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What are the number of integer solutions of $xy - 6 (x+y)=0$ with $x\leq y$ is ?

Equation $xy - 6 (x+y)=0$ can also be written as $1/x + 1/y = 1/6$

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1 Answer 1

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The equation can also be written as $(x-6)(y-6)=36$. So $x-6$ and $y-6$ are integers, not necessarily positive, whose product is $36$.

How many divisors, not necessarily positive, does $36$ have? Then you will have to take care of the $x\le y$ constraint. This can be more or less done by symmetry.

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    $\begingroup$ Ah! How did I miss that! Thanks @André Nicolas $\endgroup$
    – Zero
    Apr 15, 2013 at 19:28
  • $\begingroup$ @André Nicolas I am getting 10 solutions (7,42), (-30,5) ,(8,24) ,(-12,4),(9,18), (-6,3), (10,15) ,(-3,2), (12,12) ,(0,0) in which x<=y but answer is given 9. whats wrong!!! $\endgroup$
    – ViX28
    Mar 26, 2016 at 5:32
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    $\begingroup$ It is late here, I can check again tomorrow. But $36$ has $9$ positive divisors, and $9$ negative. Two pairs, $(6,6)$ and $(-6,-6)$ give equality. Of the remaining $16$, by symmetry half give $x\lt y$ and half give $x\gt y$. So by general considerations the number with $x\le y$ should be $2+8$. There is no need to list (but you did, and also got $10$). The only possibility that I see is that the question you were asked wants the number of solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{6}$. In that case, the solution $(0,0)$ to $xy-6(x+y)=0$ is inadmissible (division by $0$) and we get $9$. $\endgroup$ Mar 26, 2016 at 5:50
  • $\begingroup$ That's great :) I was also thinking the same. Thanks for clarifying and confirming :) $\endgroup$
    – ViX28
    Mar 26, 2016 at 7:16
  • $\begingroup$ You are welcome. $\endgroup$ Mar 26, 2016 at 7:20

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