1999 AIME Problems/Problem 3

Problem

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

Solution

If the perfect square is represented by $x^2$ , then the equation is $n^2 - 19n + 99 - x^2 = 0$ . The quadratic formula yields

$n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}$

In order for this to be an integer, the discriminant must also be a perfect square, so $4x^2 - 35 = q^2$ for some nonnegative integer $q$ . This factors to

$(2x + q)(2x - q) = 35$

$35$ has two pairs of positive factors: $\{1,\ 35\}$ and $\{5,\ 7\}$ . Respectively, these yield $9$ and $3$ for $x$ , which results in $n = 1,\ 9,\ 10,\ 18$ . The sum is therefore $\boxed{038}$ .

We have

$n^2-19n+99=x^2\Longleftrightarrow n^2-19n+99-x^2=0\Longleftrightarrow n=\frac{19\pm \sqrt{4x^2-35}}{2}$

This equation has solutions in integers if and only if for some odd nonnegative integer $q$ , $4x^2-35=q^2$ , or $(2x+q)(2x+q)=35$ . Because $q$ is odd, this makes both of the factors $2x+q$ and $2x-q$ odd,

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

Contents

Problem

Solution

Alternate Solution

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