% Hexagon source adaptation of the published Zenodo edition. % Changes are limited to packaging, bibliography rendering, navigation and archive links. % Mathematical formulas and their order are preserved; see the local integrity record. \PassOptionsToPackage{expansion=false,nopatch=footnote}{microtype} \documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{mathtools,amsthm} \usepackage{newpxtext,newpxmath} \usepackage{microtype,booktabs,array,longtable} \usepackage[margin=27mm,headheight=15pt]{geometry} \usepackage{enumitem,fancyhdr,needspace,aliascnt,float} \usepackage[hidelinks,pdfauthor={Brice Pouly},pdftitle={Reflected sign patterns of the Liouville function at every large total}]{hyperref} \usepackage[nameinlink,noabbrev]{cleveref} \pagestyle{fancy}\fancyhf{} \fancyhead[L]{\small Reflected sign patterns of the Liouville function} \fancyfoot[C]{\thepage} \renewcommand{\headrulewidth}{0.3pt} \setlength{\parindent}{1.3em} \setlength{\parskip}{2pt} \setlength{\emergencystretch}{2em} \setcounter{tocdepth}{1} \allowdisplaybreaks[2] \newtheorem{theorem}{Theorem}[section] \newaliascnt{proposition}{theorem} \newtheorem{proposition}[proposition]{Proposition} \aliascntresetthe{proposition} \newaliascnt{lemma}{theorem} \newtheorem{lemma}[lemma]{Lemma} \aliascntresetthe{lemma} \newaliascnt{corollary}{theorem} \newtheorem{corollary}[corollary]{Corollary} \aliascntresetthe{corollary} \theoremstyle{definition} \newaliascnt{definition}{theorem} \newtheorem{definition}[definition]{Definition} \aliascntresetthe{definition} \newaliascnt{inputthm}{theorem} \newtheorem{inputthm}[inputthm]{Classical input} \aliascntresetthe{inputthm} \theoremstyle{remark} \newaliascnt{remark}{theorem} \newtheorem{remark}[remark]{Remark} \aliascntresetthe{remark} \crefname{theorem}{Theorem}{Theorems} \Crefname{theorem}{Theorem}{Theorems} \crefname{proposition}{Proposition}{Propositions} \Crefname{proposition}{Proposition}{Propositions} \crefname{lemma}{Lemma}{Lemmas} \Crefname{lemma}{Lemma}{Lemmas} \crefname{corollary}{Corollary}{Corollaries} \Crefname{corollary}{Corollary}{Corollaries} \crefname{definition}{Definition}{Definitions} \Crefname{definition}{Definition}{Definitions} \crefname{inputthm}{Classical input}{Classical inputs} \Crefname{inputthm}{Classical input}{Classical inputs} \crefname{remark}{Remark}{Remarks} \Crefname{remark}{Remark}{Remarks} \numberwithin{equation}{section} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\C}{\mathbb C} \newcommand{\E}{\mathbb E} \newcommand{\1}{\mathbf 1} \newcommand{\PP}{\mathcal P} \newcommand{\BB}{\mathcal B} \newcommand{\XX}{\mathcal X} \newcommand{\UU}{\mathcal U} \newcommand{\FF}{\mathcal F} \newcommand{\rad}{\operatorname{rad}} \newcommand{\e}{\mathrm e} \newcommand{\dd}{\,\mathrm d} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\floor}[1]{\lfloor#1\rfloor} \newcommand{\bracket}[1]{\bigl[#1\bigr]_+} \newcommand{\mc}[1]{\mathcal{#1}} \title{\vspace{-1.2em}Reflected sign patterns of the Liouville function\vspace{.2em}\\ at every large total} \author{Brice Pouly} \date{25 September 2026\\[.5em]\small Developed with substantial assistance from GPT-6 Astra.} \usepackage{bookmark} \begin{document} \maketitle \begin{abstract} For signs $s,t$, let $N_{s,t}(N)$ count the ordered representations $N=n+(N-n)$ with $\lambda(n)=s$ and $\lambda(N-n)=t$. We prove that, for every fixed $\varepsilon>0$, all four counts are at least $c_\varepsilon N^{1-\varepsilon}$ at every sufficiently large integer total. The constants are ineffective through Siegel's theorem. The same conclusion holds uniformly for completely multiplicative sign functions whose positive primes satisfy $\sum_{f(p)=1}p^{-\theta}\le M$, with fixed $0<\theta<1$ and $M$. Separately, at each sufficiently large prime $p$, both patterns with product $(-1/p)$ have at least $p/(4(\log p)^{96})$ representations. All four satisfy this stronger bound in six reduced classes modulo $24$ and outside one very sparse family of primes. The proof combines finite reflection-to-dilation propagation, a short-product spectral test, reconstruction from the units, and signed counting of prime disagreements. For the uniform extension, candidates are united before a common exceptional population is removed. None of the conclusions gives a positive-density bound or cancellation of the reflected correlation at every total. \end{abstract} \par\medskip \noindent\begin{minipage}{\textwidth} \small \setlength{\parskip}{0pt} \makeatletter \renewcommand*\l@section[2]{\@dottedtocline{1}{0em}{1.7em}{#1}{#2}} \makeatother \tableofcontents \end{minipage} \par\medskip \clearpage \section{Introduction and main results}\label{sec:intro} The complete multiplicativity of the Liouville function does not make its additive observations independent. In a reflected pair $(\lambda(n),\lambda(N-n))$, the total $N$ is fixed while both summands vary. The question studied here is quantitative and pointwise in that total: how many representations of each prescribed sign type must occur? We write $\lambda(n)=(-1)^{\Omega(n)}$, so that $\lambda(p)=-1$ at every prime. For a completely multiplicative function $f:\N\to\{-1,1\}$, define \begin{equation}\label{eq:counts} N^f_{s,t}(N)=\#\{1\le n0$ there are constants $c_\varepsilon>0$ and $N_\varepsilon$ such that \begin{equation}\label{eq:main-all} N_{s,t}(N)\ge c_\varepsilon N^{1-\varepsilon} \qquad(N\ge N_\varepsilon,\ s,t\in\{-1,1\}). \end{equation} There is no restriction on the parity, prime support, or prime-power valuations of $N$. The constants and threshold are ineffective through the use of Siegel's theorem. \end{theorem} Thus $\log m(N)/\log N\to1$. The quantifier ``for each fixed $\varepsilon$'' is essential: the theorem supplies neither a rate obtained by choosing $\varepsilon$ as a function of $N$, nor a bound $N/(\log N)^A$ with fixed $A$ at all totals. A population of size $N^{1-o(1)}$ can still have density zero. The function in \cref{thm:all} is the original Liouville function. No prime values are changed, and there is no averaging over totals or over a random multiplicative model. Qualitative existence of all four patterns at every $N\ge11$ is established in Meng's recent manuscript \cite{Meng}, following the even-total result \cite{CaptainSude}. Our purpose is to obtain lower bounds for their number, not to claim the first existence result; the distinction is detailed in Section~\ref{sec:previous}. The four counts are related by one elementary identity. With $A_f(x)=\sum_{n\le x}f(n)$ and $C_f(N)=\sum_{1\le n0$. There are constants $c_{\theta,M,\varepsilon}>0$ and $N_{\theta,M,\varepsilon}$ such that \begin{equation}\label{eq:main-robust} N^f_{s,t}(N)\ge c_{\theta,M,\varepsilon}N^{1-\varepsilon} \quad\bigl(f\in\FF_{\theta,M},\ N\ge N_{\theta,M,\varepsilon},\ s,t=\pm1\bigr). \end{equation} The constants are uniform in $f$, which may vary with $N$ within the fixed budget. They remain ineffective through Siegel's theorem. The parameter $\theta$ is fixed before $N$ tends to infinity. \end{theorem} Theorem~\ref{thm:all} is the case $M=0$. We state it first because it concerns the imposed arithmetic sequence, whereas \cref{thm:robust} concerns an entire class of sequences. This extension is not obtained by subtracting a small density of errors from \eqref{eq:main-all}. For instance, changing only the prime value at $2$ gives $f(n)=\lambda(n)(-1)^{v_2(n)}$; the disagreement density with $\lambda$ is \[ \sum_{\substack{j\ge1\\j\text{ odd}}}2^{-j-1}=\frac13. \] Such a subtraction would destroy an $N^{1-\varepsilon}$ lower bound. \begin{corollary}[A counting formulation of the budget]\label{cor:primecount} Fix $0\le\beta<1$ and $C\ge0$. The conclusion of \cref{thm:robust}, with constants depending on $\beta,C,\varepsilon$, holds uniformly for functions satisfying \[ \#\{p\le x:f(p)=1\}\le Cx^\beta\qquad(x\ge2). \] \end{corollary} \begin{proof} Choose $\beta<\theta<1$. Partial summation gives \[ \sum_{p\in E(f)}p^{-\theta} \le C\theta\int_2^\infty t^{\beta-\theta-1}\dd t =\frac{C\theta\,2^{\beta-\theta}}{\theta-\beta}. \] Apply \cref{thm:robust} with this fixed budget. \end{proof} \subsection{Logarithmic losses at prime totals} For an odd prime $p$, let $\chi_p=(\cdot/p)$ and $\rho_p=\chi_p(-1)$. It is useful to separate the counts by their sign product: \begin{equation}\label{eq:productcounts} P_\rho(f;N)=\#\{1\le n0$, the quotient of $p/(\log p)^{96}$ by $p^{1-\varepsilon}$ tends to infinity. Nevertheless, \cref{thm:prime} does not replace \cref{thm:all}: its total is prime, and its all-pattern conclusion retains a possible exceptional family. The family in \eqref{eq:exceptionfamily} is a sufficient exceptional set for our argument, not a characterization of failure of the four individual lower bounds. We prove neither that it is empty nor that it is finite, and we do not construct a member. \begin{table}[H] \centering\small \begin{tabular}{p{.25\textwidth}p{.35\textwidth}p{.29\textwidth}} \toprule