Let $p, q \in \mathbb{N}$ be two prime elements and $f(x)=x^3-px^2+px-pq \in \mathbb{Z}[x]$. For all possible values of $p, q$, I have to study if in $\mathbb{Z}[x]/(f(x))$ can be exists an idempotent element $a \neq 0,1$.
If $p \neq q$, then $f(x)$ is irreducible by Eisenstein criterion thus $\mathbb{Z}[x]/(f(x))$ is a domain and such an element $a$ can't exists. If $p=q$, $f(x)=(x-p)(x^2+p)$ thus such an element can exist. Since $a^2=a$ in $\mathbb{Z}[x]/(f(x))$, then
$$a(a-1)=h(x)(x-p)(x^2+p).$$
$x-p$ and $x^2+p$ are prime elements thus the possible cases are: $x-p$ divides $a$ and $x^2+p$ divides $a-1$ or viceversa. So I search an element $a$ such that, $a=h_1(x)(x-p)$ and $a-1=h_2(x)(x^2+p)$. Is it possible to find it?
Hint: The constant terms of $h_1(x)(x-p)$ and $h_2(x)(x^2+p)$ are both divisible by $p$.