\section{Sampling the Hecke coefficients in progressions}
\label{sec:sampling}

We next combine the shifted bilinear estimate with elementary lattice
counting. The marking primes are smaller than every fixed positive power
of $x$, whereas the primes in $I_x$ are larger than $x^{1-\eps}$.
Consequently every marking weight $H$ satisfies
\begin{equation}\label{eq:mark-removal}
 H(\ell k)=H(k)\qquad(\ell\in I_x).
\end{equation}
In this section $H$ may be one array, a product of two arrays, a constant,
or a fixed linear combination of these. In particular $H$ is bounded,
with a bound depending on the fixed arrays.

\begin{proposition}\label{prop:sampling}
Let $c=\eps/64$ and $R=x^c$. For every fixed Dirichlet character $\chi$
and every $A>0$,
\begin{equation}\label{eq:sampling}
 \sum_{r\leq R}\left|
 \sum_{\substack{u\geq1\\r\mid u+1}}
 \Phi(u/x)H(u)(f_x(u)-C_x)
 \lambda_\eta(u+1)\chi(u+1)\right|\ll_A xL^{-A}.
\end{equation}
\end{proposition}

\subsection{The contribution with a large prime factor}
Expand $f_x$ and write $u=\ell k$. The support of $\Phi$ gives
$x^{\eps/2}\ll k\ll x^\eps$. Divide this range into dyadic intervals
$k\asymp T$, and write uniquely
\[
 k=d^2a,\qquad a\text{ squarefree}.
\]
Put $D=x^{\eps/32}$. For $d\leq D$, set $n=d^2\ell$. By
\eqref{eq:mark-removal} the summand becomes
\[
 H(d^2a)\ind_{d^2\mid n}\ind_{n/d^2\in I_x\ {\rm prime}}
 \Phi(an/x)\lambda_\eta(an+1)\chi(an+1),
 \qquad a\asymp T/d^2.
\]
The congruence $r\mid an+1$ implies $(a,r)=1$ and is the disjoint union
of the conditions
\[
 a\equiv b\pmod r,\qquad n\equiv-b^{-1}\pmod r,
 \qquad b\in(\Z/r\Z)^\times.
\]
There are at most $r$ such pairs. The fixed factor $\chi(an+1)$ is
handled by partitioning $a,n$ into residue classes modulo its fixed
conductor. All restrictions now separate into a bounded coefficient
in $a$ and a bounded coefficient in $n$. No factorization of
$\lambda_\eta(an+1)$ is involved.

Apply Proposition~\ref{prop:bilinear} to each pair, and sum over
$d\leq D$ and $r\leq R$. On a single $T$-interval this gives
\begin{equation}\label{eq:small-squarepart}
 \ll_\nu x^\nu R^2
 \left(xT^{-1/2}D^2+x^{7/8}T^{3/8}D^{1/4}\right).
\end{equation}
The factors in $D$ follow from $\sum_{d\leq D}d\ll D^2$ and
$\sum_{d\leq D}d^{-3/4}\ll D^{1/4}$. The smallest first-variable
scale is $x^{7\eps/16}$ up to a constant, so this application remains
in a positive-power range.

For $d>D$, an absolute bound suffices. Dropping the primality condition
on $\ell$, the number of pairs on one $T$-interval is at most
\[
 \frac{x}{T}\sum_{d>D}\frac{T}{d^2}\ll\frac{x}{D}.
\]
For each pair, the number of $r\leq R$ dividing $\ell k+1$ is at most
$\tau(\ell k+1)$. Thus \eqref{eq:hecke-bounds} and the elementary
divisor bound imply
\begin{equation}\label{eq:large-squarepart}
 \sum_{r\leq R}|\text{contribution with }d>D|
 \ll_{\nu,H}x^{1+2\nu}/D.
\end{equation}
In particular this step costs a divisor factor, rather than $R$.
The three exponents in \eqref{eq:small-squarepart} and
\eqref{eq:large-squarepart}, before logarithmic losses, are at most
\[
 1-\frac\eps4+\frac\eps{16}+\frac\eps{32}+\nu,
 \quad
 \frac78+\frac{3\eps}8+\frac\eps{128}+\frac\eps{32}+\nu,
 \quad
 1-\frac\eps{32}+2\nu.
\]
Choosing $\nu<\eps/512$ and allowing for the $O(L)$ dyadic intervals
gives a total $O(x^{1-\eps/128})$.

\subsection{Cancellation in norm classes}
The term multiplied by $C_x$ is estimated by lattice counting. The ring
$\mathcal O_F$ has class number one and six units. There is a fixed
integer $M_\eta$ such that the function
$\vartheta(z)=\eta((z))$, extended by zero at the conductor, is periodic
modulo $M_\eta\mathcal O_F$. Hence
\begin{equation}\label{eq:lattice-hecke}
 \lambda_\eta(n)=\frac16
 \sum_{\substack{z\in\mathcal O_F\\\Nm z=n}}\vartheta(z).
\end{equation}
The Dirichlet character $\chi(\Nm z)$ is periodic as well.