Conclusion & Domain & Effectivity and restriction\\ \midrule $N_{s,t}(N)\gg_\varepsilon N^{1-\varepsilon}$ & All four signs; every sufficiently large integer $N$ & Ineffective; no fixed logarithmic loss asserted\\[4pt] $N^f_{s,t}(N)\gg_{\theta,M,\varepsilon}N^{1-\varepsilon}$ & Same totals and signs; every $f\in\FF_{\theta,M}$ & Uniform in $f$; fixed $\theta<1$; ineffective\\[4pt] $N_{s,t}(p)\ge p/(4\log^{96}p)$ & Two patterns of product $\rho_p$ at every prime; all four in six classes or outside $\XX$ & Effective in principle; $\XX$ is not eliminated\\ \bottomrule \end{tabular} \caption{The three domains are not interchangeable. Every lower bound is pointwise in its stated total.}\label{tab:scope} \end{table} \subsection{Proof strategy: use the character and its disagreements}\label{sec:strategy} The argument first treats totals avoiding a fixed block of small primes. There are two branches. If both sign products have at least $N/(\log N)^{512}$ occurrences, the first moment separates the equal-sign patterns, since \[ N^f_{+,+}(N)-N^f_{-,-}(N)=A_f(N-1). \] The uniform bound $A_f(N-1)\ll_{\theta,M}N\exp(-c\sqrt{\log N})$ is negligible on this logarithmic scale. It need not be negligible relative to $N^{1-\varepsilon}$ for a fixed $\varepsilon>0$. Thus, in the remaining branch, counting products and then subtracting a first-moment error would not suffice: the construction must retain each cofactor's sign. \paragraph{From near-saturation to a positively oriented character.} When one sign product is smaller than the threshold, reflection agrees with a constant sign $\kappa$ except at very few origins. The finite descent of Section~\ref{sec:propagation} transports this error to a fixed set of prime dilations. Averaging over products of these generators reaches polylogarithmic dilations; small-denominator frequencies are paid for, including at composite totals. A high moment over small primes then confines the multiplicative Fourier mass to a small family of characters. Concentration alone is not the end of the inverse argument. The identity on short \emph{integer} products, $1=\E_{a,b\in\UU_H}f(ab)f(a)f(b)=\sum_\psi a_\psi b_\psi^2$, selects a real character with coefficient close to $+1$, using Burgess, a restricted Gram estimate and a fourth moment. Here $\UU_H$ consists of the units up to $H=\lfloor\sqrt{N-1}\rfloor$; the coefficients and the argument are given in Section~\ref{sec:spectral}. The reconstruction of Section~\ref{sec:reconstruction} then yields agreement on every prefix and small reciprocal mass of prime disagreements. In the budgeted class, the principal character is excluded. \paragraph{From disagreement primes to signed populations.} The remaining quadratic character is \emph{used}, not declared impossible. A prime $q$ with $f(q)=-1$ and $\chi(q)=1$ reverses one side of the pattern predicted by the character. Sections~\ref{sec:means}--\ref{sec:filter} count the resulting pairs with their signs intact, control overlaps, and obtain a supply of such primes from a filtered positive convolution and Siegel's theorem. Candidates from all marking primes are united \emph{before} one common exceptional population is removed. This is what leaves a power saving for every fixed $\theta<1$. \paragraph{From a convenient divisor to every total.} Among finitely many disjoint prime blocks, one carries little of the factorization of $N$. Removing its part leaves a divisor $D$ close to $N$ in exponent and coprime to that block. Simultaneous scaling of the two summands injects representations of $D$ into those of $N$ and permutes the four sign pairs. It transfers the lower bound for their minimum count, not a density; Section~\ref{sec:allproof} accounts for the exponent loss. The prime logarithmic result takes a different branch after the inverse. The values of Liouville at $2$ and $3$ restrict the possible primes; any remaining exception gives a small positive $L(1,\chi_p)$ together with a positive derivative, hence a nearby real zero. Landau--Page spaces those exceptions rather than removing them. The exponents $512$ and $96$ specify the two accuracy budgets; their role is to leave room for these different downstream steps, not to give optimized constants. \Needspace{9\baselineskip} \subsection{Relation to previous work}\label{sec:previous} \paragraph{Existence and counting.