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 $038$ .

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
All AIME Problems and Solutions

Page

Toolbox

Search

1999 AIME Problems/Problem 3

Problem

Solution

See also