How to solve this complex number question?

Question

$i=\sqrt{-1}$ and $a,b$ and $c$ are positive integers and $$c = (a+ib)^3-191i$$ is given. Find $c$.

I expanded the equation that is given and wrote that

$$ c = a^3+3ia^2b-3ab^2-ib^3-191i \\ $$

Since $c$ is a positive integer

$$ \begin{align} i(3a^2b-b^3) = 191i \\ 3a^2b-b^3 = 191 \end{align} $$

should be written. And we write $c$ as

$$ c = a^3-3ab^2 $$

After that i can't conclude anything. Hints and solutions will be appreciated.

Answer 1

Hint:

Since 191 is a prime number and $b$ is common on the left so...

Answer 2

You already wrote

$$3a^2b-b^3 = 191$$ $$a^3-3ab^2= c$$

We can factor the equations

$$b(3a^2-b^2)=191$$ $$a(a^2-3b^2)=c$$

From the first equation, $191$ is prime thus $b$ is either $191$ or $1$, so you get two sets of two equations.

When $b=1$, I get $a=8$ and $c=488$.

When $b=191$, $a$ is not an integer and we have no solutions.