Number of elements $a$ We have the ring $R=\mathbb{Z}_5[x]$ and the element $b=(x-4)(x-3)^2$ of $R$. Let $A$ the set of $a$ in $R$ such that $at=b$ has a solution $t\in R$. How can we determine the number of elements of $A$ ? 
Could you give me a hint? I don't really have an idea. 
 A: You are asking for the number of divisors of $b$. These are the polynomials of the type$$\alpha(x-4)^\beta(x-3)^\gamma,$$with $\alpha\in\mathbb{Z}_5\setminus\{0\}$, $\beta\in\{0,1\}$, and $\gamma\in\{0,1,2\}$.
A: OK, $\Bbb Z_5$ is a field, so $\Bbb Z_5[x]$ is a principal ideal domain, which implies primes and irreducibles are the same, and that we have unique factorization into primes.  Now note that $x - 3$ and $x - 4$ are irreducible, hence prime, in $\Bbb Z_5[x]$, so 
$b = (x - 3)^2(x - 4) \tag 1$
is the unique prime factorization of $b \in \Bbb Z_5[x]$.  If we first look for monic divisors of $b$,  again by uniqueness of factorization, $at = b$ must factor as (1); it follows then that the only possible monic $a$ are
$a = 1, x - 3,  (x - 3)^2,  (x - 4), (x - 3) (x - 4), (x - 3)^2(x - 4) = b; \tag 2$
BUT, you may say, we want all factors of $b$, whether monic are not; and indeed you would be correct to so interject; well the non-monic factors of $b$ are included if we take the general form $ca$, $c \in \Bbb Z_5^\times$; that is, $ca$ where $c$ is an invertible element of $\Bbb Z_5$; there are precisely $4$ such $c \in \Bbb Z_5^\times = \Bbb Z_5 \setminus \{0\}$, and since there are $6$ monic $a$, we must have
$\vert A \vert = 4 \cdot 6 = 24. \tag 3$
