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How can I Find all solutions of the diophantine equation? :

$$xy=\frac{3x+y}{2}.$$

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    Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level.2012-04-24
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    That's the [second](http://math.stackexchange.com/q/136079/742) similarly-themed question, with the same lack of background or work shown.2012-04-24
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    @Arturo: It's like they're the same person!2012-04-24
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    @TheChaz: Oh, quite so (especially since the username on the other question was changed from the generic userxxxx to "Mily"). I was trying to say "That's the second time *this user* has posted..." not to imply there was a sudden wave of people with similar manners.2012-04-24

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$$(x,y)=\left(x,\frac{3x}{2x-1}\right)$$

Hence :

$3x=k(2x-1)$

$3x=2kx-k$

$x=\frac{k}{2k-3}$ for some integer $k$

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    So... $x = 1, 2$ ??2012-04-24
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    @TheChaz right..2012-04-24
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    $x=-1,0$ works too..2012-04-24
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    This is the other way to tackle such problems. In this case, we get $y=\frac{3x}{2x-1}$. Since $x$ and $2x-1$ are relatively prime, $2x-1$ must divide $3$. That gives $4$ values for $x$.2012-04-24
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    I was under the impression that solution are sought among the positive integers, but why not find all of them?!2012-04-24
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    @TheChaz Usually, Diophantine equation assumes just any integer solution unless otherwise stated.2012-04-24
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    @pedja After 3 revisions there is still an error in the solution :P :)2012-04-24
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    @pedja Ok, now there is no. :)2012-04-24
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    @Vadim thanks..2012-04-24
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    @pedja You are welcome. But I still think that what you have after the "Hence" is not the best way to proceed... sorry. Because it is not immediately clear for what $k$ the last expression is integer. Or, maybe, in this case with small numbers it is clear enough... That's OK. But I would also suggest looking at the Andre Nicolas's comment above.2012-04-24
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Given this answer that Bill gave you just one hour ago, this would be unfair to give you another similar one for this question as well.

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    Granted, I refer to this answer in *my* answer, but I still would have thought this better as a comment.2012-04-24
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Hint: (Inspired by a similar hint from Bill Dubuque)

Rewrite the equation as $$(2 x - 1) (2 y - 3) = 3$$ and equate $2x - 1$ with the factors of $3$.

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    COME ON! Give people opportunity to think by themselves, or at least to apply what they have just learnt. :) That being said I am upvoting your solution because there are already other ready solutions.2012-04-24
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    I appreciate your enthusiasm, Vadim, but not so much the criticism of my pedagogy! Yes, this question is similar to the other, but I still can't see [how to systematically approach Diophantine equations](http://math.stackexchange.com/q/134987/7850), so I don't expect that from others...2012-04-24
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    Sorry, I had no intent to criticize your answer or especially your pedagogy. And, as I said I even upvoted it. The problem at the link you gave in the comment is a completely different story. The problem at the link in the answer is (almost) exactly the same as this one. And can be solved using the exactly same method (with a little modification which you have shown in your answer). Moreover, it was asked by the same person.2012-04-24