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How many integral solutions does the following equation have? $$x + 2y = 2xy$$

I have tried hit and trail method and I got only one solution, namley, $x=y=0$.. But Is there any other way to solve this ? If so, tell me at very basic level and also tell how to solve such questions.

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closed as off-topic by Saad, Jean-Claude Arbaut, mrf, Servaes, B. Goddard Apr 17 at 10:45

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    $\begingroup$ Please don't write in all caps. It doesn't help readability in any way, and makes it look like you are yelling at us. $\endgroup$ – Arthur Apr 17 at 7:22
  • $\begingroup$ Sorry. But could u please answer fast i am in need fast. I am not argue ing with yu anyway . I am sorry. It was my 1 st question on this site $\endgroup$ – Tim bergling Apr 17 at 7:28
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    $\begingroup$ This site doesn't work fast. Well, it does some times. But it's not something you can expect it to. There are 250 000 unanswered questions on this site. We answer the ones we feel like answering. Yelling doesn't help. Being in a hurry doesn't help. You might get an answer, or you might not. That's just how it is. $\endgroup$ – Arthur Apr 17 at 7:41
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Following the idea of @Doug M, you can actually write:

$$2y = \frac{x}{x-1} = 1+\frac{1}{x-1}$$

If $y$ is an integer, then $2y$ must be too, and $\frac{1}{x-1}$ as well. So $x-1 = \pm 1$ which leads to $x=2$ or $x=0$.

If $x=0$, $y=0$.

If $x=2$, $y=1$.

Those are the only solutions you can have.

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Here is a hint:

$x = y(2x -2)\\ y = \frac {x}{2(x-1)}$

With a little bit of thought you should be able to find an upper and lower bound of $x.$

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  • $\begingroup$ Do you not understand the algebra I did to get here, or do you not understand the implications of it? The implications. $x$ must be even, because we are dividing by 2 and getting an integer. And $(x-1)$ divides $x$ which is true if $x-1 = 1$ or $x=0$ but is it ever true otherwise? $\endgroup$ – Doug M Apr 17 at 9:14

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