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\title{A composite modulus and Frobenius digits in the\\ Heilbronn triangle construction}
\author{David Niedbala Giraudin\\
\small ORCID \href{https://orcid.org/0009-0009-1526-1178}{0009-0009-1526-1178}\\[2pt]
\small Technical assistance: Claude (Anthropic)}
\date{November 2026}
\begin{document}
\maketitle

\begin{abstract}
OpenAI proved that for every sufficiently large $n$ there are $n$ points in the
unit square all of whose triangles have area at least $c\,n^{-2+\eta}$, with an
explicit but extremely small $\eta$ (of order $10^{-235}$). Subsequent public
work lowered the cost of the construction, the latest claimed exponent being
$\eta=2/100866719\approx1.98\cdot10^{-8}$. We introduce two changes. First, the
digit base is taken to be a multiple of the label prime $r$ and the main modulus
is $h=rB^{k-1}$ instead of $B^k$; this reduces the ratio $h/\tau$ between the
modulus and the excluded determinant window from about $B$ to about $r$, which
allows the extension degree $d=3$. Second, the norm polynomial is placed in the
digits by a digit-wise convolution along the Frobenius expansion of the
exponents, with a trace digit in a polynomial basis, giving $k=1440$ positions.
The result is
\[
 \Delta(n)\ \ge\ c\,n^{-2+\eta},\qquad \eta=\frac1{177\,022}>5.64\cdot10^{-6},
\]
for every sufficiently large $n$. The finite identities are checked by exact
computation in the ancillary files.
\end{abstract}

\section{Introduction}

For a finite set $P\subset[0,1]^2$ with $|P|\ge3$ let $\Delta(P)$ be the least
area of a triangle with vertices in $P$, and let $\Delta(n)$ be the maximum of
$\Delta(P)$ over $n$-point sets. Koml\'os, Pintz and Szemer\'edi proved
$\Delta(n)\gg(\log n)/n^2$ \cite{KPS}. OpenAI~\cite{OAI} proved a power improvement
$\Delta(n)\ge c_1 n^{-2+\eta}$ with
$\eta=2/(45435k+16)$, $k=T^2+1$, $T=\binom M3$, $M=\binom{163}{41}$, so that
$\eta\approx10^{-235.5}$. Their construction samples integral columns whose
residues modulo a main modulus $h$ carry a determinant obstruction coming from a
field norm, together with an auxiliary cap modulo a second prime $q$, and
deletes the few bad triples.

Public follow-up work reduced the parameters inside the same framework:
Blinov~\cite{Blinov} lowered the extension degree to $d=5$ using row-dependent
digit lengths, label tags, a fresh-digit collision bound and a slab version of
the auxiliary cap, and the Euler team~\cite{Euler} obtained the claimed exponent
$\eta=2/100866719$ with $k=408781$ and a smaller lift box.

\begin{theorem}\label{thm:main}
There are $c>0$ and $n_0$ such that for every integer $n\ge n_0$,
\[
 \Delta(n)\ge c\,n^{-2+\eta},\qquad \eta=\frac1{177022}.
\]
\end{theorem}

The two new ingredients are the following.

\emph{Composite modulus.} Let $r$ be the label prime and choose the digit base
$B=rB_2$ with $B_2$ prime. The main modulus is $h=rB^{k-1}$. Since $B\equiv0$
modulo $r$, the part of the determinant above the digit position $k-1$ reduces
modulo $r$ to the single designated coefficient, whose residue is the field
norm. Distinct labels then exclude every representative of absolute value at
most $\tau\approx B^{k-1}/2$, while $h/\tau\approx 2r$ instead of $\approx2B$.
The price of a label collision becomes $r^{2\gamma}\cdot r\cdot r^{-d}$ and $d=3$
suffices (Section~\ref{sec:deletion}).

\emph{Frobenius digits.} The norm of the Vandermonde product,
$\Nm(V)=\prod_t V^{r^t}$, has monomials whose exponents have $r$-adic digits in
$\{0,1,2\}$, with a coefficient that factors over the digits. We place the
monomials in the digit positions by a mixed-radix convolution in which each
lower Frobenius digit is handled by a fixed three-row design of radix $6$
(Table~\ref{tab:design}), the most significant digit by a design whose rows $1$
and $3$ never reach the target together (Table~\ref{tab:top}), and the descent
to $\F_r$ by a trace digit of radix $2d-1$ in a polynomial basis. With $d=3$
this uses $k=1440$ positions.

Everything else follows the framework of~\cite{OAI} with the row-dependent
digit lengths and the slab cap of~\cite{Blinov} and the smaller lift box
of~\cite{Euler}. We give proofs of every statement that changes, citing~\cite{OAI} for the
lattice-point counts and the weight bounds of the auxiliary cap, whose proofs apply
with the changes indicated.

\paragraph{Conventions.} $A\ll B$ means $|A|\le C B$ with $C$ depending only
on the fixed parameters $d=3$ and $k=1440$, never on the growing prime $r$.
Lengths are Euclidean. All logarithms are natural.

\section{Frobenius digits}\label{sec:encoding}

\subsection{The digit design}
For $(x,y,z)\in\{0,1,2\}^3$ let $w(x,y,z)$ be the sign of the permutation
$0\mapsto x$, $1\mapsto y$, $2\mapsto z$ when $\{x,y,z\}=\{0,1,2\}$, and
$w(x,y,z)=0$ otherwise. Then, for commuting indeterminates,
\begin{equation}\label{eq:vand}
 \sum_{x,y,z}w(x,y,z)X^xY^yZ^z
 =\det\begin{pmatrix}1&1&1\\X&Y&Z\\X^2&Y^2&Z^2\end{pmatrix}
 =(Y-X)(Z-X)(Z-Y).
\end{equation}

A \emph{design} is a triple of finite lists $\mathcal O_1,\mathcal O_2,\mathcal
O_3$ of options; each option $o$ carries a value $x(o)\in\{0,1,2\}$, a height
$u(o)\in\Z_{\ge0}$ and a sign $\epsilon(o)\in\{\pm1\}$. We use the design $\mathcal
D_6$ of Table~\ref{tab:design}, with target $\kappa=5$.

