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1999 AIME Problems/Problem 3

Revision as of 11:26, 4 January 2015 by Nosaj (talk | contribs) (→‎Solution)

Problem

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


Solution (by nosaj)

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

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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