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

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

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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 24 '12 at 3:41
  • $\begingroup$ That's the second similarly-themed question, with the same lack of background or work shown. $\endgroup$ – Arturo Magidin Apr 24 '12 at 3:43
  • $\begingroup$ @Arturo: It's like they're the same person! $\endgroup$ – The Chaz 2.0 Apr 24 '12 at 3:47
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Apr 24 '12 at 3:50
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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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  • $\begingroup$ So... $x = 1, 2$ ?? $\endgroup$ – The Chaz 2.0 Apr 24 '12 at 4:03
  • $\begingroup$ @TheChaz right.. $\endgroup$ – Peđa Terzić Apr 24 '12 at 4:04
  • $\begingroup$ $x=-1,0$ works too.. $\endgroup$ – Peđa Terzić Apr 24 '12 at 4:09
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    $\begingroup$ 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 $x$. $\endgroup$ – André Nicolas Apr 24 '12 at 4:10
  • $\begingroup$ I was under the impression that solution are sought among the positive integers, but why not find all of them?! $\endgroup$ – The Chaz 2.0 Apr 24 '12 at 4:12
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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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    $\begingroup$ Granted, I refer to this answer in my answer, but I still would have thought this better as a comment. $\endgroup$ – The Chaz 2.0 Apr 24 '12 at 3:56
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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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  • $\begingroup$ 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. $\endgroup$ – Vadim Apr 24 '12 at 4:26
  • $\begingroup$ 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, so I don't expect that from others... $\endgroup$ – The Chaz 2.0 Apr 24 '12 at 4:29
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    $\begingroup$ 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. $\endgroup$ – Vadim Apr 24 '12 at 4:42

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