It is well-known that $\mathbb{Z}+\mathbb{Z}\sqrt{5}i$ is not a unique factorization domain, since, for example, $6$ has two different factorizations into irreducibles: $(1+\sqrt{5}i)(1-\sqrt{5}i)=2\cdot3$.
Let $p,q,r \in \mathbb{N}$ be odd primes (not necessarily three different). Perhaps we should also require that each of $\{p,q,r\}$ is congruent to $1$ mod $4$ (perhaps no).
Let $R_{p,q,r}:=\mathbb{Z}+\mathbb{Z}\sqrt{p}i+\mathbb{Z}\sqrt{q}j+\mathbb{Z}\sqrt{r}k$, where $i^2=j^2=k^2=-1$, $ij=k$ etc., as in the quaternions (replacing $\mathbb{R}$ by $\mathbb{Z}$).
Trivially, $R_{p,q,r}$ is an integral domain, because it is a subset of $\mathbb{C}$.
Of course,$R_{p,q,r}$ is not a UFD, since, for example, $(1-\sqrt{p}i)(1+\sqrt{p}i)=1-pi^2=1+p=1+(1+4u)=2+4u=2(1+2u)$.
What else can be said about those integral domains?
For example: I guess that $R_{p,q,r}$ is a Euclidean domain, isn't it? Is there a known proof for this or one just needs to adjust the proof for $\mathbb{Z}+\mathbb{Z}\sqrt{5}i$?
This question is relevant.
Any hints and comments are welcome!
