Pairs of integers $(a,b)$ such that $\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$ How many pairs of integers are there $(a,b)$ with $a \leq b$ such that $$\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$$
My attempt:
Clearing fractions we get $$ab = 6(a+b)$$
$$ \Longrightarrow \frac{ab}{a+b} = 6$$
Which means that $ab$ is a multiple of $a+b$ since the answer is an integer. Also $ab \equiv 0\pmod{a+b}$.
Any ideas or a hint to help me continue? Thanks.

HINT : $$ab=6(a+b)\iff (a-6)(b-6)=36.$$
