2^2 \cdot 61</math>, the factors must be <math>2</math> and <math>122</math>. Since <math>x,y > 0</math>, we have <math>y - x - 42 = 2</math> and <math>y + x + 42 = 122</math>; the latter equation implies that <math>x + y = \boxed{080}</math>.  
  
 
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.  
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Plugging into our equation, we find that <math>x=18</math>, and <math>(x,y)=(18,62)</math> indeed satisfies the original equation. <math>x+y=\boxed{080}</math>
 
Plugging into our equation, we find that <math>x=18</math>, and <math>(x,y)=(18,62)</math> indeed satisfies the original equation. <math>x+y=\boxed{080}</math>
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===Solution 5===
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Let <math>y=x+d</math> for some <math>d>0</math>, substitute into the original equation to get <math>84x + 2008 = 2xd + d^2</math>.
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All terms except for the last one are even, hence <math>d^2</math> must be even, hence let <math>d=2e</math>. We obtain <math>21x + 502 = xe + e^2</math>. Rearrange to <math>502-e^2 = x(e-21)</math>.
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Obviously for <math>0<e<21</math> the right hand side is negative and the left hand side is positive. Hence <math>e\geq 21</math>. Let <math>e=21+f</math>, then <math>f\geq 0</math>.
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We have <math>502 - (21+f)^2 = xf</math>. Left hand side simplifies to <math>61 - 42f + f^2</math>. As <math>x</math> must be an integer, <math>f</math> must divide the left hand side. But <math>61</math> is a prime, which only leaves two options: <math>f=1</math> and <math>f=61</math>.
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Option <math>f=61</math> gives us a negative <math>x</math>. Option <math>f=1</math> gives us <math>x=61/f - 42 + f = 18</math>, and <math>y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62</math>, hence <math>x+y=\boxed{080}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:42, 31 March 2009

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. Since $244 = 2^2 \cdot 61$, 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. $x+y=\boxed{080}$

Solution 5

Let $y=x+d$ for some $d>0$, substitute into the original equation to get $84x + 2008 = 2xd + d^2$.

All terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$. We obtain $21x + 502 = xe + e^2$. Rearrange to $502-e^2 = x(e-21)$.

Obviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\geq 21$. Let $e=21+f$, then $f\geq 0$.

We have $502 - (21+f)^2 = xf$. Left hand side simplifies to $61 - 42f + f^2$. As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$.

Option $f=61$ gives us a negative $x$. Option $f=1$ gives us $x=61/f - 42 + f = 18$, and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$, hence $x+y=\boxed{080}$.

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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