In answer to Pete L. Clark's question http://mathoverflow.net/questions/39510/  on Euclidean quadratic forms, I gave an example in seven variables, repeated below. Pete's Euclidean property is simply that for any point $\vec x \in \mathbf Q^7$ but $\vec x \notin \mathbf Z^7,$ we require that there be at least one $\vec y \in \mathbf Z^7$ such that $$ q(\vec x - \vec y) < 1. $$ I think my answer works (and the easier 6 variable one), based on extensive computer calculations, and Pete has been too polite to express much doubt.

Could someone please try to prove that this example works (and the 6 variable one)? It seems likely that this lies in the field [http://en.wikipedia.org/wiki/Geometry_of_numbers][1] but who can say? 


$$ q( \vec x) = x_1^2+ x_1 x_2 + x_2^2 + x_2 x_3 + x_3^2  + x_3 x_4 + x_4^2  + x_4 x_5 + x_5^2  + x_5 x_6 + x_6^2  + x_6 x_7 + x_7^2 + x_7 x_1. $$ This has the Euclidean property, its worst behavior is either when all $x_i = \frac{1}{4}$ or when all $x_i = \frac{3}{4},$ with ``Euclidean minimum'' equal to $\frac{7}{8}.$
Notice that with $\vec x = \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right),$ the integer lattice points $\vec y$ such that $ q( \vec x - \vec y)=\frac{7}{8} $ include $\vec y = \left( 0,0,0,0,0,0,0\right)$ and all seven cyclic permutations (including the identity) of
$\vec y = \left( 0,1,0,1,0,1,0\right),$ another seven for
$\vec y = \left( 1,0,0,0,0,0,0\right),$ another seven for
$\vec y = \left( 1,0,1,0,0,0,0\right),$ finally seven for
$\vec y = \left( 1,0,0,1,0,0,0\right),$ a total of 29 lattice points on the ellipsoid, of 128 in the standard unit 7-cube. The Gram matrix for the form is
$$  Q \; \; = \; \; 
\left(  \begin{array}{ccccccc}
  1 & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}\\\
  \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 \\\
  0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0  \\\
  0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0   \\\
  0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0    \\\
  0 &  0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2}     \\\
   \frac{1}{2} &   0 &  0 & 0 & 0 & \frac{1}{2} & 1
\end{array} 
  \right)  ,  $$
which has determinant $\frac{1}{32}$ and characteristic polynomial 
$$ \left( \frac{1}{64} \right) \left(x - 2 \right) \left(8 x^3 - 20 x^2 + 12 x - 1  \right)^2. $$ So the ellipsoids described are not oblate spheroids, there is less symmetry than that.


  [1]: http://en.wikipedia.org/wiki/Geometry_of_numbers