<s>[INST] Pairs $(x,y)$ such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$. How many possible pairs of integers $x$ and $y$ exist such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$?

Since $y| 3x-1$, there exists some $k$ such that $3x-1 = ky$.
On the other hand since $x| 3y-1$ we have $3y-1 = lx$ for some $l$.
i.e. $9y-3 = l(3x) = l(ky+1)$ which implies $(9-kl)y = l+3$.

I am not able to procced further.
Any hints will be helpful. Thanks.
 [/INST] Multiplying the two divisibility relations together gives
$$xy\mid(3x-1)(3y-1)=9xy-3(x+y)+1\implies xy\mid3(x+y)-1$$
Now neither $x$ nor $y$ can be zero, since the original relations would then force $3x-1=3y-1=0$ which is impossible for integral $x,y$, so the quotient $\frac{3(x+y)-1}{xy}$ exists and is an integer with absolute value at least $1$. If we then assume without loss of generality that $|x|\ge|y|$, $-6\le y\le5$:

This means we can get all solutions by checking each possibility for $y$ in turn, which is easy since $x$ must then be a factor of the fixed integer $3y-1$. The solutions with $|x|\ge|y|$ are
$$(x,y)\in\{(-8, -5), (-13, -4), (-7, -2), (7, -2), (-4, -1), (-2, -1), (-1, -1), (1, -1), (2, -1), (4, -1), (-2, 1), (-1, 1), (1, 1), (2, 1), (-5, 2), (5, 2), (11, 4), (7, 5)\}$$
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