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 $1/x + 1/y = 1/6$

enter image description here


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.

  • $\begingroup$ Ah! How did I miss that! Thanks @André Nicolas $\endgroup$ – Zero Apr 15 '13 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 '16 at 5:32
  • $\begingroup$ 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 $9$. $\endgroup$ – André Nicolas Mar 26 '16 at 5:50
  • $\begingroup$ That's great :) I was also thinking the same. Thanks for clarifying and confirming :) $\endgroup$ – ViX28 Mar 26 '16 at 7:16
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Mar 26 '16 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.