$$ (m+1)x^2 - (2m+1)x - 2m = 0 $$ $$ m \in \mathbb{R}-\{-1\} $$
Find $ m \in \mathbb{Z} $ for which $ x_1 $ and $ x_2 $ (the solutions of equation, the roots) are integers ($x_1,x_2 \in \mathbb{Z}$)
I tried to make it as follows but I am unsure if it is correct. Please tell me if I am wrong (| means divide - I don't know exactly english terms and symbols, please ask me if you don't understand):
If $m\in\mathbb{Z}$ then:
$ m+1 \in \mathbb{Z} \\ -2(m+1) \in \mathbb{Z} \\ -2m \in \mathbb{Z} $
$ \Rightarrow x_{1,2} = \frac{p}{q}, where \ p|(-2m) \ and \ q|(m+1)$
$ but \ x_{1,2} \in \mathbb{Z} \Rightarrow (m+1)|(-2m) $
$ \Rightarrow \frac{2m}{m+1} \in \mathbb{Z} \Rightarrow \frac{m+m+1-1}{m+1} = 1 + \frac{m-1+1-1}{m+1} = 2 - \frac{2}{m+1} $
$ D_2 = \{\pm1, \pm2\} $
$ 1.\ m+1 = -1 \Rightarrow m = -2 $
$ 2.\ m+1 = 1 \Rightarrow m = 0 $
$ 3.\ m+1 = -2 \Rightarrow m = -3 $
$ 4.\ m+1 = 2 \Rightarrow m = 1 $
And after that I verified all $ m \in \{-3,-2,0,1\} $ in the equation above and then finded only $ m \in \{-2,0\} $ as for $ x_{1,2} \in \mathbb{Z} $
Please tell me if the demonstration before is enough and if is there any other method of finding m.
Thank you!
PS: Please ask me all you don't understand and help me to correct the question, if I am wrong. Also, I think title is not describing well what I am asking but I don't know what to say about. Thank you!
The following is an alternative procedure to arrive at your solution. If $x_{1}$ and $x_{2}$ are the solutions of the equation
\begin{eqnarray} (m+1)x^{2}-(2m+1)x-2m=0, \end{eqnarray}
then, we have
\begin{align} x_{1}+x_{2}=\frac{2m+1}{m+1}=1+\frac{1}{m+1},\\ \\ x_{1}x_{2}=\frac{-2m}{m+1}=-2+\frac{2}{m+1}. \end{align}
From the above equations, it is clear that if $x_{1}$ and $x_{2}$ have to be integers, the only possibilities for $m$ are $m=0$ and $m=-2$.