# Number of integer solutions of $xy - 6 (x+y)=0$

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

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.
• 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 $9$. – André Nicolas Mar 26 '16 at 5:50