Take $A,B,C,D$ pairwise coprime with $$n<A,B,C,D<2n$$ $$
n/4<|A−B|,|C−D|,|A−C|,|A−D|,|B−C|,|B−D|$$ and consider the space of solutions to $ACa+ADb+BCc+BDd=0$ spanned by $3\times 4$ matrix
$$N=\begin{bmatrix}
-D&C&0&0\\
-B&0&A&0\\
0&0&-D&C
\end{bmatrix}.$$


Denote the $\Bbb Q$-linear space spanned by rows of $N$ by $T\_{A,B,C,D}\subseteq\Bbb Q^4$ and denote the set of non-zero $\Bbb Z$ vectors in $T\_{A,B,C,D}$ by $T\_{A,B,C,D}[\Bbb Z]^\star$.


Consider the quantity $\mu(A,B,C,D)=\min\_{v\in T\_{A,B,C,D}[\Bbb Z]^\star}\|v\|\_\infty$ where $\|v\|\_\infty$ is largest coordinate by magnitude of vector $v$. It is clear $\mu(A,B,C,D)\leq|a|,|b|,|c|,|d|$ at any $a,b,c,d$ with $ACa+ADb+BCc+BDd=0$ and $|a|+|b|+|c|+|d|\neq0$.



> 
> 1. Is there a name for $\mu(A,B,C,D)$?
> 2. How is $\mu(A,B,C,D)$ distributed as a function of $A,B,C,D$ chosen with the constraints above (at least consider $A,B$ and $C,D$ each a coprime pair and $n<A,B,C,D<2n$) and what is its average value?
> 
> 
> 


Simulations and heuristics suggest a value between $\Omega(n^{1/2})$ and $\Omega(n^{2/3})$ with $\Omega(n^{2/3})$ being the most likely possibility (at least probability $>1/n^\epsilon$ at any fixed $\epsilon>0$) of lower bound at least when $A,B,C,D$ are prime and hence is also the expected value at least when $A,B,C,D$ are prime.


Note here I am interested in *lower bound* in probabilities and expected value. There are worse case counterexamples. 



On distribution of size of integer points in a subspace associated to a linear diophantine equation
We will assume that $n<A,B,C,D<2n$ and $\gcd(A,B)=\gcd(C,D)=1$.
For $x,y,z\in\mathbb{Q}$, we write $v(x,y,z)$ for the vector
$$\pmatrix{\frac{x}{C}&\frac{y}{C}&\frac{z}{C}}N\in\mathbb{Q}^4.
$$


**Lemma 1.** For $x,y,z\in\mathbb{Q}$, it holds that
$v(x,y,z)\in\mathbb{Z}^4$
iff $x,y,z\in\mathbb{Z}$ and the following conditions hold:
$$\begin{eqnarray\*}
 xD+yB\equiv 0\pmod{C},\\
 zD-yA\equiv 0\pmod{C}.\\
 \end{eqnarray\*}
$$


**Proof.** A direct calculation shows
$$v(x,y,z)
 =\pmatrix{\frac{-yB-xD}{C}&x&\frac{yA-zD}{C}&z},
$$
which immediately shows the implication from right to left.
Assume then that the vector belongs to $\mathbb{Z}^4$.
Then obviously $x,z\in\mathbb{Z}$.
It follows that $yA,yB\in\mathbb{Z}$,
which implies $y\in\mathbb{Z}$ by the assumption $\gcd(A,B)=1$.


We will call $x,y,z$ such as above *witnesses* for
$\mu(A,B,C,D)\le\|v(x,y,z)\|\_\infty$.


**Lemma 2.** For $x,y,z\in\mathbb{Q}$, it holds that
$$\frac{\max(|x|,|y|,|z|)}{4}\le\|v(x,y,z)\|\_\infty\le 4\max(|x|,|y|,|z|).
$$
**Proof.** Write $v(x,y,z)=\pmatrix{u&x&v&z}$.
For the first inequality, it suffices to note that
$$|y|=\frac{|vC+zD|}{A}\le\frac{C}{A}|v|+\frac{D}{A}|z|\le 2(|v|+|z|)\le 4\|v(x,y,z)\|\_\infty.
$$
For the second inequality, we get
$$|u|\le\frac{B}{C}|y|+\frac{D}{C}|x|\le 2(|x|+|y|)\le 4\max(|x|,|y|,|z|)
$$
and similarly the same bound for $v$.


**Theorem 1.** There is a constant $K$ (independent of $n$) such that for all choices of $A,B,C,D$ as above, it holds that $\mu(A,B,C,D)<Kn^{2/3}$.


**Proof.** For $y\in\mathbb{N}$, let $f(y)$ and $g(y)$ be the unique integers such that $f(y)D+yB\equiv 0\pmod{C}$, $g(y)D-yA\equiv 0\pmod C$, and $0\le f(y),g(y)<C$. They are unique and exist because $\gcd(C,D)=0$. Let $M={}^\lceil\root 3\of{2n}\,{}^\rceil$. By the usual pigeonhole argument, there are $y\_1,y\_2$ such that $0\le y\_1<y\_2\le M^2$, $|f(y\_1)-f(y\_2)|\le M^2$, and $|g(y\_1)-g(y\_2)|\le M^2$. Now set
$$\begin{eqnarray\*}
 x&=&f(y\_1)-f(y\_2),\\
 y&=&y\_1-y\_2,\\
 z&=&g(y\_1)-g(y\_2).\\
 \end{eqnarray\*}
$$
By Lemma 1 and the definition of $f$ and $g$, it follows that $x,y,z$ are witnesses for
$\mu(A,B,C,D)\le\|v(x,y,z)\|\_\infty$.
By Lemma 2, this implies
$$
 \mu(A,B,C,D)\le 4\max(|x|,|y|,|z|)\le 4M^2.
$$


**Theorem 2.** There is a constant $L$ such that for $A,B,C,D$ chosen randomly from all possibilities satisfying the assumptions given at the beginning of this answer, the probability of $\mu(A,B,C,D)$ being below a given limit $m$ is less than $Lm^3/n^2$.


**Proof.** First, fix $C$ and $D$ satisfying the assumptions, and an integer $m>1$. There is a constant $L\_1>0$ such that there are at least $L\_{1}n^2$ choices for $A,B$. Assume $A,B$ are chosen so that $\mu(A,B,C,D)<m$. Let $x,y,z$ witness this. Then $|x|,|z|<m$. Moreover, $|y|<4m$ by Lemma 2. So, consider $y$ such that $-4m<y<4m$. Let $k=\gcd(y,C)$, and let $y'=y/k,C'=C/k$. For every pair $x',z'$, there are unique $A',B'$ such that $0\le A',B'<C'$ and the following hold:
$$
 \begin{eqnarray\*}
 x'D+y'B'\equiv 0\pmod {C'},\\
 z'D-y'A'\equiv 0\pmod {C'}.\\
 \end{eqnarray\*}
$$
Moreover, for each pair $A',B'$ there are $k^2$ pairs $A,B$ such that $0\le A,B<C$ and $A\equiv A'\pmod {C'}$, $B\equiv B'\pmod {C'}$. There are no more than $2\lceil m/k\rceil$ choices for $x'$ such that $|kx'|<m$. So, for each choice of $y$, there are fewer than $5m^2$ possibilities for the pair $A,B$. Altogether, there are fewer than $10m^3$ choices for $A,B$ such that $\mu(A,B,C,D)<m$, out of at least $L\_{1}n^2$. So, we may choose $L=10/L\_{1}$.


