<s>[INST] On largest box not containing integer vector solutions I am trying to understand the largest cube in $\Bbb Z^4$ around origin not containing integer vector $(a,b,c,d)$ solutions to $$aBD+bBC+cAD+dAC=0$$ where $gcd(AB,CD)=gcd(AC,BD)=1$ when $n<A,B,C,D<2n$ and $n/8<|A-B|,|C-D|,|A-C|,|A-D|,|B-C|,|B-D|<n/4$ holds?
Is the box $c'n^{2/3}$ in length around origin for some constant $0<c'$? That is a vector $(a,b,c,d)\in\Bbb Z^4$ with $|a|,|b|,|c|,|d|<c'n^{2/3}$ exists?
This is where I get the $2/3$ value.
$$aBD+bBC+cAD+dAC=0$$ 
$$B(aD+bC)=-A(cD+dC)$$ 
$$cD+dC=\ell B$$
$$aD+bC=-\ell A$$
If size of $\ell$ is $n^\eta$ then size of $\ell B$ is size of $cD+dD$ is $n^{1+\eta}$. We have $n^{2\eta}$ values in $cd$ and $n^{2\eta}$ values in $ab$ and we want to hit $\ell B$ and $\ell A$ (a total of $2n^{\eta}$ possibilities) of a total of $n^{2(1+\eta)}$ possibilities. The expected intersection is 
$$\frac{n^{2\eta}n^{2\eta}2n^{\eta}}{n^{2(1+\eta)}}=\frac{2n^{3\eta}}{n^2}$$
If $3\eta>2$ or $2/3<\eta$ we have expected intersection $\gg1$. 
 [/INST] October 4, 2017
You had a question for a few hours about doubling the dimension; maybe that was yesterday.  We are looking for a basis for the lattice inside $\mathbb Z^8$ of vectors orthogonal to
$$ (EBD, EBC, EAD, EAC,FBD, FBC, FAD, FAC )  $$ 
where we define integers such that
$$ MB - NA =1,  $$
$$ JD-KC=1.  $$
A 7 by 8 matrix with rows giving a basis is
$$
R =
\left(
\begin{array}{rrrrrrrr}
  A & 0  & -B  & 0 & 0 & 0 & 0 & 0 \\
  0& -A  &0  & B &0  & 0 & 0 &0 \\
  0& 0 &0  &0  & A &0 & -B  & 0  \\
  0& 0 & 0 &0  & 0 & -A  &0  & B \\
  MC& -MD & -NC  & ND  & 0 & 0 & 0 & 0\\
 0 & 0 & 0 & 0 & MC & -MD & -NC & ND \\
  FMJ & -FMK  & -FNJ  & FNK  & -EMJ  & EMK  & ENJ  & -ENK 
\end{array}
\right)
$$
The resulting Gram matrix is
$$
G = R R^T =
\scriptsize
\left(
\begin{array}{ccccccc}
  A^2 + B^2 & 0  & 0  & 0 & (AM+BN)C & 0 & (AM+BN)FJ  \\
  0& A^2 + B^2  &0  & 0 &  (AM+BN)D  & 0 & (AM+BN)FK  \\
  0& 0 & A^2 + B^2  &0  & 0 &  (AM+BN)C & -(AM+BN)EJ    \\
  0& 0 & 0 & A^2 + B^2  & 0 &  (AM+BN)D  &-(AM+BN)EK   \\
  (AM+BN)C& (AM+BN)D & -NC  & ND  & (M^2 + N^2)(C^2 + D^2) & 0 &  (M^2 + N^2)(JC+KD)F\\
 0 & 0 & (AM+BN)C & (AM+BN)C & 0 &(M^2 + N^2)(C^2 + D^2) & -(M^2 + N^2)(JC+KD)E  \\
   (AM+BN)FJ & (AM+BN)FK  & -(AM+BN)EJ  &-(AM+BN)EK  & (M^2 + N^2)(JC+KD)F  &-(M^2 + N^2)(JC+KD)E   &  (M^2 + N^2)(J^2 + K^2)(E^2 + F^2)
\end{array}
\right)
$$
The amazing thing is all the cancellation as
$$  \det G = (A^2 + B^2)(C^2 + D^2)(E^2 + F^2)  $$
As before, the smallest element of the indicated quadratic form is no more than an explicit constant times
$$  \left( \det G \right)^{1/7}.  $$
Each element of the vector in $\mathbb Z^8$ that achieves that quadratic form value is no larger than a constant times
$$  \left( \det G \right)^{1/14}.  $$
It is pretty clear that this process continues; repeat to get, in dimension 16,
$$  \det G = (A^2 + B^2)(C^2 + D^2)(E^2 + F^2)(H^2 + I^2)  $$
with exponents given as $1/15$ then $1/30.$
Here are pages 136-137 of Cassels, Rational Quadratic Forms, which you should purchase. 

</s>