In a ring $\mathbb{Z}[\sqrt{k}]$ with positive non-square $k$, all associates obviously have the same norm as each other. But is the converse true, i.e. if two elements have the same norm, are they necessarily associates?
In other words, if $a+b\sqrt{k}$ and $c+d\sqrt{k}$ are elements such that $a^2-b^2k=c^2-d^2k$, does there necessarily exist a unit $u+v\sqrt{k}$ in $\mathbb{Z}[\sqrt{k}]$ such that $c+d\sqrt{k}=(a+b\sqrt{k})(u+v\sqrt{k})$?
OK, so it turns out to be easy to find counterexamples to answer my own question: conjugates certainly have the same norm, but need not be associates.
I was particularly interested in primes. Here is a numerical example: $(5,2)$ and its conjugate $(5,-2)$ are both prime in $\mathbb{Z}[\sqrt{2}]$, since their norm is $5^2-2\times2^2=17$, which is prime. If we try dividing $(5,2)$ by $(5,-2)$ (which would give a unit in $\mathbb{Z}[\sqrt{2}]$ if they are associates), we get $(33+20\sqrt{2})/17$, which is not in $\mathbb{Z}[\sqrt{2}]$.