As an example: $$x_{n+1}=(3x_n+b)\quad mod\quad 20 \quad (1)$$ Here we can choose b and $x_0$ arbitrarily. Instead I wrote an equivalent form of the above equation:$$3x_n+b\equiv x_{n+1}\pmod {20} \quad (2)$$
The solution to the second equation I find as: $$x_n = 7(x_{n+1}-b)\quad mod\quad20 \quad (3)$$.
Further to find the periode i take the first $x_{n+1}$ which is equal to $x_0$: $$x_{n+1} = x_0 = (3x_n+b) \quad mod \quad 20 \quad (4)$$
I substitute equation (3) in equation (4) and find:
$$x_0 = (((x_0-b) \quad mod \quad 20) \quad +b) mod \quad 20$$.
I got stuck here. My plan is to solve the last equation for $x_0$ and conclude something.
I already tried to gain some idea via coding it in python, so far i only found only twee different periode for this problem, nl. 2 and 4.
I ain't quite sure if my strategy gonna work or maybe I'm in wrong path.