A Priori proof that Covering Radius strictly less than $\sqrt 2$ implies class number one 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. In the language of (positive) integral lattices, the condition is that the covering radius be strictly smaller than $\sqrt 2.$
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://alpha.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, based on taking the sum of a given lattice with a 2-dimensional Lorentzian lattice. From page 378 in SPLAG first edition, two lattices are in the same genus if and only if their sums with the same 2-dimensional Lorentzian lattice are integrally equivalent.  I hope someone posts a fuller answer, otherwise I'll be spending the next six months trying to complete the sketch myself. The references: Lattices like the Leech lattice, J.Alg. Vol 130, No. 1, April 1990, p.219-234, then earlier The Leech lattice, Proc. Royal Soc. London A398 (1985) 365-376. Quoted:

But there is a good way to
show that some genus of lattices L has class number 1 if its
covering
radius is small. What you do is look at the sum of L and a
2-dimensional
Lorentzian lattice. Then other lattices in the genus correspond
to some
norm 0 vectors in the Lorentzian lattice. On the other hand,
if the
covering radius is small enough one can use this to show that a
fundamental domain of the reflection group of the Lorentzian
lattice has
only one cusp, so all primitive norm 0 vectors are conjugate
and therefore
there is only 1 lattice in the genus of L. This works nicely when
L is the
E8 lattice for example.
There are some variations of this. When the covering radius is
exactly
sqrt 2 the other lattices in the genus correspond to deep holes,
as in the
Leech lattice (Conway's theorem). The covering radius sqrt 2
condition
corresponds to norm 2 reflections, and more generally one
can consider
reflections corresponding to vectors of other norms; see
http://dx.doi.org/10.1016/0021-8693(90)90110-A
for details

 A: EDIT, Tuesday, July 26. I have convinced myself, with my own C++ programs, that the sum of five squares does satisfy the "easier" Borcherds-Allcock condition, although it fails Pete L. Clark's criterion. Four squares works,  I believe that six and seven squares will work, while eight squares will be "borderline," with the worst behavior happening at the point with all coordinates being 1/2.  Now, as Prof. Nebe has designed a new algorithm for covering radius, and Magma uses this now, I can at least hope that she gets interested enough to design an algorithm for this hybrid condition. Otherwise it is going to take me forever to find all of the "odd" lattices that succeed. Not at all by the way, the original impetus for Prof. Nebe's involvement was a request by Richard Parker for the positive integral lattices P for which it would be possible to calculate Aut(P + U) where U is the unimodular two-dimensional "even" Lorentzian lattice... 
I finally figured out that Nebe gets every lattice to be "even" by doubling all coefficients, this particularly happening to the "odd" lattices such as the sum of $k$ squares. If we were to leave the odd lattices as they are, her condition (and Pete's) would be that the covering radius be strictly less than 1! The reality check is that both Nebe and Pete include the sum of three squares, but both exclude the sum of five squares. For comparison, see LINK
Before continuing, let me point out that this strong  condition happens only a few times, $$ x^2, 3 x^2, \; x^2 + y^2, x^2 + 2 y^2, 2 x^2 + 2 x y + 3 y^2, \; x^2 + y^2 + z^2, x^2 + 2 y^2 + 2 y z + 2 z^2,$$ in dimension no larger than 3.
Note that the earlier proof for "even" lattices does, in fact, apply to the "odd" lattices.
Both Richard Borcherds and his student, Daniel Allcock of U. T. Austin, pointed out that I did not have quite the "correct" condition. In the Duke Math. J., vol. 103 (2000) pp. 303-333, see LINK
section 6, Allcock discusses what he calls a "well-covering," which is a covering by balls of varied radius, where the radius of the ball around a given lattice point is decided by the norm of the lattice point. In particular, he says (but not in the paper) that the correct condition for ``odd'' lattices is that $\mathbb R^n$ is covered by {\bf open} balls of radius 1 around all lattice points of odd norm, then radius $\sqrt 2$ around lattice points of even norm, where the norm of some $x \in \Lambda$ is $x \cdot x.$ This is called a "strict well-covering." This condition is milder than Pete's.
Given such a positive  odd lattice $\Lambda$ with a strict well-covering as described, form the integral Lorentzian lattice $L =  \Lambda \oplus U.$ Elements are of the form 
$ (\lambda, m,n)   $ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by
$$   (\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product 
$$ (\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1  \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$
Note that the sublattice of all points with even norm has index 2, and that norm 0 vectors in $L =  \Lambda \oplus U$ can only be created 
from such points with even norm.