\begin{table}[h]
\centering
\begin{tabular}{@{}ll@{}}
\toprule
row & options $(x,u,\epsilon)$\\
\midrule
1 & $(0,1,+)$,\ $(0,3,+)$,\ $(1,4,-)$,\ $(2,0,+)$\\
2 & $(0,1,+)$,\ $(1,2,+)$,\ $(2,0,-)$\\
3 & $(0,3,-)$,\ $(0,1,+)$,\ $(1,4,+)$,\ $(2,0,+)$\\
\bottomrule
\end{tabular}
\caption{The digit design $\mathcal D_6$ (target $\kappa=5$).}
\label{tab:design}
\end{table}

\begin{lemma}\label{lem:design}
The design $\mathcal D_6$ has the following properties.
\begin{enumerate}
\item[(D1)] For every $(x,y,z)\in\{0,1,2\}^3$,
\[
 \sum_{\substack{o_i\in\mathcal O_i,\ (x(o_1),x(o_2),x(o_3))=(x,y,z)\\
 u(o_1)+u(o_2)+u(o_3)=5}}
 \epsilon(o_1)\epsilon(o_2)\epsilon(o_3)=w(x,y,z).
\]
\item[(D2)] For all options, $|u(o_1)+u(o_2)+u(o_3)-5|\le5$.
\item[(D3)] All heights are at most $4$.
\end{enumerate}
\end{lemma}

\begin{proof}
A finite check of the $4\cdot3\cdot4=48$ option triples; see
\texttt{anc/check\_design.py}. For example, for $(x,y,z)=(0,1,2)$ the option
triples have heights $(1,2,0)$ and $(3,2,0)$; only the second sums to $5$, with
sign $+$. For $(0,0,0)$ the triples summing to $5$ are $(1,1,3)$ with sign $-$
and $(3,1,1)$ with sign $+$, which cancel.
\end{proof}

For the most significant digit we use the design $\mathcal T_7$ of
Table~\ref{tab:top}, with target $7$.

\begin{table}[h]
\centering
\begin{tabular}{@{}ll@{}}
\toprule
row & options $(x,u,\epsilon)$\\
\midrule
1 & $(0,4,+)$,\ $(1,1,-)$,\ $(2,0,+)$\\
2 & $(0,5,-)$,\ $(0,6,-)$,\ $(1,3,-)$,\ $(2,1,+)$\\
3 & $(0,5,-)$,\ $(0,4,+)$,\ $(1,2,-)$,\ $(2,0,-)$\\
\bottomrule
\end{tabular}
\caption{The top design $\mathcal T_7$ (target $7$).}
\label{tab:top}
\end{table}

\begin{lemma}\label{lem:top}
The design $\mathcal T_7$ satisfies (D1) with $5$ replaced by $7$, and moreover:
\begin{enumerate}
\item[(T1)] no option of row $1$ and option of row $3$ have heights summing to $7$;
\item[(T2)] the options of height $0$ in rows $1$ and $3$ have value $x=2$, and
row $2$ has no option of height $0$.
\end{enumerate}
The rows of $\mathcal D_6$ have options of height $0$ only with value $x=2$.
\end{lemma}

\begin{proof}
A finite check; see \texttt{anc/check\_design.py}. For (T1), the heights in rows
$1$ and $3$ are $\{0,1,4\}$ and $\{0,2,4,5\}$, whose sums avoid $7$.
\end{proof}

\subsection{Field, labels and the norm}
Fix $d=3$. Let $r>6d$ be a prime, $\F=\F_r$, and $K=\F_{r^d}=\F[\theta]$ with
$\theta$ a root of an irreducible cubic. Write $c_\mu(\xi)$, $\mu=0,1,2$, for
the coordinates of $\xi\in K$ in the basis $1,\theta,\theta^2$, and $\Tr$ for
the trace $K\to\F$. For $e=(e_0,e_1,e_2)\in\{0,1,2\}^3$ put
$\xi^{[e]}=\xi^{e_0+e_1r+e_2r^2}=\prod_t(\xi^{r^t})^{e_t}$. For labels
$\xi_1,\xi_2,\xi_3\in K^\times$ let $V=(\xi_2-\xi_1)(\xi_3-\xi_1)(\xi_3-\xi_2)$ and
$\Nm(V)=\prod_{t<d}V^{r^t}\in\F$. Since $d$ is odd, $\Nm(-v)=-\Nm(v)$, so
$\Nm(V)$ is alternating in the labels; it is nonzero when the labels are
distinct.

\subsection{Positions and residues}
Use the mixed radix with digit weights
\[
 W_{\mathrm{tr}}=1,\qquad W_0=5,\qquad W_1=30,\qquad W_2=180,
\]
that is, a trace digit of radix $2d-1=5$, two Frobenius digits of radix $6$
(digits $t=0,1$, design $\mathcal D_6$) and a most significant Frobenius digit
($t=2$, design $\mathcal T_7$). Write $\mathcal O_i^{(t)}$ for the options of row
$i$ in the design of digit $t$. For a row $i$ and an option word
$o=(o_0,o_1,o_2)$ with $o_t\in\mathcal O_i^{(t)}$ put
\[
 x(o)=(x(o_0),x(o_1),x(o_2)),\quad \epsilon(o)=\prod_t\epsilon(o_t),\quad
 H(o)=\sum_{t=0}^{2}u(o_t)W_t .
\]
The \emph{terms} of the three rows, their \emph{positions} and their
\emph{functions} $K\to\F$ are:
\begin{center}
\begin{tabular}{@{}llll@{}}
\toprule
row & term & position & function \\
\midrule
1 & $(o,\mu)$, $0\le\mu\le2$ & $\mu+H(o)$ & $\epsilon(o)\,c_\mu(\xi^{[x(o)]})$\\
2 & $(o,\nu)$, $0\le\nu\le2$ & $\nu+H(o)$ & $\epsilon(o)\,c_\nu(\xi^{[x(o)]})$\\
3 & $(o,s)$, $0\le s\le4$ & $4-s+H(o)$ & $\epsilon(o)\,\Tr(\theta^s\xi^{[x(o)]})$\\
\bottomrule
\end{tabular}
\end{center}
Put
\[
 X^*=4\cdot W_{\mathrm{tr}}+5W_0+5W_1+7W_2=1439,\qquad k=X^*+1=1440 .
\]
The largest position is $2+4\cdot5+4\cdot30+4\cdot180=862$ in row $1$,
$2+2\cdot5+2\cdot30+6\cdot180=1152$ in row $2$ and $4+4\cdot5+4\cdot30+5\cdot180=1044$
in row $3$, all below $k-1=1439$. For a label $\xi\in K^\times$, a row $i$ and a
position $0\le v<k$, let $\rho_{i,v}(\xi)\in\F$ be the sum of the functions of
all row-$i$ terms at position $v$ (zero if there are none), except that we add
the \emph{control}
\begin{equation}\label{eq:control}
 \rho_{2,0}(\xi)=\varphi(\xi)\Nm(\xi)^2,\qquad \varphi(\xi)=c_0(\xi);
\end{equation}
by (T2) row $2$ has no term at position $0$. By (T2) and
Lemma~\ref{lem:top}, the only terms at position $0$ are, in row $1$, the word of
value $(2,2,2)$ with $\mu=0$, of function $c_0(\xi^{[2,2,2]})=c_0(\Nm(\xi)^2)=\Nm(\xi)^2$,
and, in row $3$, the word of value $(2,2,2)$ with $s=4$, of function
$-\Tr(\theta^4\Nm(\xi)^2)$. Hence
\begin{equation}\label{eq:column0}
 (\rho_{1,0},\rho_{2,0},\rho_{3,0})(\xi)=\Nm(\xi)^2\,\bigl(1,\ \varphi(\xi),\ -\Tr(\theta^4)\bigr).
\end{equation}

