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$$ (m+1)x^2 - (2m+1)x - 2m = 0 $$ $$ m \in \mathbb{R}-\{-1\} $$

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!

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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 $m=-2$.

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  • $\begingroup$ Thank you very much! One more question: is my solving correct or do I need to add more infos? $\endgroup$ – MM PP Jul 13 '16 at 9:56
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    $\begingroup$ I actually traced half the way in your solution before presenting mine :) So, I think your solution is ok, but just that it took you a lot of effort to plug in all the values and check. $\endgroup$ – PN Karthik Jul 13 '16 at 10:02

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