We choose a particular set of roots (elements of norm 2) beginning with any  $\lambda \in \Lambda$ of even norm by
$$ \tilde{\tilde{\lambda}} =   \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$ 
However, starting with points  $\beta \in \Lambda$ of odd norm, we create root-ones by
$$ \tilde{\beta} =   \left( \beta, 1,  \frac{ 1 - \beta^2}{2} \right),$$ which are elements of norm 1 in $L.$
Note that when an English soccer goalkeeper boots the ball down the center of the pitch and attack remains central, this is referred to as root-one football. I'm sure I have that right.
Let the group $ R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\tilde{\lambda}}$ and all the
$ \tilde{\beta}$ and by $\pm 1.$
Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$ s_r (z) = z - ( r \cdot z) r.$$
Also $ s_r^2(z) = z.$ 
Given any root-one $ \tilde{\beta} \in L,$ meaning $ \tilde{\beta}^2 = 1,$ we get the reflection $$ s_{\tilde{\beta}} (z) = z - 2(  \tilde{\beta} \cdot z)  \tilde{\beta}.$$
Also $ s_{\tilde{\beta}}^2(z) = z.$ 
Given some primitive null vector
$$ z = (\xi, a,b),$$
so that $ 2 a b = - \xi^2,$ and $b = \frac{- \xi^2}{2 a} \in \mathbb Z.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$
In the first case, suppose $|b| < |a|.$ Then $ | 2 a b| = \xi^2 < 2 a^2,$ so in fact
$$ \left( \frac{\xi}{a} \right)^2 < 2.$$
We choose the root $ \tilde{\lambda} =   \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and
$$  s_{\tilde{\lambda}} (z) = z - (  \tilde{\lambda} \cdot z)  \tilde{\lambda} =   (\xi, -b,-a).$$
Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that
$ b = \frac{- \xi^2}{2a}.$ Now we have
$ 2 a^2 \leq  | 2 a b| = \xi^2,$ so $ \left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some 
nonzero vector $\lambda \in \Lambda$ such that the rational number
$$   \left( \frac{\xi}{a} - \lambda \right)^2 < 2.   $$
If $\lambda$ has even norm, everything proceeds as before. 
However, it is possible the only point close enough is  $\beta \in \Lambda$ of odd norm. We create the root-one 
$ \tilde{\beta} =   \left( \beta, 1,  \frac{ 1 - \beta^2}{2} \right).$
By Allcock's strict well-covering condition,
$$   \left( \frac{\xi}{a} - \beta \right)^2 < 1.   $$
For convenience we write
$$  a' = a \left( \frac{\xi}{a} - \beta \right)^2  $$
From the equation $ 2 z \cdot \tilde{\beta} = a - a',$ we see that $a' \in \mathbf Z.$
Well, we have   $$  s_{\tilde{\beta}} (z) =  (\xi - ( a - a') \beta, a',b'), $$ where
$$ b' = b - ( a - a') \left(  \frac{1 - \beta^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left(  \frac{1 - \beta^2}{2} \right) .$$    
Now, if $a'=0,$ then $ s_{\tilde{\beta}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to
$w = (0, 0,1),$ with
$E(w) = \Lambda.$
If, instead, $a' \neq 0,$ note that our use of the strict well-covering  condition shows that
$ | a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and
$| a - a'| < |a|.$  From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \beta)^2,$ so
$b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\beta^2 \geq 1,$ so
$$  \frac{1 - \beta^2}{2} \leq 0,   $$ 
and $  ( a - a') \left( \frac{1 - \beta^2}{2} \right)$ has the same sign as $b$ and $b'.$
 From
$$ b' = b - ( a - a') \left( \frac{ 1 -\beta^2}{2} \right)$$ 
we conclude that $| b'| \leq |b|.$
So, these steps also have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w = (0, 0,1).$
A: This is the more important case of "even" lattices, where all inner products are integral and all vector norms are even. We follow pages 131-134 in Wolfgang Ebeling, Lattices and Codes, available for sale at LINK 
\$45.00 for paperback.
Given an even positive lattice $\Lambda$ with covering radius below $\sqrt 2,$ form the integral Lorentzian lattice $L =  \Lambda \oplus U.$ Elements are of the form 
$ (\lambda, m,n)   $ where $\lambda \in \Lambda, \; m,n \in \mathbb Z.$ The norm on $L$ is given by
$$   (\lambda, m,n)^2 = \lambda^2 + 2 mn.$$ We infer the inner product 
$$ (\lambda_1, m_1,n_1) \cdot (\lambda_2, m_2,n_2) = \lambda_1  \cdot \lambda_2 + m_1 n_2 + m_2 n_1.$$
We choose a particular set of roots (elements of norm 2) beginning with any $\lambda \in \Lambda$ by
$$ \tilde{\lambda} =   \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$ 
Let the group $ R \subseteq \mbox{Aut}(L)$ be generated by reflections in all the $\tilde{\lambda}$ and by $\pm 1.$
Given any $ l \in L,$ a primitive null vector, we have $l \cdot l = 0,$ and so $l \in l^\perp.$ Furthermore, if 
$k \in l^\perp$ as well, then $ (k + l)^2 = k^2.$ As a result, we may form the lattice spanned $\langle l \rangle$ by $l$ itself, then form another lattice with norm, $$ E(l) = l^\perp / \langle l \rangle.$$ It took me a bit of doing to confirm (a calculus exercise along with Cauchy-Schwarz) that $E(l)$ is positive definite, all norms are positive except for the $0$ class.