\begin{lemma}[Encoding]\label{lem:encoding}
For all labels $\xi_1,\xi_2,\xi_3\in K^\times$,
\begin{equation}\label{eq:encoding}
 \sum_{p_1+p_2+p_3=k-1}\det\bigl(\rho_{i,p_i}(\xi_j)\bigr)_{i,j=1}^3=6d\,\Nm(V)=18\,\Nm(V)
\end{equation}
in $\F$, the sum running over $0\le p_i<k$. In particular the left side is
nonzero when the labels are distinct.
\end{lemma}

\begin{proof}
\emph{Selection.} Consider three terms $(o^1,\mu)$, $(o^2,\nu)$, $(o^3,s)$ of
rows $1,2,3$ with positions summing to $X^*$. Put
$\delta_{\mathrm{tr}}=\mu+\nu-s\in[-4,4]$ and
$\delta_t=u(o^1_t)+u(o^2_t)+u(o^3_t)-\kappa_t$ with $\kappa_0=\kappa_1=5$,
$\kappa_2=7$; by (D2), $\delta_0,\delta_1\in[-5,5]$. Then
$0=\delta_{\mathrm{tr}}+5\delta_0+30\delta_1+180\delta_2$. Reduction modulo $5$
gives $\delta_{\mathrm{tr}}=0$, then reduction modulo $6$ gives $\delta_0=0$, then
$\delta_1=0$, and finally $\delta_2=0$. Conversely these equalities give the
target. So the contributing triples of terms are exactly those with $\mu+\nu=s$
and, in each digit $t$, heights summing to $\kappa_t$.

\emph{The control.} The control \eqref{eq:control} contributes only through
triples with $p_2=0$ and $p_1+p_3=X^*$, where $p_1,p_3$ carry terms of rows $1$
and $3$ (otherwise a residue is $0$). For such terms put
$\delta'_{\mathrm{tr}}=\mu+(4-s)-4\in[-4,2]$ and
$\delta'_t=u(o^1_t)+u(o^3_t)-\kappa_t$, with $\delta'_0,\delta'_1\in[-5,3]$
because the heights of $\mathcal D_6$ in rows $1$ and $3$ lie in $[0,4]$ (D3). As above $\delta'_{\mathrm{tr}}=\delta'_0=\delta'_1=\delta'_2=0$, so the top
heights of rows $1$ and $3$ would sum to $7$, contradicting (T1). Hence the
control does not contribute.

\emph{Expansion.} By multilinearity of the determinant in its rows, the left
side of \eqref{eq:encoding} equals
\[
 \sum\epsilon(o^1)\epsilon(o^2)\epsilon(o^3)
 \det\begin{pmatrix}
 c_\mu(\xi_j^{[a]})\\ c_\nu(\xi_j^{[b]})\\ \Tr(\theta^s\xi_j^{[c]})
 \end{pmatrix}_{j=1,2,3},
\]
over the contributing triples, where $a=x(o^1)$, $b=x(o^2)$, $c=x(o^3)$. Fix
$(a,b,c)$. By (D1) for $\mathcal D_6$ and $\mathcal T_7$ applied digit by digit, the
total sign of the option words with these values and digit sums $\kappa_t$ is
$\prod_tw(a_t,b_t,c_t)$. For fixed
$(a,b,c)$ the sum over $\mu+\nu=s$ is, expanding the determinant over
permutations $\sigma$,
\[
 \sum_\sigma\sgn(\sigma)\sum_{s}\sum_{\mu+\nu=s}c_\mu(\alpha)c_\nu(\beta)
 \Tr(\theta^s\gamma)
 =\sum_\sigma\sgn(\sigma)\Tr(\alpha\beta\gamma),
\]
with $\alpha=\xi_{\sigma(1)}^{[a]}$, $\beta=\xi_{\sigma(2)}^{[b]}$,
$\gamma=\xi_{\sigma(3)}^{[c]}$, because
$\alpha\beta=\sum_{\mu,\nu}c_\mu(\alpha)c_\nu(\beta)\theta^{\mu+\nu}$ and the trace is
$\F$-linear. Summing over $(a,b,c)$ and using that Frobenius is a ring
endomorphism together with \eqref{eq:vand},
\[
\begin{aligned}
 \sum_{a,b,c}\prod_t w(a_t,b_t,c_t)\,X^{[a]}Y^{[b]}Z^{[c]}
 &=\prod_{t<d}\ \sum_{x,y,z}w(x,y,z)(X^{r^t})^x(Y^{r^t})^y(Z^{r^t})^z\\
 &=\prod_{t<d}V(X,Y,Z)^{r^t}=\Nm V(X,Y,Z).
\end{aligned}
\]
Hence the left side is
\[
 \sum_\sigma\sgn(\sigma)\Tr\bigl(\Nm V(\xi_{\sigma(1)},\xi_{\sigma(2)},\xi_{\sigma(3)})\bigr)
 =\sum_\sigma\sgn(\sigma)^2\Tr(\Nm V)=6\,\Tr(\Nm V)=6d\,\Nm V,
\]
since $\Nm V$ is
alternating and lies in $\F$. Finally $6d=18$ is invertible because $r>18$.
\end{proof}

The identity is checked by exact computation for random labels in
\texttt{anc/check\_encoding.py}.

