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: $n^{2}+n+\left(212-m^{2}\right)=0$. Use the quadratic formula: $n=\frac{-1+\sqrt{1-4\left(212-m^{2}\right)}}{2}$. Since $n$ is clearly an integer, $1-4\left(212-m^{2}\right)=4 m^{2}-847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.  Let $x=\sqrt{4 m^{2}-847}$; note that this means $n=\frac{-1+x}{2}$. It can be rewritten as $x^{2}=4 m^{2}-847$, so $4 m^{2}-x^{2}=847$. Factoring the left side by using the difference of squares, we get $(2 m+x)(2 m-x)=847=7 \cdot 11^{2}$. Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2 m+x)-(2 m-x)=2 m+x-2 m+x=2 x$. 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 $2 x$. 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 $255$. 
The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$? 
Options:
  * $$255$$ 
  * $$256$$ 
  * $$257$$ 
  * $$258$$ 
  * $$259$$ 