Solve $ab = 2(a+b)$ using modular arithmetic 
I have the following equation which I want to solve where $a,b \in \mathbb{N}$:
\begin{align*}ab &= 2(a+b) \\ \iff ab &= 2a + 2b \\ \iff ab - 2a &= 2b \\ \iff a(b-2) &= 2b \\ \iff a &= \frac{2b}{b-2}\end{align*}

Therefore I must solve for $a,b$ such that all pairs $\displaystyle(a,b) = \bigg(\frac{2b}{b-2},b\bigg)$ where $a,b\in \mathbb{N}$
I must now find, using modulo properties, the values of $a,b \in \mathbb{N}$ which satisfies the equation.
Clearly $(0,0)$ is a solution, but $0 \not\in \mathbb{N}$. We cannot have $b=1$ since then $a<1$ and $b =2$ is a horizontal asymptote.
Notice, however that $ \displaystyle a = \frac{2b}{b-2} = 2 + \frac{4}{b-2}$. Thus for $a$ to be an integer, we require $(b-2)|4$.
Hence we must have that $4 \equiv 0 \mod{(b-2)}$.
Thus we have the following equations:
$b-2 = 1 \implies b = 3 \implies a = 3$
$b-2 = 2 \implies b = 4 \implies a = 4$
$b-2 = 4  \implies b = 6 \implies a = 3$
Thus our solutions are given by $$\{ (6,3), (4,4), (3,6) \} $$
Is the steps and reasoning used correct?

HINT
$$a =\dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$$
That means, for $a$ to be an integer we must have $(b-2) | 4$
So $b-2$ has to be a factor of $4$.