\section{Digit columns and the composite modulus}\label{sec:digits}

\subsection{Parameters}
Let $r>18$ be a prime and put
\begin{equation}\label{eq:params}
 M=4800k^2r^3,\qquad L_1=L_2=r,\qquad L_3=rM,\qquad A_0=6k^2L_1L_2L_3=6k^2r^3M .
\end{equation}
Let $B_2$ be a prime with $Q<B_2\le2Q$, where $Q=600k^2r^2M$ (it
exists for large $r$ by Bertrand's postulate), and set
\begin{equation}\label{eq:modulus}
 B=rB_2,\qquad h=rB^{k-1}=r^kB_2^{k-1},\qquad \tau=\lfloor B^{k-1}/2\rfloor .
\end{equation}
Then $B=rB_2>600k^2r^3M=100A_0$, so $B-1\ge100A_0$; also $L_3<B_2$, $B_2\ne r$, and
\begin{equation}\label{eq:theta}
 \theta_0:=\frac{B_2r}{M^2}\le\frac{2(600k^2r^2M+1)r}{M^2}\le\frac12 .
\end{equation}
Moreover $B\asymp r^6$, with constants depending only on $k$.

\subsection{Random columns}
A column is drawn as follows. Choose a label $\xi$ uniformly in $K^\times$. For rows
$i=1,2$ and every position $0\le v<k$, let $d_{i,v}$ be the unique integer in
$[1,r]$ congruent to $\rho_{i,v}(\xi)$ modulo $r$. For row $3$ and every $v$,
choose $d_{3,v}$ uniformly from the $M$ integers in $[1,rM]$ congruent to
$\rho_{3,v}(\xi)$, independently. Put $c_i=\sum_{v<k}d_{i,v}B^v\bmod h$.
Different columns use independent labels and digits. Since $B\equiv0$ modulo
both primes, $c_i\equiv d_{i,0}$ modulo $r$ and modulo $B_2$. In particular
$c_1\equiv\Nm(\xi)^2\not\equiv0\pmod r$ by \eqref{eq:column0}, and
$c_1\equiv d_{1,0}\in[1,r]$ is a unit modulo $B_2$. By \eqref{eq:column0} the third row of
$C\bmod r$ is a multiple of the first, so $C\bmod r$ always has rank at most two
and $r\mid\det C$; this is harmless, since the obstruction below only uses
$\mathrm{Hi}\not\equiv0\pmod r$ and the divisor $E$ cancels in
\eqref{eq:bad}.

\begin{lemma}[Obstruction]\label{lem:obstruction}
Let $C$ be the residue matrix of three such columns. If the three labels are
distinct, then no integer in $[-\tau,\tau]$ is congruent to $\det C$ modulo $h$.
\end{lemma}

\begin{proof}
Let $\widehat C$ be the integer matrix with entries $\sum_vd^{(j)}_{i,v}B^v$ and
put $F(Y)=\det\bigl(\sum_vd^{(j)}_{i,v}Y^v\bigr)=\sum_e\gamma_eY^e$. For each
exponent $e$ and permutation, the positions in two rows determine the third,
so $|\gamma_e|\le A_0$. Write
$F(B)=L_0+B^{k-1}\,\mathrm{Hi}$ with $L_0=\sum_{e<k-1}\gamma_eB^e$ and
$\mathrm{Hi}=\sum_{e\ge k-1}\gamma_eB^{e-k+1}$. Since $B-1\ge100A_0$,
$|L_0|\le A_0(B^{k-1}-1)/(B-1)\le\frac{1}{100}(B^{k-1}-1)$. Since $B\equiv0\pmod r$,
$\mathrm{Hi}\equiv\gamma_{k-1}\pmod r$, and since every digit is congruent to
its residue, Lemma~\ref{lem:encoding} gives
$\gamma_{k-1}\equiv18\Nm(V)\not\equiv0\pmod r$. Every integer $t\equiv\det
C\equiv F(B)\pmod h$ has the form $t=L_0+B^{k-1}(\mathrm{Hi}+rm)$ with
$m\in\Z$, and $\mathrm{Hi}+rm\ne0$. Hence
$|t|\ge B^{k-1}-|L_0|>\frac{99}{100}B^{k-1}>\tau$.
\end{proof}

\subsection{Smith forms modulo $h$}
For a prime $p\mid h$ let $h_p=p^{k_p}$ be the $p$-part of $h$ ($k_r=k$,
$k_{B_2}=k-1$). By the Chinese remainder theorem
$\Z/h\Z=\Z/h_r\Z\times\Z/h_{B_2}\Z$ and
$\mathrm{SL}_3(\Z/h\Z)=\prod_p\mathrm{SL}_3(\Z/h_p\Z)$. Since $C_{11}$ is a unit
modulo both primes, \cite[Lemma 4.1]{OAI} applies to $C\bmod h_p$ and gives
$C\equiv P_p\operatorname{diag}(1,p^{b_p},p^{e_p})Q_p$ with $0\le b_p\le e_p\le k_p$,
where $b_p\ge j$ if and only if all $2\times2$ minors of $C$ vanish modulo $p^j$.
Put $D=\prod_pp^{b_p}$, $E=\prod_pp^{e_p}$ and $I=DE$. Let $\Lambda=\Lambda_C$ be
the set of $v\in\Z^3$ whose reduction modulo $h$ lies in the row span of $C$.

