Let $K$ be an imaginary quadratic field and $O\_K$ be its ring of integers. We say $O\_K$ is norm Euclidean if the norm is a [Euclidean function.](https://en.wikipedia.org/wiki/Euclidean_domain#Definition)  It is known from the classification of imaginary quadratic fields with class number 1 that $O\_K$ is Euclidean if and only if it is norm Euclidean. Is there a more straightforward proof of this fact?



Imaginary quadratic fields: Euclidean if and only if norm Euclidean
Assume that $O\_K$ is Euclidean. The Motzkin Sets $\_j$ are defined by
$E\_0 = \{0\}$, $E\_1 = E\_0 \cup O\_K^\times$, $E\_2$ is the set of elements of $O\_K$ such that each residue class is represented by an element in $E\_1$ etc.
$O\_K$ is Euclidean if every element of $O\_K$ is in some $E\_i$ (see Sect. 2.3 
[here](http://www.rzuser.uni-heidelberg.de/~hb3/publ/survey.pdf)). Since $E\_1$ has $3$ elements, $O\_K$ must have an element of norm $\le 3$, which implies that the discriminant of $K$ is $\ge -11$. All these rings are norm-Euclidean, and the converse is trivial. 


This is classical and has nothing to do with the solution of the class number $1$ problem.


