AMC10 2012 A

Answer (A): The sum of the first $k$ positive integers is $\frac{k(k+1)}{2}$. Therefore the sum of the first $k$ even integers is $2+4+6+\cdots+2k = 2(1+2+3+\cdots+k)=2\cdot\frac{k(k+1)}{2}=k(k+1).$ The sum of the first $k$ odd integers is $(1+2+3+\cdots+2k)-(2+4+6+\cdots+2k)=\frac{2k(2k+1)}{2}-k(k+1)=k^2.$ The given conditions imply that $m^2-212=n(n+1)$, which may be rewritten as $n^2+n+(212-m^2)=0$. The discriminant for $n$ in this quadratic equation is $1-4(212-m^2)=4m^2-847$, and this must be the square of an odd integer. Let $p^2=4m^2-847$, and rearrange this equation so that $(2m+p)(2m-p)=847$. The only factor pairs for 847 are $847\cdot1$, $121\cdot7$, and $77\cdot11$. Equating these pairs to $2m+p$ and $2m-p$ yields $(m,p)=(212,423)$, $(32,57)$, and $(22,33)$. Note that the corresponding values of $n$ are found using $n=\frac{-1+p}{2}$, which yields 211, 28, and 16, respectively. The sum of the possible values of $n$ is 255.