imc

The sum of the first $m$ odd integers is given by $m^2$. The sum of the first $n$ even integers is given by $n(n+1)$. Thus, $m^2=n^2+n+212$. Since we want to solve for n, rearrange as a quadratic equation: $n^2+n+(212-m^2)=0$. Use the quadratic formula: $n=\frac{-1+\sqrt{1-4(212-m^2)}}{2}$. Since $n$ is clearly an integer, $1-4(212-m^2)=4m^2-847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer. Let $x=\sqrt{4m^2-847}$; note that this means $n=\frac{-1+x}{2}$. It can be rewritten as $x^2=4m^2-847$, so $4m^2-x^2=847$. Factoring the left side by using the difference of squares, we get $(2m+x)(2m-x)=847=7\cdot 11^2$. Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2m+x)-(2m-x)=2m+x-2m+x=2x.$ We have three pairs of factors, $847\cdot 1,121\cdot 7,$ and $77\cdot 11$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2x$. Thus the possibilities for $x$ are $423$, $57$, and $33$. Now plug in these values into the equation $n=\frac{-1+x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $\boxed{255}$.