We are given the following equation: $(m+1)x^2-(2m+1)x-2m=0$, where $m\neq-1$.
We have to find all integers $m$ so that the equation above has integer solutions.
I know that $m=0$ and $m=-2$ satisfy the conditions, but is there a way to find $m$ without trying random values, as I did?
Let $\alpha,\beta\in\mathbb Z$ be the solutions of $(m+1)x^2-(2m+1)x-2m=0$.
Then, by Vieta's formula, $$\alpha+\beta=-\frac{-(2m+1)}{m+1}=2-\frac{1}{m+1}$$ Now $\frac{1}{m+1}$ has to be an integer, so $m+1=\pm 1\implies m=0,-2$.