Solve the Diophantine equation $a+2b=2ab$, where $(a,b)$ are positive integers. Let $a$ be a positive integer. Show that $\gcd(a, a-1) = 1.$
Let $d$ be the greatest common divisor of $a$ and $a-1.$ I.e. $\gcd(a, a-1) = d$
Therefore, $d$ must divide $a-(a-1),$ following the rule that if any number, when some number d divides two numbers, then d must also divide the difference. 
In other words, $d\mid (a-(a-1)).$
If we simplify,  d must divide $a-a+1= 1$.
I have shown that the $\gcd(a, a-1) = 1,$ since d = 1. QED.

b) Use the result of part a) to solve the Diophantine equation $a+2b=2ab,\,$ where $a, b \in \mathbb Z$.
Let $2b = 2k,$ where $k$ is an integer and therefore $2b$ is even.
Case 1: $a$ is even
If $a$ is even, then $a = 2k,$ where k is an integer.
$2k + 2k = 2(2k)(1)$
$4k = 4k$
Both sides equal each other...
This is where I stopped. I realized I wasn't using the previous proof to solve this equation. This is where I need help.

Note we can rewrite your equation as
$$a + 2b = 2ab \iff a = 2b (a - 1) $$
Both sides of the equation are integers so it follows that $a $ divides $2b(a -1) $. But because of your lemma we know $a $ and $a-1$ are coprime so $a \mid 2b \rightarrow 2b = ak $ for some integer $k $. Substituting we get that
$$a = 2b (a - 1) \iff a = ak(a - 1) \iff 1 = \frac {ak(a-1)}{a} \iff 1 = k(a-1)$$
Because both $a $ and $k $ are positive integers it follows that $a - 1 = 1 \rightarrow a = 2$ and $k = 1 \rightarrow b = 1$.
We divided by $a $ in our calculations so $a \not= 0$. A second of thought shows $a = b = 0$ would also be a possible solution.
Another way of going about it, without the lemma $gcd(a, a-1) = 1$:
$2b $ is always an even number and so is $2ab $ so it follows, from the equality $a + 2b = 2ab $ that $a $ is even.
If $a $ is even, then $a = 2k $ for some integer k. Substituting in first equality we get $2k + 2b = 2 (2k)b \iff 2 (k + b) = 4kb \iff k + b = 2kb $
Now we divide both sides by $b $ and $k $ separately to get
$$\frac{k + b}{b} = 2k \iff \frac{k}{b} + 1 = 2k$$
$$ \frac{k + b}{k} = 2b \iff \frac{b}{k} + 1 = 2b$$
From the left sides of the right equalities you get that both $\frac{b}{k}$ and $\frac{k}{b}$ must be integers so you conclude $b = k $.
Plugging in you get that 
$$ k + b = 2kb \iff 2k = 2k^2 \iff k = k^2 \iff k = 0, 1$$
If you pick $k=0$ then $b = a = 0$. If you pick $k=1$ then $b = 1$ and $a = 2k = 2$.
