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. 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?
1 Answer
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). 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 normEuclidean, and the converse is trivial.
This is classical and has nothing to do with the solution of the class number $1$ problem.

1$\begingroup$ Motzkin has a short five page paper, titled the Euclidean algorithm, with a proof of the classical fact. His paper can be found in the references of the linked paper above. $\endgroup$ Jun 6, 2016 at 23:29