Perfect Square by Prabhav Bansal

We want $n^2+n+41=m^2$ for some integer $m\geq{0}$ . Multiplying by 4, completing the square, and rearranging the terms we find $163=4m^2-(2n+1)^2$ $=(2m-2n-1)(2m+2n+1)$ . Since 163 is prime, one factor must be $\pm{163}$ and the other $\pm{1}$ . The sum of the factors is $4m=164$ so $m=41$ . Substituting into the original equation, we find $n=40$ and $n=-41$ , with their sum being $\boxed{-1}$ .