\section{Subset-sum anticoncentration}\label{sec:anticoncentration}

The analytic input is the following theorem of Pham and Sauermann.

\begin{theorem}[{\citep[Theorem~1.3]{Pham2026}}]\label{thm:scalar}
There is an absolute constant $C\ge1$ such that the following holds.
Let $p$ be prime, let $Y\subseteq\Fp$ have size $N\ge2$, and let $Z$ be a uniformly random $v$-element subset of $Y$, where
\begin{equation*}
C\log N\le v\le10^{-3}N/\log N.
\end{equation*}
Then
\begin{equation*}
\sup_{z\in\Fp}\Pr\left(\sum_{y\in Z}y=z\right)
\le\frac1p+\frac{C}{N\sqrt v}.
\end{equation*}
\end{theorem}

Zero is allowed in $Y$.
Fix $\beta>0$ and put
\begin{equation}\label{eq:probability-scale}
L=\lfloor m^{1/3}\rfloor,
\qquad \eta=\min\{\beta,1/6\}.
\end{equation}
The cube-root scale balances the scalar estimate with the count of clustered collision witnesses.

\begin{lemma}\label{lem:scalar-range}
For each fixed $0<a\le1$, there is a threshold $m_a$, depending only on $a$ and $C$, with the following property.
Suppose $m\ge m_a$, $p\ge m^{1+\beta}$, and $am\le N\le m$.
If $Z$ is a uniform $L$-subset of an arbitrary $N$-element set $Y\subseteq\Fp$, then
\begin{equation}\label{eq:scalar-atom}
\sup_z\Pr\left(\sum_{y\in Z}y=z\right)
\le K_a m^{-1-\eta},
\qquad K_a=1+\frac{\sqrt2 C}{a}.
\end{equation}
\end{lemma}

\begin{proof}
Choose $m_a$ so that, for every $m\ge m_a$,
\begin{equation*}
m\ge\max\{8,2/a\},\qquad
2C\log m\le m^{1/3},\qquad
1000\log m\le a m^{2/3}.
\end{equation*}
Such a threshold exists and is independent of $\beta$.
Then $N\ge2$, $L\ge m^{1/3}/2$, and
\begin{align*}
C\log N&\le C\log m\le L \\
&\le m^{1/3}\le10^{-3}am/\log m \\
&\le10^{-3}N/\log N.
\end{align*}
The last inequality uses $N\ge am$ and $0<\log N\le\log m$.
Theorem~\ref{thm:scalar} therefore bounds the largest atom by
\begin{equation*}
m^{-1-\beta}+\frac{\sqrt2 C}{a}m^{-1-1/6}
\le K_a m^{-1-\eta}.\qedhere
\end{equation*}
\end{proof}