For every modulus $q$ and every $b\bmod q$, the complete sum of
$\vartheta(z)\chi(\Nm z)$ subject to $\Nm z\equiv b\pmod q$ is zero,
when taken modulo a common multiple of all the fixed moduli and $q$.
To see this, lift a unit $z_0$ on which $\vartheta$ is nontrivial to the
larger finite ring. Such a lift exists by surjectivity of reduction on
the unit groups of these quotient rings. Then
\[
 v=z_0/\overline{z_0},\qquad
 \Nm v=1,\qquad
 \vartheta(v)=\vartheta(z_0)^2\ne1.
\]
Multiplication by $v$ preserves every norm condition and every conductor
exclusion, and multiplies the sum by $\vartheta(v)$. This argument also
applies when $b$ is not a unit modulo $q$.

Tile the Eisenstein lattice by complete residue cells of side $O(q)$.
The radial smooth function in \eqref{eq:lattice-hecke} is supported in a
disk of radius $O(\sqrt x)$ and changes by $O(q/\sqrt x)$ across a cell.
Subtracting its value at one point of each complete cell, and bounding
the boundary cells directly, gives
\begin{equation}\label{eq:norm-progression}
 \left|\sum_{n\equiv b\, (q)}
 \lambda_\eta(n)\chi(n)\Phi((n-1)/x)\right|
 \ll_{\eta,\chi,\Phi}q\sqrt x+q^2.
\end{equation}

\subsection{A square-mean truncation of the marking weights}
We record the truncation estimate that also enters the positive sieve.

\begin{lemma}\label{lem:array-truncation}
Let $H$ be as above, and let $h_0=\lfloor L^{1/100}\rfloor$.
There is a divisor polynomial
\[
 H_0(u)=\sum_d c_d\ind_{d\mid u},\qquad
 \max_{c_d\ne0}d=x^{o(1)},\qquad\sum_d|c_d|=x^{o(1)},
\]
with squarefree support, and a nonnegative function $E_H$ such that
$|H-H_0|\leq E_H$ and
\begin{equation}\label{eq:array-truncation}
 \sum_{u\asymp x}E_H(u)^2
 \ll_H xh_0^{O_H(1)}\frac{C_H^{h_0}}{h_0!}+x^{o(1)}.
\end{equation}
\end{lemma}

\begin{proof}
Expand each array by choosing its represented prime in each interval and
then expanding the remaining damping factors in the divisibility
indicators. If several arrays occur, group equal chosen primes first;
the resulting monomials are indexed by the union of their prime labels.
Truncate the expansion of the unrepresented factors at degree $h_0$.
The Bonferroni inequalities bound the absolute error by a fixed power
of $1+\Omega_0$, times a constant multiple of
$\binom{\Omega_0}{h_0}$; here $\Omega_0$ counts the distinct prime hits in
the union of the fixed marking intervals. This is the represented-label
expansion used in~\cite[proof of Lemma~7.2]{PD}.

For completeness, its square has the required small mean. Let $V_0$ be
the bounded sum of reciprocal primes in those intervals. In the
independent divisor model, group the two $h_0$-element label sets by the
size $j$ of their intersection. Their total expectation is at most
\[
 \sum_{j=0}^{h_0}
 \frac{V_0^{2h_0-j}}{j!(h_0-j)!^2}
 \leq\frac{[2\max(V_0,V_0^2)]^{h_0}}{h_0!}.
\]
The fixed powers coming from represented labels introduce only
$h_0^{O_H(1)}$ and a fixed exponential factor. Every union of labels
in this calculation has degree at most $2h_0+O_H(1)$ and hence product
$x^{o(1)}$. Counting its divisibility condition by $O(x/d)+O(1)$
proves~\eqref{eq:array-truncation}. The total number and coefficient
mass of the retained terms and of these error terms are $x^{o(1)}$,
which also bounds the sum of rounding errors.
\end{proof}

Apply this lemma to the contribution in Proposition~\ref{prop:sampling}
without $f_x$. Its summed truncation error is at most a constant times
\[
 \sum_{u\asymp x}\tau(u+1)|\lambda_\eta(u+1)|E_H(u).
\]
Cauchy's inequality, \eqref{eq:hecke-bounds}, and
$\sum_{n\asymp x}\tau(n)^4\ll xL^{15}$ show that this is $O_A(xL^{-A})$
for every $A$. For a retained $d$, the conditions $d\mid u$ and
$r\mid u+1$ either are inconsistent or specify one residue class for
$n=u+1$ modulo $rd$. Equation~\eqref{eq:norm-progression} and the
coefficient-mass bound therefore give
\[
 x^{1/2+2c+o(1)}+x^{3c+o(1)}
\]
after summing $r\leq R$. This is a power saving. Combining it with the
large-prime-factor estimate proves Proposition~\ref{prop:sampling}.