It turns out that each of Pete L. Clark's "euclidean" quadratic forms, as long as it has coefficients in the rational integers $\mathbb Z$ and is positive, is in a genus containing only one equivalence class of forms. Please see these for background:

Intuition for the last step in Serre's proof of the three-squares theorem

Is the square of the covering radius of an integral lattice/quadratic form always rational?

Must a ring which admits a Euclidean quadratic form be Euclidean?

http://www.math.rwth-aachen.de/~nebe/pl.html

http://www.math.rwth-aachen.de/~nebe/papers/CR.pdf

I have been trying, for some months, to find an a priori proof that Euclidean implies class number one. I suspect, without much ability to check, that any such Euclidean form has a stronger property, if it represents any integral form (of the same dimension or lower) over the rationals $\mathbb Q$ then it also represents it over $\mathbb Z.$ This is the natural extension of Pete's ADC property to full dimension. Note that a form does rationally represent any form in its genus, with Siegel's additional restriction of "no essential denominator." If the ADC property holds in the same dimension, lots of complicated genus theory becomes irrelevant.

EDIT: Pete suggests people look at §4.4 of http://math.uga.edu/~pete/ADCFormsI.pdf .

EDIT 2: It is necessary to require Pete's strict inequality, otherwise the Leech lattice appears.

So that is my question, can anyone prove a priori that a positive Euclidean form over $\mathbb Z$ has class number one?

EDIT 3: I wrote to R. Borcherds who gave me a rough idea. Let me at least repeat that Pete's Euclidean property is usually referred to as an integral lattice having a covering radius less than $\sqrt 2.$