The smallest volume possible for a lattice with integer distances? Let $\Lambda \subset\mathbb{R}^n$ be a lattice satisfying $\|x-y\|_2^2 \in \mathbb{Z}$ for all $x,y\in\Lambda$. How small can $\text{vol}(\Lambda)=\det(\Lambda)$ be?
For example, in dimension $2$, the hexagonal lattice has smallest volume of any integral lattice, equal to $\frac{\sqrt{3}}{2}$. In higher dimensions, rescaling an even unimodular lattice by a factor of $\sqrt{2}$ yields a lattice with integral distances with $\text{vol}(\Lambda) = 2^{-\frac{n}{2}}$.
What is the smallest volume possible as $n\rightarrow\infty$?
 A: Note that $\sqrt2\Lambda$ is even, and $\mathrm{vol}(\sqrt2\Lambda)^2=\det(A)$, where $A$ is the Gram matrix of $\sqrt2\Lambda$. Thus $\mathrm{vol}(\Lambda)=2^{-n/2}\sqrt{\det(A)}$, so the smallest possible volume in dimension $n$ is determined by the smallest determinant of an even $n$-dimensional lattice.
The smallest determinant question is answered in the lemma of section 5 in Conway & Sloane's Lattices with Few Distances. The answer is $8$-periodic in $n$, given by




$n\ (\mathrm{mod}\ 8)$
$0$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
$8$




$\Lambda_0$
$\{0\}$
$A_1$
$A_2$
$D_3$
$D_4$
$D_5$
$E_6$
$E_7$
$E_8$


$\det A$
$1$
$2$
$3$
$4$
$4$
$4$
$3$
$2$
$1$




where the minimum for a particular $n$ is achieved (usually not uniquely!) by the orthogonal sum of $\Lambda_0$ with some number of copies of $E_8$.
That one cannot do better follows from the classification of even forms of small determinant, SPLAG 15.8, table 15.4: if $\det A<4$ then the mod-8 signature is $\pm (\det A-1)$.
Returning to the original question, this shows
\begin{align}
\liminf_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda)  =1,\\
\limsup_{n\to\infty}\ 2^{n/2}\mathrm{vol}(\Lambda)  =2.
\end{align}
A: After scaling your lattice by $\sqrt{2}$, the Gram matrix has integral entries, so the absolute value of its determinant, being a nonzero integer, is at least $1$. This gives a lower bound of $2^{-n/2}$.