\begin{lemma}\label{lem:smith}
$[\Z^3:\Lambda]=I$, $E\Z^3\subseteq\Lambda$, and the orbit
$\mathcal O_C=\{GC:G\in\mathrm{SL}_3(\Z/h\Z)\}$ satisfies
$|\mathcal O_C|\ge(21/64)^2h^8/(D^3E^2)$. For uniform $G$, $GC$ is uniform on
$\mathcal O_C$, and all elements of $\mathcal O_C$ have the row span and the
determinant of $C$. Moreover, for $x\in\Z^3$ write $x\cdot\Lambda=g\Z$; for every $m\mid h$, $m\mid g$
if and only if $Cx\equiv0\pmod m$; and for $m=\prod_pp^{j_p}$ with $j_p\le e_p$ the
proportion of residues $x$ modulo $m$ with $Cx\equiv0\pmod m$ is
$\prod_pp^{-j_p-\max(j_p-b_p,0)}$.
\end{lemma}

\begin{proof}
For $p\in\{r,B_2\}$ put
\[
 \Lambda_p=\{v\in\Z^3:\ v\bmod h_p\in\text{row span of }C\bmod h_p\}.
\]
By the Chinese remainder theorem, $\Lambda=\Lambda_r\cap\Lambda_{B_2}$, and $\mathcal O_C$ is the product of the orbits of $C\bmod h_r$ and
$C\bmod h_{B_2}$. For each
prime $p$, \cite[Lemma 4.1 and Lemma 4.2]{OAI}, which are stated for a prime power
modulus and need one unit entry, give: the row span modulo $h_p$ has
$h_p^3/p^{b_p+e_p}$ elements, so $[\Z^3:\Lambda_p]=p^{b_p+e_p}$;
$p^{e_p}\Z^3\subseteq\Lambda_p$; and
$|\mathcal O_{C\bmod h_p}|\ge(21/64)\,h_p^8/p^{3b_p+2e_p}$. In the diagonal form,
the proportion of $x$ modulo $p^{j}$, $j\le e_p$, with $Cx\equiv0\pmod{p^j}$ is
$p^{-j-\max(j-b_p,0)}$, as in the proof of \cite[Lemma 7.3]{OAI}. Since the two
indices are coprime,
\[
 [\Z^3:\Lambda]=\prod_p[\Z^3:\Lambda_p]=I,\qquad
 E\Z^3\subseteq\bigcap_pp^{e_p}\Z^3\subseteq\Lambda,\qquad
 |\mathcal O_C|=\prod_p|\mathcal O_{C\bmod h_p}|,
\]
and the proportions multiply. A uniform $G\in\mathrm{SL}_3(\Z/h\Z)$ has uniform components, so $GC$ is uniform
on $\mathcal O_C$, and left multiplication by $G$ preserves the row span and the
determinant. Finally, for $m\mid h$ the reduction of $\Lambda$ modulo $m$ is the row span
of $C\bmod m$, because $\Lambda\supseteq h\Z^3$; hence $m\mid g$, that is $x\cdot v\equiv0
\pmod m$ for all $v\in\Lambda$, holds if and only if $x$ is orthogonal modulo $m$ to the
rows of $C$, that is $Cx\equiv0\pmod m$.
\end{proof}

\subsection{The divisor moment}

\begin{lemma}\label{lem:moment}
Fix the three labels. Then
\[
 \E[D\mid\text{labels}]\le 2+6r\,\mathbf 1\{\operatorname{rank}(C\bmod r)\le1\}.
\]
\end{lemma}

\begin{proof}
Write $b=b_{B_2}$ and $b'=b_r$. Since $p^\beta\le\sum_{i=0}^\beta p^i$,
\[
 D\le\sum_{j\ge0}\sum_{j'\ge0}B_2^j\,r^{j'}\,\mathbf 1\{b\ge j,\ b'\ge j'\}.
\]
Condition on the first column and on the first entries of the second and third
columns. As $C_{11}$ is a unit modulo $p$, $b_p\ge j$ implies the two
congruences $C_{3v}\equiv C_{31}C_{11}^{-1}C_{1v}\pmod{p^j}$, $v=2,3$, whose
right sides are now fixed. We use only these two bottom entries, which have
fresh independent digits.

\emph{The prime $B_2$.} Modulo $B_2^j$ only the digits at positions $<j$ of an
entry matter. The congruence modulo $B_2$ fixes $d_{3,0}$ modulo $B_2$, hence
fixes $d_{3,0}$ since $1\le d_{3,0}\le rM<B_2$. Inductively, given the lower
digits, the congruence modulo $B_2^{u+1}$ fixes $d_{3,u}B^u=r^uB_2^ud_{3,u}$ modulo
$B_2^{u+1}$, hence $d_{3,u}$ modulo $B_2$ (as $r$ is invertible modulo $B_2$), hence $d_{3,u}$. Each digit has $M$
equally likely values, so the event $b\ge j$ fixes the bottom digits at
positions $<j$ of both entries and has probability at most $M^{-2j}$.

\emph{The prime $r$.} The event $b'\ge1$ is the deterministic event that $C\bmod
r$ has rank at most one. Write $d_{3,u}=\rho_{3,u}+rm_u$; the integers $m_u$ range
over $M$ consecutive values, and $r\mid M$, so $m_u\bmod r$ is uniform. Since
$B^u=r^uB_2^u$, an entry is $\sum_ud_{3,u}r^uB_2^u=\sum_u\rho_{3,u}r^uB_2^u+\sum_um_ur^{u+1}B_2^u$.
Its residue modulo $r^{t+1}$ ($1\le t\le k-1$) is therefore the residue modulo $r^{t+1}$ of a
quantity determined by the residues $\rho_{3,u}$ and the $m_u$ with $u\le t-2$, plus
$m_{t-1}B_2^{t-1}r^t$. As $B_2$ is invertible modulo $r$, given the congruence of the
entry modulo $r^t$ and all $m_u$ with $u\le t-2$, the congruence modulo $r^{t+1}$ (whose
right side is fixed) holds for exactly one value of $m_{t-1}\bmod r$, hence with
conditional probability $1/r$. The same holds for the second entry, with its own
independent $m_{t-1}$.

