Is it possible to solve for $a, b \in \mathbb{N}$?

Question

I need to solve the following equation so that both $a$ and $b$ are natural numbers.

$$ab - 2a = 2b$$

I must also prove that the solutions found are the only ones possible.

Is it possible to do so, and if yes how can I do it? Can this be solved as some type of diophantine equation?

Answer 1

Hint:

$$ab-2a=2b\iff (a-2)(b-2)=4$$

Answer 2

Hint $\ b\neq 2\,\Rightarrow\ a = \dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$

Answer 3

$(0,0)$ is a solution.

It can't be that $b = 1$, neither can it be that $b = 2$. Assume $b > 2$.

Considering the equation modulo $b - 2$, we get that $b - 2$ divides $4$. We conclude that $b - 2$ is either $1$, or $2$, or $4$. So $b = 3$ or $b = 4$ or $b = 6$. It happens that all of these give solutions.

Therefore the solutions are $(0,0)$, and the pairs for which $b \in \{3,4,6\}$.