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Recently I found a theorem about Bézout domains (BD) and it says that an integral domain $D$ is a PID (principal ideal domain) iff $D$ is both an UFD (unique factorization domain) and a BD.

So I was wondering if there is an analogous result if we replace PID by euclidean domain (ED). More exactly, is there a theorem that says something like: $$ED \iff UFD + (\text{some special domain})\; \text{or}\\ ED \iff BD + (\text{some special domain})\;?$$

Well, after some googling I haven't found anything and it seems, as Hans Lundmark points out, that there is no such a characterization of euclidean domains. Is this related to the fact that UFDs and BDs have characterizations by ideal theoretic conditions whereas euclidean domains, as follows from this question, do not have it?

Thanks in advance.

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    Not very likely. At least none of the commonly encountered types of integral domains will work (Dedekind, Prüfer, Noetherian, ACCP, Krull, GCD, Schreier, atomic, integrally closed, ...). – 2017-02-11

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