\emph{Joint bound.} Here $0\le j\le k-1$ and $0\le j'\le k$. The event $b\ge j$ depends only
on the bottom digits at positions $<j$ and fixes them; it costs at most $M^{-2j}$, and
the congruences modulo $r^{t}$ with $t\le j$ then cost nothing more. Next condition, in
the order $t=j+1,\dots,j'-1$, on the congruence modulo $r^{t+1}$: it involves
$m_{t-1}$ with $t-1\ge j$, which is uniform modulo $r$ and independent of the event
$b\ge j$ and of all digits at positions $\le t-2$, so by the previous paragraph each step
costs a factor $1/r$ in each of the two entries. Therefore
\[
 \Pr(b\ge j,\ b'\ge j')\le \mathbf 1\{j'=0\ \text{or rank}(C\bmod r)\le1\}\,
 M^{-2j}\,r^{-2\max(j'-1-j,0)} .
\]
The terms with $j'=0$ contribute at most $\sum_j(B_2/M^2)^j\le2$. If
$\operatorname{rank}(C\bmod r)\le1$, the terms with $j'\ge1$ contribute at
most
\[
 \sum_j(B_2/M^2)^j\Bigl(\sum_{j'=1}^{j+1}r^{j'}+\sum_{s\ge1}r^{j+1+s}r^{-2s}\Bigr)
 \le\sum_j(B_2/M^2)^j\,3r^{j+1}=3r\sum_j\theta_0^j\le6r,
\]
using \eqref{eq:theta}.
\end{proof}

Let $\mathcal J$ be the event that some integer in $[-\tau,\tau]$ is congruent to
$\det C$ modulo $h$.

\begin{corollary}\label{cor:collision}
For three independent columns, $\E[D\mathbf 1_{\mathcal J}]\le33r^{-3}$.
\end{corollary}

\begin{proof}
By Lemma~\ref{lem:obstruction}, $\mathcal J$ requires two equal labels. Exactly two
labels coincide with probability at most $3/(r^3-1)\le4r^{-3}$. By
\eqref{eq:column0} the columns of $C\bmod r$ are
$\Nm(\xi_j)^2(1,\varphi(\xi_j),-\Tr\theta^4)^{\mathsf T}$, two of them equal; the
rank is at most one only if the third label has the same value of $\varphi$,
which has conditional probability at most $(r^2-1)/(r^3-2)\le1/r$ (the number of
$\xi\ne\xi_1$ with $\varphi(\xi)=\varphi(\xi_1)$ is $r^2-1$ or $r^2-2$). By
Lemma~\ref{lem:moment} this case contributes at most
$4r^{-3}(2+6r\cdot r^{-1})=32r^{-3}$. Three equal labels have probability at
most $2r^{-6}$ and contribute at most $2r^{-6}(2+6r)$.
\end{proof}

\section{The auxiliary modulus}\label{sec:aux}

Put $H=2h$ and choose a prime $q$ with $h^6<q\le2h^6$. Let
$w=\lfloor q/(1000H^2)\rfloor$. Choose a nonsquare $\nu\in\F_q$ and put
\[
 S=\{(x,y,x^2-\nu y^2):0\le x<w,\ y\in\F_q\},\qquad s=|S|=wq .
\]
This is the slab cap of~\cite{Blinov}. A line meets the graph of
$x^2-\nu y^2$ in at most two points (the restriction is a quadratic with
nonzero leading coefficient $x_0^2-\nu y_0^2$ for a direction with
$(x_0,y_0)\ne0$, and vertical lines meet once), so $S$ has no three collinear
points. Since $q\ge2000H^2$,
\begin{equation}\label{eq:qs}
 s\ge\frac{q^2}{2000H^2},\qquad \frac{q^3}{s}\le2000H^2q=8000h^2q .
\end{equation}
The allowed shifts $\mathcal A=\{a\in\F_q^3:\|ma_1\|_q>3Hw\text{ for }1\le
m\le3H\}$, the random set $\mathcal V=G_q(a+S)$ with $a$ uniform in $\mathcal A$ and $G_q$
uniform in $\mathrm{GL}_3(\F_q)$, and the weight
\[
 W(A)=(q^3/s)^3\Pr(\text{all columns of }A\bmod q\text{ lie in }\mathcal V)
\]
are as in \cite[Section 5]{OAI}; the same analysis for the slab is written out in
\cite[Section 3.4]{Euler}. Lemma 5.2 of \cite{OAI} is stated for
$S\subset\{0,\dots,w-1\}^3$, but its proof uses the box only through the first
coordinate. Indeed: (a) a uniform shift violates the condition for a given $m$ with
probability $(6Hw+1)/q$, so $\Pr(a\notin\mathcal A)\le3H(6Hw+1)/q<1/2$, which does not
involve $S$; (b) if $a+v=0$ with $v\in S$, then $\|a_1\|_q=\|v_1\|_q\le w-1$, contrary to
$m=1$; (c) a relation $\sum_ix_iu^{(i)}=0$ with $u^{(i)}=G_q(a+v^{(i)})$, $0<|x|\le H$
and $m=\sum x_i\ne0$ gives $ma_1=-\sum_ix_iv^{(i)}_1$, of absolute value at most
$3H(w-1)$ because $0\le v^{(i)}_1<w$, again a contradiction; (d) if $m=0$ the relation
contradicts the affine independence of distinct points of $\mathcal V$, which follows from the
line property. The proof of \cite[Proposition 5.3]{OAI} uses only the affine invariance
of the line property, the transitivity of the affine group on affinely independent
triples, and $|S|=s$. We obtain:
$|\mathcal A|\ge q^3/2$; $0\notin\mathcal V$; a relation $\sum x_iu^{(i)}=0$ in
$\F_q^3$ with $0<|x|\le H$ is impossible for three distinct points of $\mathcal V$; and
\begin{equation}\label{eq:weights}
 W(A)\ll\begin{cases}
 1&\text{if the columns of }A\bmod q\text{ are affinely independent},\\
 q^3/s&\text{if }\operatorname{rank}(A\bmod q)\ge2,\\
 (q^3/s)^2&\text{always}.
 \end{cases}
\end{equation}
If $W(A)>0$ then no column of $A\bmod q$ is zero, and the columns are either
affinely independent or two of them are equal.

