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:

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

The discriminant must also be a perfect square ( $y^2$ ), so $x^2 - 35 = y^2$ . This factors to:

$(x + y)(x - y) = 35$

$35$ has two pairs of factors: $\{1,\ 35\}$ and $\{5,\ 7\}$ . Respectively, these yield 18 and 6 for $x$ , which results in $n = 1,\ 9,\ 10,\ 18$ . The sum is therefore $038$ .

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
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1999 AIME Problems/Problem 3

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