Problem: Find all ordered pairs of positive integers $(a,b)$ such that $\frac 1a + \frac1b = \frac 3{2018}$.
My beginning solution: Using basic algebra and factorization, we can see that the equation reduces to $$2 \cdot 1009 \cdot (a+b) = 3 \cdot a \cdot b$$
From this, we can get several key pieces of information:
i) At least one of $a,b$ must be divisible by $2$,
ii) At least one of $a,b$ must be divisible by $1009$, and
iii) $a+b$ must be divisible by $3$.
However, I don't know how to effectively utilize this information. The solutions that I have found so far have been $(1009,2018)$ and $(2018,1009)$, and I think that these are the only ones, but how do I prove this result?
Any help would be greatly appreciated, but I ask that you give HINTS ONLY.
Thank you in advance.