Root systems and sums of squares It is easy to see that the quadratic form for the root system $A_n$ is a sum of $n+1$ squares of integral linear forms:
$$q_{A_n} = 2 \sum_{i=1}^n x_i^2 - 2 \sum_{i=1}^{n-1} x_i x_{i+1} =
x_1^2 + (x_1-x_2)^2 + \dots (x_{n-1}-x_n)^2 + x_n^2$$
It is equally easy to see that $q_{D_n}$ is a sum of $n$ squares. It is a little harder to see but I think is true that $q_{E_n}$ ($n=6,7,8$) is not a sum of $\ge n$ squares of integral forms.
Question: is this a standard fact, well-known to experts? Is there a standard reference? (I hate to reinvent a bycicle.) And has this fact been used for something interesting? (I have an interesting application in mind, so I am looking for connections...)
 A: If a quadratic form in $n$ variables is the sum of the squares of
$n$ integer linear forms, it's the sum of the squares of $n$ rational linear forms.
Thus it's equivalent
as a rational quadratic form to $x_1^2+\cdots+x_n^2$. In particular
its discriminant is a square. This rules out $E_6$ and $E_7$.
The case of $E_8$ is trickier, as it is equivalent over the rationals
to $x_1^2+\cdots+x_8^2$. This time you have to show there's no
equivalence over the integers, but one form takes solely even values
and the other doesn't.
Added
The first time round I didn't clock the $\ge n$ condition. But there's certainly
a way to put an upper bound on the number $m$ of linear forms one needs.
I'll stick to the $E_8$ form. One can think of this as describing a lattice
$L$ in Euclidean space. This lattice $L$ is self-dual, unimodular and even.
Its shortest vectors form the set of 240 roots $R$. This set of roots
has the nice property (I think it may be called the eutactic property
or the perfect property; all this is in Martinet's book on lattices)
that $x\mapsto\sum_{y\in R}(x\cdot y)^2$ is proportional to the quadratic
form $x\mapsto x\cdot x$. I think actually $\sum_{y\in R}(x\cdot y)^2=30x\cdot x$
as $30=240/8$.
Now an integer linear form is a linear form taking integer
values on the lattice $L$ and so is $x\mapsto x\cdot z$ for some $z$ in the dual
of $L$, so here $z\in L$. If $x\cdot x=\sum_{j=1}^m(x\cdot z_j)^2$
for $z_j\in L$ then
$$480=2|R|=\sum_{y\in R}y\cdot y=\sum_{j=1}^m\sum_{y\in R}(y\cdot z_j)^2
=30\sum_{j=1}^m z_j\cdot z_j.$$
Each $z_j\cdot z_j\ge 2$ so we must have $m=8$ and each $z_j\cdot z_j=2$.
You should check my numbers... For $E_6/E_7$ the dual lattice is different
from the original but I'm sure they still have the eutactic(?) property. Any way
one can get an effective upper bound on $m$, probably not much bigger than $n$.
A: Just saw this thanks to a "Related" link from
Question 154928.
Yes, it is known that the $E_6$ form cannot be written as a sum of
integral squares, and thus (by specialization) that the same is true of
$E_7$ and $E_8$; moreover the representations of $D_n$ ($n>2$) and $A_n$
as the sum of $n$ and $n+1$ squares respectively are the only ways
(up to isomorphism) to write these forms as the sum of any number of
nonzero squares, with the exception of $D_3 \cong A_3$ which has
both a three-square and a four-square representation.
Or at least it is known once one makes Will Jagy's
key observation that writing a form as a sum of $m$ integral squares is
tantamount to embedding the corresponding lattice into ${\bf Z}^m$.
(As it happens I was just asked a few days ago whether any integral
positive-definite lattice can be embedded in some ${\bf Z}^m$,
so this question was very familiar.)
I don't know a reference, but the result is not hard starting from the
Coxeter diagrams
and the fact that the vectors of norm $2$ in ${\bf Z}^m$ are exactly
the vectors $e+e'$ for some $\pm$ unit vectors $e,e'$ with $e' \neq \pm e$.
For $A_n$ we need a sequence of $n$ such vectors any two of which
are orthogonal except that consecutive vectors have inner product $-1$.
The first two must be $e'-e, \, e''-e'$ with $e,e',e''$ orthogonal
unit vectors.  The third could be either $e'''-e''$ or $e+e'$.
The latter choice does not extend to $A_4$,
and the former extends uniquely to $e''''-e'''$, and then by induction to
$\{ e^{(i)} - e^{(i-1)} \}_{i=1}^n$ with all $n+1$ unit vectors orthogonal.  
For $D_n$ we need an $A_{n-1}$ configuration together with a norm-$2$ vector
orthogonal to all but the second vector, with which it has inner product $-1$.
For $n=3$ we've done this already because the $D_3$ and $A_3$ diagrams
are isomorphic.  For $n=4$ either of our two $n=3$ solutions extends uniquely
and both give $e'-e, e''-e', e'''-e'', e+e'$.  For all $n > 4$ the unique
$A_{n-1}$ diagram extends uniquely, again with extra vector $e+e'$.
We can now obtain the impossibilty of the $E_6$ configuration
by trying to extend either $D_5$ at a short end or $A_5$ at the middle vertex
(or by trying to overlap $D_5$ with $A_5$ or with another $D_5$).
Since $E_7$ and $E_8$ contain $E_6$, they are impossible too.  
Hence none of the $E_n$ lattices are contained in any ${\bf Z}^m$, 
whence the corresponding quadratic forms are not sums of integral squares, 
QED.
A: If a quadratic form $x^TAx$ (for lattice $L$) in $n$ variables is expressible as a sum of $m (\ge n)$ squares of integral linear forms
then $x^TAx=||Fx||^2$, where  $F$ is an $m \times n$ integral matrix and the columns of $F$
gives an explicit embedding of $L$ as a sublattice of $\mathbb{Z}^m$. For $E_8$ and $m=8$, this will mean
$E_8=\mathbb Z^8$ since both have the same volume, which is not possible. Also from $A=F^TF$ and the Cauchy-Binet formula,
$\det A$ is a sum of  $m \choose n$ integral squares which are squared volume of the projections.
Since $\det E_8=1$, there can only be a single term 1 and the rest are zero and this means $E_8$ is embedded inside some
$\mathbb Z^8$ inside $\mathbb Z^m$ so this reduces to the case $m=8$. For $E_6,E_7$, the case $m=n$ can be ruled out since $\det A$ is
not a square as Robin Chapman noted. For $m>n$, since $2=1+1,6=1+1+4=1+1+1+1+1+1$ are the only partition of $\det A$ a a sum of squares, 
many projections have to be zero which means large $m$ can be reduced to smaller $m$. Can this be used to reduce to the case $m=n$ ?     
A: Fairly recent work by Ellenberg and Venkatesh, later improved by Schulze-Pillot, show that it is reasonable to hope that $n+3$ squares of linear forms suffice. See Theorem 11, bottom of pdf page 8
http://arxiv.org/abs/0804.2158 
Of the hypotheses involved, the more serious is that of sufficiently large minimum, as your quadratic forms have very small minima. Well, if you have favorite expressions for the $E_n$ quadratic forms, let me know, I can probably program something definitive up to $n+3.$
