Can someone confirm my reasoning?
If I understood this correctly, this is a special case of a linear Diophantine equation $ax +by = c$ where $c = ab$ and $x,y = 2018.$
In this case, an integer solution exists if $c$ is a multiple of $\gcd(x,y)$. In this case, $c = \gcd(x,y)\cdot 1$. Therefore, integer solutions certainly exist.
Here, certainly we cannot consider $a =b = 0$ as a solution, since it invalidates our original equation in the title.
I tried to rearrange the equation to $$(7a-2018)(7b-2018) = 1009^2 \cdot 2^2$$ and considered this equation $\mod 7$. I noticed each factor is congruent to $5 \mod 7$ and the solution is congruent to $4 \mod 7$.
The difficulty came when I checked to see if any factor pair arrangement, $(2, 2^2\cdot 1009^2), (2^2, 1009^2), (2^2\cdot 1009, 1009), (1, 2^2 \cdot 1009^2)$ is congruent to $5\mod 7$, but none of them were. That is to say, $7a-2018 = 2$ and its pair $7b-2018 = 2^2 1009^2$ are both congruent to $5\mod 7$, etc. However, none of them are.
How am I meant to conclude that these have no pairs $(a,b)\in \mathbb{N}$ satisfy the above equation? Is there an easier lemma, theorem or way of thinking about this that can generalize this to any $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ where $a,b,c,d \in \mathbb{N}$?
