We need some notation before we state the problem. Let $K$ be a quadratic number field, $d$ its discriminant. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $2$. Let $D$ its discriminant. Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$. By this question, $R = [1, \omega]$, where $\omega = \frac{(D + \sqrt D)}{2}$. Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$. We denote $\sigma(\alpha)$ by $\alpha'$ for $\alpha \in R$. We denote $\sigma(I)$ by $I'$ for an ideal $I$ of $R$. Let $f$ be the order of $\mathcal{O}_K/R$ as a $\mathbb{Z}$-module. Then $D = f^2d$ (see this question). Let $p$ be a prime number such that gcd$(p, f) = 1$. Then $pR$ is regular by this question. Hence $pR$ is uniquely decomposed as a product of regular prime ideals by this question. So it is natural to ask how it is decomposed. I came up with the following proposition.
Proposition Let $K, R$, etc. be as above. Let $p$ be an odd prime number such that gcd$(p, f) = 1$.
Case 1 $D$ is divisible by $p$.
$P = [p, \omega]$ is a prime ideal and $pR = P^2$. Moreover $P = P'$.
Case 2 gcd$(D, p) = 1$ and $D$ is quadratic residue modulo $p$.
Let $b, b'$ be solutions of $(2x + d)^2 \equiv d$ (mod $p$) such that $b - b'$ is not divisible by $p$. Then $P = [p, b + \omega]$ and $P' = [p, b' + \omega]$ are distinct prime ideals and $pR = PP'$.
Case 3 gcd$(D, p) = 1$ and $D$ is quadratic non-residue modulo $p$.
$pR$ is a prime ideal.
Method of my proof I used the result of this question.
My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.