1999 AIME 第 3 题

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 $18 + 10 + 9 + 1 = 38.$