Given any root $r \in L,$ meaning $r^2 = 2,$ we get the reflection $$ s_r (z) = z - ( r \cdot z) r.$$
Also $ s_r^2(z) = z.$
It is an exercise to show that, when $ s_r(z) = y,$ then $E(z)$ and $E(y)$ are isomorphic. 
There is rather more than first appears to the question:
Lorentzian characterization of genus 
In particular, every even lattice in the same genus as  $\Lambda$ occurs as some  $E(u),$ where
$u  \in L$ is a primitive null vector. Furthermore, taking $w = (0, 0,1),$ we find that
$E(w) = \Lambda.$
So, all we really need to do to prove class number one is show that every primitive null vector can be taken to $w = (\vec{0}, 0,1),$ 
by a sequence of operations in $R.$
Given some primitive null vector
$$ z = (\xi, a,b),$$
so that $ 2 a b = - \xi^2.$ Note that, in order to have a primitive null vector, if one of $a,b$ is 0, then $\xi = 0$ and the other one of $a,b$ is $\pm 1.$
In the first case, suppose $|b| < |a|.$ Then $ | 2 a b| = \xi^2 < 2 a^2,$ so in fact
$$ \left( \frac{\xi}{a} \right)^2 < 2.$$
We choose the root $ \tilde{\lambda} =   \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and
$$  s_{\tilde{\lambda}} (z) = z - (  \tilde{\lambda} \cdot z)  \tilde{\lambda} =   (\xi, -b,-a).$$ 
Therefore, we may always force the second case, which is $|a| \leq |b|.$ We assume that $a \neq 0,$ so that
$ b = \frac{- \xi^2}{2a}.$ Now we have
$ 2 a^2 \leq  | 2 a b| = \xi^2,$ so $ \left( \frac{\xi}{a} \right)^2 \geq 2.$ From the covering radius condition, there is then some 
nonzero vector $\lambda \in \Lambda$ such that the rational number
$$   \left( \frac{\xi}{a} - \lambda \right)^2 < 2.   $$
We form the root from this  $\lambda,$ as in
$$ \tilde{\lambda} =   \left( \lambda, 1, 1 - \frac{\lambda^2}{2} \right).$$
For convenience we write
$$  a' = \frac{a}{2} \left( \frac{\xi}{a} - \lambda \right)^2  $$
From the equation $z \cdot \tilde{\lambda} = a - a',$ we see that $a' \in \mathbb Z.$
Well, we have   $$  s_{\tilde{\lambda}} (z) =  (\xi - ( a - a') \lambda, a',b'), $$ where
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right) = \frac{- \xi^2}{2a} - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right).$$    
Now, if $a'=0,$ then $ s_{\tilde{\lambda}} (z) = (0,0,\pm 1),$ and an application of $\pm 1 \in R$ takes us to
$w = (0, 0,1),$ with
$E(w) = \Lambda.$
If, instead, $a' \neq 0,$ note that our use of the covering radius condition shows that
$ | a'| < | a|,$ while $a,a'$ share the same $\pm$ sign, that is their product is positive. So $a - a'$ shares the same sign, and
$| a - a'| < |a|.$  From $2 a b = - \xi^2$ we know that $b$ has the opposite sign. But $2 a' b' = - (\xi - ( a - a') \lambda)^2,$ so
$b'$ has the opposite sign to $a'$ and $a-a'$ and the same sign as $b.$ Now, $\lambda^2 \geq 2,$ so
$$ 1 - \frac{\lambda^2}{2} \leq 0,   $$ 
and $  ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$ has the same sign as $b$ and $b'.$
 From
$$ b' = b - ( a - a') \left( 1 - \frac{\lambda^2}{2} \right)$$ 
we conclude that $| b'| \leq |b|.$
So, these steps have $|a| + |b|$ strictly decreasing, until such time that one of them becomes 0, and we have arrived at $w.$ So, actually, the covering radius hypothesis implies that all primitive null vectors are mapped to $w$ by a finite sequence of reflections, so that all the $E(z)$ are in fact isomorphic to $E(w) = \Lambda.$ That is, all lattices in the genus are in fact isomorphic, and the class number is one. 