\section{Weighted lattice counts}\label{sec:counts}

Fix $C$ with the data $D,E,I,\Lambda$ of Lemma~\ref{lem:smith}, and an integer
$N\ge h$ with $hq\mid N$. Let
$\mathcal B=([0,N)^2\times[N,2N))\cap\Z^3$ and $\pi(u)=(u_1/u_3,u_2/u_3)$.

\begin{proposition}\label{prop:count}
For every integer $t$ with $|t|\le\tau$, the sum of $W(A)$ over integer
matrices $A$ with columns in $\mathcal B$, rows in $\Lambda$, pairwise distinct
projected columns and $\det A=t$ is $\ll(\log2N)^2N^6/I^2$.
\end{proposition}

\begin{proof}
If $t\ne0$, then $0<|t|\le\tau<q$, so $A\bmod q$ has rank three and
$W(A)\ll1$; every row has length $<4N$ and $\lambda_3(\Lambda)\le E\le h<4N$,
so \cite[Proposition 2.5]{OAI} with $X=4N$ gives the bound.

Let $t=0$. We follow the proof of \cite[Proposition 7.1]{OAI}. The matrix has
rational rank two, and its primitive null vector $x$ has all coordinates
nonzero and $|x|\ll N^2$. Put $\Lambda_x=\Lambda\cap x^\perp$, $x\cdot\Lambda=g(x)\Z$
and $J_x=I|x|/g(x)$. Then $g(x)\mid E$, so $g(x)^3/I\le E^2/D\le h^2$, and
$\Lambda_x$ reduces onto the plane $\Pi_x\subset\F_q^3$ orthogonal to $x$
because $q\nmid E$ \cite[Lemma 7.2]{OAI}. For a sublattice of index $m$ in
$\Lambda_x$, the number of rank-two row triples of length $\le4N$ is
$\ll N^6/(mJ_x)^3$.

\emph{Moment.} For $Y\ge h$, $\sum_{Y\le|x|<2Y}g(x)^3\ll Y^3I$. Indeed
$g^3\le\sum_{m\mid g}m^3$ over $m=\prod_pp^{j_p}$ with $j_p\le e_p$, and by
Lemma~\ref{lem:smith} the vectors with $m\mid g(x)$ form a union of residue
classes modulo $m$ of proportion $\prod_pp^{-j_p-\max(j_p-b_p,0)}$; since
$m\le E\le h\le Y$ each class has $\ll(Y/m)^3$ vectors of length $<2Y$. Hence
the sum is
$\ll Y^3\prod_p\sum_{j\le e_p}p^{2j-\max(j-b_p,0)}\ll Y^3\prod_pp^{b_p+e_p}=Y^3I$,
each inner sum being dominated by its last term.

Split the null vectors into $O(\log2N)$ dyadic shells $Y\le|x|<2Y$; for fixed
$x$ and index $m$ the count is $\ll N^6g(x)^3/(m^3I^3Y^3)$. The cases are those
of \cite[Proposition 7.1]{OAI}, with \eqref{eq:qs} in place of
$q^3/s\ll qh^{12}$:
\begin{itemize}
\item affinely independent residues: $|x|>H$, so $Y>H/2=h$; with $m=1$,
$W\ll1$ and the moment bound, the shell contributes $\ll N^6/I^2$;
\item an equal pair $i,j$ with $x\bmod q\notin\F_q(\mathbf e_i-\mathbf e_j)$:
rows lie in an index-$q$ sublattice, there are $\ll Y^3$ normals, and with
$g^3\le Ih^2$ and $W\ll(q^3/s)^2$ the shell contributes
$\ll(N^6/I^2)\,h^2q^{-3}(q^3/s)^2\ll(N^6/I^2)h^6/q\le N^6/I^2$;
\item an equal pair with $x\bmod q\in\F_q(\mathbf e_i-\mathbf e_j)$: then
$Y>q/2$ and there are $\ll Y^3/q^2$ normals; rank $\ge2$ modulo $q$ gives
$\ll(N^6/I^2)h^2q^{-2}(q^3/s)\ll(N^6/I^2)h^4/q$, and rank $\le1$ (rows in one
of $q+1$ lines, each of index $q$) gives
$\ll(N^6/I^2)h^2q^{-4}(q^3/s)^2\ll(N^6/I^2)h^6/q^2$.
\end{itemize}
Since $q>h^6$ all contributions are $\ll N^6/I^2$ per shell.
\end{proof}

\section{Sampling, deletion and the exponent}\label{sec:deletion}

\subsection{Sampling}
Let $s_0=479$ and
\begin{equation}\label{eq:N}
 N=\frac{(hq)^4}{B^{s_0}}=r^{4k-s_0}B_2^{4(k-1)-s_0}q^4 .
\end{equation}
This is an integer divisible by $hq$, and $N\ge h$. Choose $G_h$ uniformly in
$\mathrm{SL}_3(\Z/h\Z)$ and $\mathcal V$ as in Section~\ref{sec:aux}, independently.
Given these shared choices, each sample draws a fresh column $c$ as in
Section~\ref{sec:digits}, a fresh uniform element of $\mathcal V$, and a uniform
$u\in\mathcal B$ with $u\equiv G_hc\pmod h$ and $u\bmod q$ equal to that element.
Each prescribed residue pair has exactly $(N/(hq))^3$ lifts. The exact lifting
identity \cite[Lemma 6.2]{OAI} holds verbatim: conditional on $C$ and the
labels,
\[
 \Pr(A\in\mathcal E\mid C)=\frac{h^9}{N^9|\mathcal O_C|}\sum_{A\in\mathcal E,\
 A\bmod h\in\mathcal O_C}W(A).
\]
A triple is \emph{bad} if its projections are pairwise distinct and
$|\det A|\le\tau$. Since $2\tau<h$, at most one determinant value in
$[-\tau,\tau]$ is possible, and none unless $\mathcal J$ occurs. By
Lemma~\ref{lem:smith}, Proposition~\ref{prop:count} and
Corollary~\ref{cor:collision},
\begin{equation}\label{eq:bad}
 \Pr(\text{bad})\ll(\log2N)^2\frac{h}{N^3}\E[D\mathbf 1_{\mathcal J}]\ll(\log2N)^2\frac{h}{N^3r^3}.
\end{equation}
For two samples, $\Pr(\pi(u)=\pi(v))\ll(hq)^6N^{-3}$ \cite[Lemma 6.1]{OAI}.

