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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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.


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 $x.$

  • $\begingroup$ 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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