2012 AMC 10A Problem 22

The first $m$ positive odd integers sum to $m^2$, and the first $n$ positive even integers sum to $n(n+1)$. Thus $m^2=n(n+1)+212$. As a quadratic in $n$, this has discriminant $1-4(212-m^2)=4m^2-847$, which must be an odd square. Let $p^2=4m^2-847$. Then $(2m+p)(2m-p)=847$. The positive factor pairs of $847$ are $847\cdot1$, $121\cdot7$, and $77\cdot11$. They give $p=423,57,33$, respectively. Because $n=\dfrac{-1+p}{2}$, the possible values of $n$ are $211,28,16$. Their sum is $255$. Thus, A is the correct answer.