1999 AIME Problems/Problem 3

Problem

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

Solution

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.

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1999 AIME Problems/Problem 3

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Solution

Alternate Solution

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