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

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== Problem ==
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== Problem ==  
 
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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== Solution 1==
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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>
 
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== Solution ==
 
 
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,  
 
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,  
  
 
:<math>n=\frac{19\pm \sqrt{4x^2-35}}{2}</math>
 
:<math>n=\frac{19\pm \sqrt{4x^2-35}}{2}</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{038}</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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==Alternate Solution==
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==Solution 2==
  
 
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.
 
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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We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math>
  
 
== See also ==
 
== See also ==

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)
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