The number of integer solutions of the following equation

$(x \leq y)$

$xy -6(x+y) = 0$

I have concluded that $xy$ is a multiple of $6$ and that $x$ is negative and $y$ is positive.

After that, I obtained these solutions with trial and error: $(-3,2) , (-6,3), (-12,4), (-18,4)$.

Now the obvious question is how long do I continue? And isn't there any better process than this clumsy mess?

And @Dhruv in the comments below proves my analysis wrong as well.

  • 2
    $\begingroup$ $(x-6)(y-6)=36$. $\endgroup$ – almagest Mar 26 '16 at 13:15
  • $\begingroup$ Are you sure $(-18, 4)$ is a solution? $\endgroup$ – TheRandomGuy Mar 27 '16 at 6:22
  • $\begingroup$ Ciuld you explain why you reasoned that $x$ is negative and $y$ is positive? $\endgroup$ – TheRandomGuy Mar 27 '16 at 6:25
  • 1
    $\begingroup$ What about the solution $(12, 12)$? $\endgroup$ – TheRandomGuy Mar 27 '16 at 6:26
  • $\begingroup$ @Druv I agree that my analysis is wrong. Appreciate it if you could further point me in the right direction. $\endgroup$ – smkj33 Mar 28 '16 at 14:39

This was a fun question, thank you :)

First you factor using Simon's favorite factoring trick (explanation below):

Multiplying out $$xy-6(x+y)=0,$$ you get:

Adding $-6\cdot-6=36$ to both sides, you end up with: $$xy-6x-6y+36=36\implies (x-6)(y-6)=36.$$ Now factor $36$ for the posisble values of $(x-6)$ and $(y-6)$.

Hint: $36$ can be broken up as $1\cdot36$, $2\cdot18$, $3\cdot12$, $4\cdot9$, and $6\cdot6$ (and their negative counterparts).

Another hint: since $x\leq y$, we know that $(x-6)\le(y-6)$.

Another hint: plug in the factors of $36$ for $(x-6)$ and $(y-6)$:

Solution: We count the amount of listings above to get our answer $\boxed{10}.$
If we want the individual values of $x$ and $y$, we plug them into the equations above, for example, for $(-9)(-4)$, we would do:
$x-6=-9\implies x=-3$ and $y-6=-4\implies y=2$.
However, your assumption about a positive $y$ and negative $x$ is false. For example, for $(6)(6)$, we get $x=y=0$, where neither $y$ is positive nor $x$ is negative.

Hope this helps :)


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