\subsection{Deletion}
Fix $\gamma=354042/354043<1$, put $a_r=r^\gamma\sqrt{N^3/\tau}$,
$n_r=\lfloor a_r\rfloor$, and take $2n_r$ samples. Let $Z$ count the pairs with
equal projections and the bad triples. Then
\[
 \frac{\E Z}{n_r}\ll n_r\frac{(hq)^6}{N^3}+n_r^2(\log2N)^2\frac{h}{N^3r^3}.
\]
For the first term, $N^{3/2}=(hq)^6B^{-3s_0/2}$ and $\tau\ge B^{k-1}/3$, so
\[
 n_r\frac{(hq)^6}{N^3}\le\frac{r^\gamma B^{3s_0/2}}{\sqrt\tau}
 \le\sqrt3\,r^\gamma B^{(3s_0-(k-1))/2}=\sqrt3\,r^\gamma B^{-1}\to0,
\]
because $3s_0-(k-1)=1437-1439=-2$ and $B\gg r^6$. For the second term,
$n_r^2\le r^{2\gamma}N^3/\tau$ and $h/\tau\le3r$, so it is
$\ll(\log2N)^2r^{2\gamma-2}\to0$, since $\log2N\ll\log r$ and $\gamma<1$.
Hence $\E Z<n_r$ for all large $r$. Deleting one index from each counted pair or
triple leaves at least $n_r$ samples whose projections are distinct and whose
triples satisfy $|\det A|>\tau$. Every coordinate $u_3$ lies in $[N,2N)$, so the
projective area identity gives
\[
 \Delta(P_r)\ge\frac{\tau}{16N^3},\qquad n_r^2\Delta(P_r)\ge\frac{a_r^2}{4}\cdot\frac{\tau}{16N^3}=\frac{r^{2\gamma}}{64}
\]
for a set $P_r$ of $n_r$ points.

\subsection{The exponent}
We have $h=rB^{k-1}$, $h^6<q\le2h^6$, $B\asymp r^6$ and $\tau\asymp B^{k-1}$.
Hence
\[
 a_r=r^\gamma\frac{(hq)^6}{B^{3s_0/2}\sqrt\tau}\asymp
 r^{\gamma}\,h^{42}B^{-3s_0/2-(k-1)/2}
 \asymp r^{\gamma+42}B^{\frac{83}{2}(k-1)-\frac32s_0}\asymp r^{\alpha},
\]
\[
 \alpha=\gamma+42+249(k-1)-9s_0=\gamma+42+249\cdot1439-4311=\gamma+354042 .
\]
So $a_0 r^\alpha\le n_r\le b_0r^\alpha$ for constants $a_0,b_0>0$ and all large
primes $r$, and $n_r^2\Delta(P_r)\ge r^{2\gamma}/64\ge c\,n_r^{2\gamma/\alpha}$.
With $\gamma=354042/354043$,
\[
 \frac{2\gamma}{\alpha}=\frac{2\cdot354042/354043}{354042\cdot354044/354043}
 =\frac{2}{354044}=\frac1{177022}=\eta .
\]

\subsection{Every cardinality}
Given a large $n$, let $y=(n/a_0)^{1/\alpha}$ and choose a prime $r\in(y,2y]$. Then
$n\le n_r\le C_0n$ with $C_0=2^\alpha b_0/a_0$. Keeping $n$ points of $P_r$ does not
decrease the minimum area, so
$\Delta(n)\ge\Delta(P_r)\ge c\,n_r^{-2+\eta}\ge c\,C_0^{-2}n^{-2+\eta}$. This proves
Theorem~\ref{thm:main}. \qed

\section{Remarks}

\paragraph{Where the gain comes from.} In~\cite{OAI,Blinov,Euler} the modulus is
$h=B^k$ and the obstruction excludes $[-\tau,\tau]$ with $h/\tau\asymp B$; the
collision probability must beat $B\,r^{2\gamma}$, which forces $d=5$ even with
the fresh-digit refinement. With $B\equiv0\pmod r$ the information
``$\gamma_{k-1}\not\equiv0\pmod r$'' already excludes the window for the
smaller modulus $rB^{k-1}$, so only $r\,r^{2\gamma}$ must be beaten. The digit
lengths, the cap and the lift box are unchanged from~\cite{Blinov,Euler}.

\paragraph{The value of $\gamma$.} Any $\gamma<1$ works in the deletion step;
the choice $\gamma=354042/354043$ gives the clean value $1/177022$, and the
exponents obtained approach $2/354043$. With $\gamma$ this close to $1$ the second term in the
deletion step is small only once $(\log r)^2r^{-2/354043}$ is, so $n_0$ is astronomically
large; this does not affect the theorem.

\paragraph{Checks.} The ancillary files contain: the design check
(\texttt{check\_design.py}); the exact identity \eqref{eq:encoding} and the
column \eqref{eq:column0} for random labels with $d=3$, $r=23$
(\texttt{check\_encoding.py}); an end-to-end test that
builds the actual integer digit columns with the parameters
\eqref{eq:params}--\eqref{eq:modulus} for $r=23$ and verifies
Lemma~\ref{lem:obstruction} (\texttt{check\_obstruction.py}); and the exponent
arithmetic (\texttt{exponent.py}).

\paragraph{Status.} This note has not been refereed.

\begin{thebibliography}{9}
\bibitem{OAI} OpenAI, \emph{A power improvement in the Heilbronn triangle lower
bound}, OpenAI Math Release preprint, September 25, 2026; source commit
1581d4cb.
\bibitem{Blinov} I.~Blinov, \emph{Heilbronn triangle bounds: finite packings
and improved scales}, GitHub research draft (littleBro/heilbronn-triangle-bounds),
October 2026.
\bibitem{Euler} Euler team (canghaimeng), \emph{A smaller lift box in the
degree-five Heilbronn construction}, GitHub
(canghaimeng/openai-math-improvement, bound-191), October 18, 2026.
\bibitem{KPS} J.~Koml\'os, J.~Pintz and E.~Szemer\'edi, A lower bound for
Heilbronn's problem, \emph{J. London Math. Soc.} (2) 25 (1982), 13--24.
\end{thebibliography}
\end{document}