Let $K$ be a real quadratic number field of prime discriminant $D$. We define for $\mathfrak a$ being a fractional ideal of $K$, $n \in \mathbb Z$ and $b \in \mathbb N$ $$R(\mathfrak a,n,b) := \# \{ x \in \mathfrak a/b\mathfrak a : N(x)/N(\mathfrak a) \equiv n \pmod b \}.$$ A few facts I discovered:
- Because $D$ is prime, we have a unit of $\varepsilon \in \mathcal O_K$ of negative norm, hence $R(\mathfrak a,n,b)=R(\mathfrak a,-n,b)$.
- For fixed $\mathfrak a$ and $n$ the function $b \mapsto R(\mathfrak a,n,b)$ is multiplicative.
- For fractional ideals $\mathfrak a,\mathfrak b$ of the same ideal class we have $R(\mathfrak a,n,b)=R(\mathfrak b,n,b)$.
My conjecture: The value $R(\mathfrak a,n,b)$ does not depend on $\mathfrak a$. Hence, we have $R(\mathfrak a,n,b)=R(\mathfrak b,n,b)$ for all fractional ideals $\mathfrak a,\mathfrak b$.
Any ideas how to prove that? Because of my third fact the statement is only substantial if the class number $h_K>1$.