What is sum of integer values of $m$ so that $\frac1x+\frac1{x+1}=\frac1m$ has answer? 
What is sum of integer values of $m$ so that $\frac1x+\frac1{x+1}=\frac1m$ has answer?
$1)\text{zero}\qquad\qquad2)1\qquad\qquad3)7\qquad\qquad4)-10$

This is a high school level problem so I think by answer it means $x\in\mathbb{R}$. Here is my answer:
We have $m\neq0$ and $x\neq0,-1$
$$\frac{2x+1}{x^2+x}=\frac1m\;\Rightarrow\; x^2+x=2mx+m\;\;\Rightarrow\;x^2+(1-2m)x-m=0$$
We have$\;\Delta=(1-2m)^2+4m=4m^2+1$. Which is always positive. Hence there are always two values for  $x$ satisfies the original equation regardless of value of $m$  (if I plug in $x=0,-1$ in above equation I get $m=0$ which is a contradiction). And answer is "zero".
Is my solution right?

The answer must be $0$ (if there are at most finitely many such $m$) by symmetry.  If $(x, x+1)$ yields a solution $m$, then $(-(x+1), -x)$ also yields the solution $-m$.
