2017 AIME II 第 6 题

Set $n^2 + 85n + 2017 = m^2$ and multiply by $4$ so the left side becomes $(2n + 85)^2 + 843$

Factor $4m^2 - (2n + 85)^2$ as a difference of squares; $843 = 3 \cdot 281$ has only two factorizations

Suppose $n^2 + 85n + 2017 = m^2$ for a positive integer $m.$ Multiplying by $4$ and completing the square gives $(2n + 85)^2 + 843 = 4m^2,$ so $$\begin{aligned} &(2m - 2n - 85)(2m + 2n + 85) \\ &= 843 = 3 \cdot 281, \end{aligned}$$ where $281$ is prime. Both factors are positive with the second one larger, so either $2m - 2n - 85 = 1$ and $2m + 2n + 85 = 843,$ or $2m - 2n - 85 = 3$ and $2m + 2n + 85 = 281.$

The first system gives $m = 211$ and $n = 168,$ and indeed $168^2 + 85 \cdot 168 + 2017$ $= 44521 = 211^2.$ The second gives $m = 71$ and $n = 27,$ with $27^2 + 85 \cdot 27 + 2017$ $= 5041 = 71^2.$

The requested sum is $168 + 27 = 195.$