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

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Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]].
 
Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]].
  
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== Solution (by nosaj)==
 
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We have <math> n^2-19n+99 =m^2</math>, where <math>m</math> is a positive integer. Rearranging gives us
 
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<cmath>n^2-19n+(-m^2+99)=0.</cmath>
 
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Applying the quadratic formula yields
 
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<cmath>n=\frac{19 \pm \sqrt{361-4(1)(-m^2+99)}}{2}.</cmath> Now, we simplyfy:
 
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<cmath>n=\frac{19 \pm \sqrt{4m^2-35}}{2}.</cmath>
 
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In order for <math>n</math> to be an integer, the discriminant <math>4m^2-35</math> must be a perfect square. In other words,
 
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<cmath>4m^2-35=t^2,</cmath>
 
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where <math>t</math> is a positive integer. Rearranging gives us
 
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<cmath>4m^2-t^2=35.</cmath> Aha! Difference of squares.
 
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<cmath>(2m+t)(2m-t)=35.</cmath>
 
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Remember that both of these factors are integers, so we have a very limited amount of choices. Either
 
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<cmath>2m+t=35 \\ 2m-t=1</cmath> or <cmath>2m+t=7 \\ 2m-t=5</cmath>
 
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These two systems yield <math>m=9</math> and <math>m=3</math>. Plugging <math>m</math> back in gives us <math>n=9</math>, <math>n=10</math>, <math>n=18</math>, <math>n=1</math>. Therefore, our answer is <math>9+10+18+1=\boxed{38}</math>
 
  
 
== Solution ==
 
== Solution ==

Revision as of 16:55, 4 January 2015

Problem

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


Solution

If $n^2-19n+99=x^2$ for some positive integer $x$, then rearranging we get $n^2-19n+99-x^2=0$. Now from the quadratic formula,

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

Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$. Rearranging gives $(2x+q)(2x-q)=35$. Thus $(2x+q, 2x-q)=(35, 1)$ 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
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All AIME Problems and Solutions

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