1999 AIME Problem 3

Question

Find the sum of all positive integers

n

for which

n^2 - 19n + 99

is a perfect square.

Answer

Multiply by

4

and complete the square:

4(n^2 - 19n + 99)

= (2n - 19)^2 + 35

If the expression equals

k^2,

then

(2k)^2 - (2n - 19)^2 = 35;

factor

35

as a difference of squares in every way

Suppose

n^2 - 19n + 99 = k^2.

Multiplying by

4

and completing the square gives

(2n - 19)^2 + 35 = (2k)^2,

so

\begin{aligned} &\bigl(2k - (2n - 19)\bigr) \\ &\quad {}\cdot \bigl(2k + (2n - 19)\bigr) \\ &= 35. \end{aligned}

The two factors sum to

4k \gt 0,

so both are positive: the factor pairs are

(1, 35),

(5, 7),

(7, 5),

and

(35, 1).

Subtracting the first factor from the second gives

2(2n - 19) = 34,

2,

-2,

or

-34,

so

n = 18,

10,

9,

or

1.

Each indeed makes the expression a perfect square (

81,

9,

9,

81

), and the sum is

18 + 10 + 9 + 1 = 38.

Problem 3 in other years