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

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== Problem ==
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== Problem ==  
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Find the sum of all [[positive integer]]s <math>\displaystyle n</math> for which <math>\displaystyle n^2-19n+99</math> is a [[perfect square]].
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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]].
  
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== 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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== Solution 1==
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If <math>n^2-19n+99=x^2</math> for some positive integer <math>x</math>, then rearranging we get <math>n^2-19n+99-x^2=0</math>. Now from the quadratic formula,
  
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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>n=\frac{19\pm \sqrt{4x^2-35}}{2}</math>
  
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:<math>(2x + q)(2x - q) = 35</math>  
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Because <math>n</math> is an integer, this means <math>4x^2-35=q^2</math> for some nonnegative integer <math>q</math>. Rearranging gives <math>(2x+q)(2x-q)=35</math>. Thus <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{38}</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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==Solution 2==
  
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==Alternate Solution==
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Suppose there is some <math>k</math> such that <math>x^2 - 19x + 99 = k^2</math>. Completing the square, we have that <math>(x - 19/2)^2 + 99 - (19/2)^2 = k^2</math>, that is, <math>(x - 19/2)^2 + 35/4 = k^2</math>. Multiplying both sides by 4 and rearranging, we see that <math>(2k)^2 - (2x - 19)^2 = 35</math>. Thus, <math>(2k - 2x + 19)(2k + 2x - 19) = 35</math>. We then proceed as we did in the previous solution.
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==Solution 3==
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When <math> n \geq 12 </math>, we have
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<cmath> (n-10)^2 < n^2 -19n + 99 < (n-8)^2. </cmath>
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So if <math> n \geq 12</math> and <math> n^2 -19n + 99 </math> is a perfect square, then
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<cmath> n^2 -19n + 99 = (n-9)^2 </cmath>
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or <math> n = 18 </math>.
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For <math> 1 \leq n < 12 </math>, it is easy to check that <math> n^2 -19n + 99 </math> is a perfect square when <math> n = 1, 9 </math> and <math> 10 </math> ( using the identity <math> n^2 -19n + 99 = (n-10)^2 + n - 1.) </math>
  
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Suppose there is some <math>k</math> such that <math>x^2 - 19x + 99 = k^2</math>. Completing the square, we have that <math>(x - 19/2)^2 + 99 - (19/2)^2 = k^2</math>, that is, <math>(x - 19/2)^2 + 35/4 = k^2</math>. Multiplying both sides by 4 and rearranging, we see that <math>(2k)^2 - (2x - 19)^2 = 35</math>. Thus, <math>(2k - 2x + 19)(2k + 2x - 19) = 35</math>. We then proceed as we did in the previous solution.
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We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:18, 29 January 2021

Problem

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


Solution 1

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 $1+9+10+18=\boxed{38}$.

Solution 2

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.


Solution 3

When $n \geq 12$, we have \[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\]

So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then \[n^2 -19n + 99 = (n-9)^2\]

or $n = 18$.

For $1 \leq n < 12$, it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$

We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$

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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