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Difference between revisions of "1999 AIME Problems/Problem 3"

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== Solution ==
 
== Solution ==
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If the perfect square is represented by <math>x^2</math>, then the [[equation]] is <math>n^2 - 19n + 99 - x^2 = 0</math>. The [[quadratic formula]] yields
 
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:<math> n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math>
 
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In order for this to be an [[integer]], the [[discriminant]] must also be a perfect square, so <math>4x^2 - 35 = q^2</math> for some [[nonnegative]] integer <math>q</math>. This [[factoring | factors]] to
 
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:<math>(2x + q)(2x - q) = 35</math>
 
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<math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>\boxed{038}</math>.
 
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We have  
 
We have  
  
 
:<math>n^2-19n+99=x^2\Longleftrightarrow n^2-19n+99-x^2=0\Longleftrightarrow n=\frac{19\pm \sqrt{4x^2-35}}{2}</math>
 
:<math>n^2-19n+99=x^2\Longleftrightarrow n^2-19n+99-x^2=0\Longleftrightarrow n=\frac{19\pm \sqrt{4x^2-35}}{2}</math>
  
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This equation has solutions in integers if and only if for some odd nonnegative integer <math>q</math>, <math>4x^2-35=q^2</math>, or <math>(2x+q)(2x+q)=35</math>. Because <math>q</math> is odd, this makes both of the factors <math>2x+q</math> and <math>2x-q</math> odd,
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This equation has solutions in integers if and only if for some odd nonnegative integer <math>q</math>, <math>4x^2-35=q^2</math>, or <math>(2x+q)(2x-q)=35</math>. Because <math>q</math> is odd, this makes both of the factors <math>2x+q</math> and <math>2x-q</math> odd, so <math>(2x+q, 2x-q)=(35, 1)</math> or <math>(7,5)</math>, giving <math>x=3</math> or <math>9</math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{038}</math>.
  
 
==Alternate Solution==
 
==Alternate Solution==

Revision as of 16:31, 13 May 2012

Problem

Find the sum of all positive integers $\displaystyle n$ for which $\displaystyle n^2-19n+99$ is a perfect square.

Solution

We have

$n^2-19n+99=x^2\Longleftrightarrow n^2-19n+99-x^2=0\Longleftrightarrow n=\frac{19\pm \sqrt{4x^2-35}}{2}$

This equation has solutions in integers if and only if for some odd nonnegative integer $q$, $4x^2-35=q^2$, or $(2x+q)(2x-q)=35$. Because $q$ is odd, this makes both of the factors $2x+q$ and $2x-q$ odd, so $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$, giving $x=3$ or $9$. This gives $n=1, 9, 10,$ or $18$, and the sum is $1+9+10+18=\boxed{038}$.

Alternate Solution

Suppose there is some $k$ such that $x^2 - 19x + 99 = k^2$. Completing the square, we have that $(x - 19/2)^2 + 99 - (19/2)^2 = k^2$, that is, $(x - 19/2)^2 + 35/4 = k^2$. Multiplying both sides by 4 and rearranging, we see that $(2k)^2 - (2x - 19)^2 = 35$. Thus, $(2k - 2x + 19)(2k + 2x - 19) = 35$. We then proceed as we did in the previous solution.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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