$a=\frac{3b}{b-3}$ Find all values of $b$ where $a$ is a positive integer.

Question

Here is the problem:$$ a = \frac{3 b}{b-3} $$ $a$ and $b$ are positive integers. You have to find all the possible values of $b$ where $a$ is a positive integer. Once all the values of $b$ has been found, you have to prove that they are the only possible values of $b$.

What I've attempted

I've found 2 values for a and b: (4, 12) and (6, 6). I got this mostly by luck, which is not what I'm aiming for. I've also wrote a code which confirmed that these were the only combinations up to $b=100,000$, however I would like a mathematical method to solve these kind of questions.

Since $a$ is an integer, this implies that $3b$ is divisible by $b-3$. I wrote expressed this as $3 b = (b-3)\cdot m$, where $m$ is a natural number. I tried use this to prove that the values I found above were the only values possible, however to no avail.

Could someone please explain how to approach and solve problems like this (where you have to find all possible values and prove that they are the only possible values)? Thank you for your time.

Answer 1

Hint:

$$\dfrac{3b}{b-3}=\dfrac{3(b-3)+9}{b-3}=3+\dfrac9{b-3}$$

Now what are Aliquot Divisors of $9?$

Note $a,b>0$

Answer 2

Hint:

First we observe that $b > 3$ follows that the numerator is greater than $9$. Then we do the following: $$\frac{3b}{b - 3} = \frac{3b}{b - 3}\times 1 = \frac{3b}{b - 3}\times \frac{b + 3}{b + 3}$$ We do this because $b + 3$ is the conjugate of $b - 3$. $$\frac{3b}{b - 3} = \frac{3b(b + 3)}{(b - 3)(b + 3)} = \frac{3b^2 + 9}{b^2 - 9}$$

$$\therefore a(b^2 - 9) = ab^2 - 9a = 3b^2 + 9 \Leftrightarrow a = \frac{b^2(a - 3) - 9}{9} = \frac{1}{9}b^2(a - 3) - 1$$

This implies that $9\mid b^2(a - 3)$ but pursuant to the examples of $(a, b)$ you have found, it seems like $9\mid b^2$, though this is just a mere assumption. Hope this makes the problem easier :)