Let $x$ be any solution to an equation of the form $x^2 + ax + b = 0$, where $a$ and $b$ are integers.
We define any particular set/ring $\textbf Z[x]$ to be all numbers of the form $c + dx$, where $c$ and $d$ are integers. We note that there are two $x$'s. I know from experience that appending one to the integers automatically brings in the other as the two solutions are related by the equation $x = -a - x'$. This leads to the definition of the conjugate for this arbitrary collection of "Quadratic Rings" which each consist of "Quadratic Integers".
Definition: The Conjugate of a Quadratic Integer $z = c + dx$ is $\bar{z} = c - d(a+x)$
The multiplication of any element and its conjugate is the norm of the Quadratic Integer. This is formalized below.
Definition: The Norm of a Quadratic Integer $z = c + dx$ is $N(z) = z\bar{z} = c^2 - cda - axd^2 - d^2x^2$.
^^Hopefully I did that multiplication right.
So this leads me to a conjecture based on an earlier question I had that was for the case of $\textbf Z[\sqrt{3}]$:
Conjecture: If there exists quadratic integers $z$ and $y$ and an ordinary integer $c$, such that $N(z) = N(y)c$, then there exists a quadratic integer $w$ such that $N(w) = c$.
I am seeking a proof of the conjecture that isn't too complex. Rigorous is fine and formality is as well. The problem is that there isn't a whole ton of algebraic number theory knowledge that I have. Primarily I know of integers of rational sets (such as $\textbf Q[x]$), Euclidean domains, modular arithmetic, and what is in this question. So primarily thinking that this question has to do with either solving equations or factoring but I can't quite put my finger on it. I see no reason for it to not be true in the general case, but if it isn't I would like it if someone could say when it is true and perhaps prove that particular case.
Also, I am aware that the norm and conjugate operations distribute across multiplication. Beyond that, I've never looked beyond the $\textbf Z[\sqrt{3}]$ in this rigorous of a sense.
This is not true in general. For example consider the case $a=0, b=4$, so $R = \Bbb Z[2i]$
$N(c+dx) = |c+2di|^2 = c^2+4d^2$.
$x = 2i$ has norm $4$, $2+x = 2+2i$ has norm $8$, but there is no element of norm $2$ (because that would require $d=0$ and then $c^2=2$)
However, your conjecture is true if $R$ is integrally closed.