TITLE: Can every symmetric, unimodular and positive definite $G\in\mathbb{Z}^{n\times n}$ be written as $G=U^TU$?
QUESTION [5 upvotes]: Let $G\in\mathbb{Z}^{n\times n}$ be symmetric, unimodular and positive definite. Does there exist a unimodular matrix $U\in\mathbb{Z}^{n\times n}$ such that $G=U^TU$? I now that the result is true if $n=2$, but I have a feeling that it fails if $n$ gets larger.

REPLY [6 votes]: No. This is essentially asking whether every unimodular lattice is isometric to $\Bbb Z^n$.
This is false from $n=8$ onwards, and the $n=8$ counterexample is the $E_8$ root
lattice. I think an appropriate $G$ is
$$\pmatrix{2&0&0&0&0&0&0&1\\0&2&1&1&1&1&1&-2
\\0&1&2&1&1&1&1&-2
\\0&1&1&2&1&1&1&-2
\\0&1&1&1&2&1&1&-2
\\0&1&1&1&1&2&1&-2
\\0&1&1&1&1&1&2&-2
\\1&-2&-2&-2&-2&-2&-2&4
}.$$
I should add that this matrix is, or at least is intended to be $VV^T$ where
$$V=\pmatrix{1/2&1/2&1/2&1/2&1/2&1/2&1/2&1/2\\
0&1&0&0&0&0&0&-1\\
0&0&1&0&0&0&0&-1\\
0&0&0&1&0&0&0&-1\\
0&0&0&0&1&0&0&-1\\
0&0&0&0&0&1&0&-1\\
0&0&0&0&0&0&1&-1\\
0&0&0&0&0&0&0&2}$$
so that being unimodular and positive definite is clear. Also it cannot
have the form $UU^T$ for $U$ a unimodular integer matrix, since such
a matrix would have at least one odd diagonal entry.