The question is in the title, but employs some private terminology, so I had better explain.


Let $R$ be an integral domain with fraction field $K$, and write $R^{\bullet}$ for $R \setminus \{0\}$. For my purposes here, a **norm** on $R$ will be a function $| \ |: R^{\bullet} \rightarrow \mathbb{Z}^+$
such that for all $x,y \in R^{\bullet}$, $|xy| = |x||y|$ and $|x| = 1$ iff $x \in R^{\times}$. Also put $|0| = 0$.


[**Edit**: I forgot to mention that the norm map extends uniquely to a group homomorphism $K^{\bullet} \rightarrow \mathbb{Q}^{> 0}$.]


The norm is **Euclidean** if for all $x \in K \setminus R$, there exists $y \in R$ such that $|x-y| < 1$.


Now let $q(x) = q(x\_1,\ldots,x\_n)$ be a (nondegenerate) quadratic form over $R$, which I mean in the relatively naive sense of just an element $R[x\_1,\ldots,x\_n]$ which is homogeneous of degree $2$. Supposing that a norm $| \ |$ on $R$ has been fixed, I say that the quadratic form $q$ is **Euclidean** if for all $x \in K^n \setminus R^n$, there exists $y \in R^n$ such that $0 < |q(x-y)| < 1$.


(If $q$ is anistropic, then $x \in K^n \setminus R^n$, $y \in R^n$ implies that $x-y \neq 0$, so $|q(x-y)| \neq 0$, and the condition simplifies to $|q(x-y)| < 1$.)


As an example, the sum of $n$ squares form is Euclidean over $\mathbb{Z}$ iff $1 \leq n \leq 3$.


It is easy to see that the ring $R$ (with its fixed norm) is Euclidean iff the quadratic form $q(x) = x^2$ is Euclidean (iff the quadratic form $q(x,y) = xy$ is Euclidean), so the concept of a Euclidean quadratic form is indeed some kind of generalization of that of a Euclidean ring.


Conversely, the following is an obvious question that I have not been able to answer.



> 
> Suppose that a normed ring $R$ admits some Euclidean quadratic form $q$. Is $R$ then necessarily a Euclidean domain? In other words, can there be any Euclidean quadratic forms if $q(x) = x^2$ is not Euclidean?
> 
> 
> 


If the answer happens to be "no", I would of course like to know more: can this happen with an anisotropic form $q$? What can one say about a ring $R$ which admits a Euclidean quadratic form?


I have some suspicions that a domain which admits a Euclidean quadratic form is at least a PID. Indeed, let $q$ be a quadratic form over $R$. By an $R$-linear change of variables we may write it as $a\_1 x\_1^2 + \sum\_{j=2}^n a\_{1j} x\_1 x\_j + \sum\_{i=2}^n \sum\_{j=1}^n a\_{ij} x\_i x\_j$ with $a\_1 \neq 0$. Then if I take my vector $x$ to be $(x\_1,0,\ldots,0)$ for $x\_1 \in K \setminus R$, then the Euclidean condition implies the following: there exist $y\_1,z \in R$ such that $|a(x\_1-y\_1)^2 - z| < 1$. This is reminiscent of the Dedekind-Hasse property for a norm which is known to imply that $R$ is a PID. In fact, it is much stronger in that the $a \in R$ is fixed (and the $z$ is not arbitrary either). Unfortunately in place of an arbitrary element $x$ of $K$ we have a certain square, so this does not match up with the Dedekind-Hasse criterion...but it seems to me somewhat unlikely that a domain which is not a PID would satisfy it.


For more information on Euclidean forms, please feel free to consult


[http://alpha.math.uga.edu/~pete/ADCforms.pdf](http://alpha.math.uga.edu/%7Epete/ADCforms.pdf)




---


**Added**: a new draft which takes into account comments of J. Hanke and F. Lemmermeyer is available at


[http://alpha.math.uga.edu/~pete/ADCformsv2.pdf](http://alpha.math.uga.edu/%7Epete/ADCformsv2.pdf)


I warn that the new Section 2.1 showing that (primitive) binary Euclidean forms correspond to Euclidean ideal classes in quadratic orders in the sense of Lenstra -- as pointed out by Franz Lemmermeyer in his answer -- is rather miserably written at the moment, but at least it's there.



Must a ring which admits a Euclidean quadratic form be Euclidean?
Following Pete's request, I give the following as a second answer. 


Take $R = {\mathbb Z}[\sqrt{34}]$ and $q(x,y) = x^2 - (3+\sqrt{34})xy+2y^2$; observe that the discriminant of $q$ is the fundamental unit $\varepsilon = 35 + 6 \sqrt{34}$ of $R$, and that its square root generates $L = K(\sqrt{2})$ since $2\varepsilon = (6+\sqrt{34})^2$. Then q is Euclidean over $R$ since the ring of integers in $L = K(\sqrt{2})$ is generated over $R$ by the roots of $q$, and since $L$ is Euclidean by results of J.-P. Cerri (see Simachew's 
 [A Survey On Euclidean Number Fields](http://www.algant.eu/documents/theses/simachew.pdf)). But $R$ is not principal ($L/K$ is an unramified quadratic extensions), so the answer to your question, if I am right, is negative.


