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 $\gg1$. 
 A: September 18.
I wrote my test programs with fixed integers $A,B,C,D,$ positive and pairwise coprime. I called the variables $w,x,y,z.$  The homogeneous linear diophantine equation to be considered is
$$  BDw + BC x + AD y + ACz = 0  $$ 
The article that showed me how to find a basis for the lattice was L. W. Griffiths, A Note on Linear Homogeneous Diophantine Equations, Bulletin AMS 1946.
Short version, Lois Wilfred Griffiths was a student of L. E. Dickson. Her 1927 Ph.D. dissertation was "Certain quaternary quadratic forms and diophantine equations by generalized quaternion algebras." 
As $\gcd(A,B) = 1,$ we solve
$$  MB-NA = 1.  $$
Then, a three by four matrix whose rows give a basis of the indicated lattice is
$$
\left(
\begin{array}{cccc}
-A & 0 & B & 0 \\
0 & A & 0 & -B  \\
MC & -MD & -NC & ND 
\end{array}
\right)
$$
Next, a (real) unit vector in the 3-space containing the  lattice, that is orthogonal to the first two rows, is
$$ \left( \frac{1}{\sqrt {A^2 + B^2} \sqrt{C^2 + D^2}} \right) \left( BC, -BD, AC, -AD \right). $$
The  length of   $(MC, -MD, -NC, ND)$ projected onto the unit vector above is exactly
$$   \sqrt { \frac{C^2 + D^2}{A^2 + B^2}  }.  $$ As a result, the fundamental parallelopiped has volume precisely 
$$ \sqrt {A^2 + B^2} \sqrt{C^2 + D^2} $$
Then, the Gram matrix obtained by taking the basis matrix times its transpose has determinant 
$$ \left( A^2 + B^2 \right) \left( C^2 + D^2 \right) $$
From formula (12) on page 92 of Dickson's Modern Elementary Theory of Numbers, we find that the squared norm of the shortest nonzero vector in the lattice is no larger than
$$ \frac{4}{3} \; \; \sqrt[3]{ \left( A^2 + B^2 \right) \left( C^2 + D^2 \right)} $$
The absolute values of $w,x,y,z$ can therefore be no larger than 
$$ \frac{2}{\sqrt 3} \; \; \sqrt[6]{ \left( A^2 + B^2 \right) \left( C^2 + D^2 \right)} $$
If, as originally stipulated, we have some positive $n$ with
$$  n \leq A,B,C,D \leq 2n,$$ we get
$$       |w|,|x|,|y|,|z| \leq    \frac{2 \cdot 16^{1/6}}{\sqrt 3} \; \; n^{2/3}                $$
September 19.
Worth recording how we know we have found a basis for the lattice. 
$$  BDw + BC x + AD y + ACz = 0  $$
$$  MB-NA = 1.  $$
$$  C(Bx + Az) + D(Bw + Ay) = 0.  $$
$$  C(Bx + Az) =  -D(Bw + Ay) .  $$
$$ \gcd(C,D) = 1. $$
$$    C | (Bw + Ay) $$
Let $r \in Z,$
$$  Bw + Ay = Cr, \; \; \;  \mbox{therefore} \; \; Bx + Az = -Dr.  $$
$$  Bw + Ay = C ( MB-NA)r,$$
$$  B(w - MCr) = -A(y+NCr).  $$
$$ \gcd(A,B)=1, A | (w - MCr), \mbox{define} \; \; p, $$
$$  w - MCr = -Ap, \; \; \;  \mbox{therefore} \; \; y+NCr = Bp.  $$
Return to
$$  Bx + Az = -Dr,  $$
$$  Bx + Az = -D( MB-NA)r,  $$
$$  B(x + MDr) = -A(z-NDr)  $$
$$ A | (x + MDr) , \mbox{define} \; \; q, $$
$$  x + MDr = Aq, \; \; \;  \mbox{therefore} \; \; z-NDr = -Bq.  $$
Together,
$$  w = MCr  -Ap, \; \;  x = - MDr + Aq, \; \;  y= -NCr + Bp, \; \;z=NDr  -Bq. $$
$$  (w,x,y,z) = p(-A,0,B,0) + q (0,A,0,-B) + r (MC,-MD, -NC,ND).  $$




A: 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. 