} Mangerel's published nonextremality theorem gives $|C_\lambda(N)|2$, and the positive-positive pattern at $2p$ for every prime $p>3$ (Theorems~1.1--1.2 there). Meng \cite[Theorem~A]{Meng} proves all four patterns for every $N\ge2$ outside the sharp exceptional set $\{2,3,4,5,6,9,10\}$. His Theorem~B gives both equal-sign patterns at each prime $p\ge5$. These are unconditional existence results with explicit small exceptions. Meng explicitly separates them from quantitative counting bounds in his introduction and Section~9. The point of \cref{thm:all} is instead the lower bound $N^{1-\varepsilon}$ for each fixed $\varepsilon>0$ at \emph{every} sufficiently large total. It is ineffective and does not replace the sharp finite-range existence statements. The additional quantifier in \cref{thm:robust} is uniformity over an entire class of functions, which may vary with $N$ under the fixed budget. Neither theorem imports the recent existence results or their formalizations as a proof input. \paragraph{Conditional quantitative results.} Mangerel's work on Shusterman's problem \cite{MangerelShusterman} proves under GRH the negative-negative pattern at all sufficiently large even totals (Theorem~1.2), and all four patterns outside a fixed friable domain (Theorem~1.3). Its Proposition~1.4 gives, under GRH, a lower bound of the form $p\exp(-C(\log\log p)^6)$ for every pattern at sufficiently large primes. For fixed $\varepsilon,C>0$, \[ p^{1-\varepsilon}=o\!\left(p\exp(-C(\log\log p)^6)\right),\qquad p\exp(-C(\log\log p)^6)=o\!\left(\frac p{(\log p)^{96}}\right). \] Thus our all-total power bound is not quantitatively larger than that conditional prime bound. Its distinguishing features are the absence of GRH and of restrictions on the total. The stronger logarithmic bound in \cref{thm:prime} has its own, explicitly smaller all-pattern domain. The reflection-to-dilation cocycle and Pierce-descent framework have a direct antecedent in \cite{MangerelShusterman}. Section~\ref{sec:propagation} proves the propagation used here in full; neither the GRH-dependent result nor the analytic smooth-number propagation of that work is invoked as a black box. \Needspace{7\baselineskip} \paragraph{Exact and approximate rigidity.} Mangerel's Gauss-sum rigidity theorem \cite[Theorem~1.1 and Corollary~1.2]{MangerelRigidity}, first available as a preprint in February 2025, already relates approximate dilation symmetry to a real character and, for Liouville, to a real zero near $1$. Its prime-dilation hypotheses reach $p^c$, with $c>1/4$. Here the hypothesis is a much more accurate reflected saturation, and complete multiplicativity on every short integer product is essential. Comparing the cutoffs alone would therefore misstate the relationship. His Lemma~3.6 is also a specific antecedent for passing from integer disagreements to reciprocal prime disagreement; Section~\ref{sec:reconstruction} states the differences in our unit-to-prefix reconstruction. Meng's exact prime-modulus classification \cite[Theorem~7.2]{Meng} assumes $f(2)=f(3)=-1$, and also $f(5)=-1$ for a missing equal-sign pattern. Under the stated exclusions it forces agreement with the Legendre symbol below $p/2$. Our \cref{thm:inverse} instead treats quantitative near-saturation at composite as well as prime totals, without prescribing those three prime signs, and allows a principal character or a proper conductor. These are different hypotheses and conclusions; neither comparison is a blanket strengthening of the other. Version~v3 of \cite{Clankers} gives exact extremizer classifications at odd prime powers in Section~5.3; the proper-conductor example in \cref{rem:proper} belongs to that class. Section~7.2 gives an approximate-dilation rigidity and pattern-scarcity statement at prime moduli, without GRH in that generic statement. Its Liouville separation and quantitative dilation-defect results in Sections~7.3--7.4 explicitly assume GRH. In particular, the scarcity estimate is an upper bound for the patterns forbidden by an approximately imitated character, not an unconditional lower bound for every pattern. These compared statements do not give the all-total conclusion of \cref{thm:all} or the uniform class of \cref{thm:robust}. \paragraph{Other quantifiers and analytic context.} Krishnamoorthy \cite[Theorem~2]{Krishnamoorthy} bounds the number of totals with a large normalized binary convolution: \[ \#\{N\le X:|C_\lambda(N)|>\delta N\} \ll_A\delta^{-2}X/(\log X)^A. \] An exceptional-set estimate of this kind does not by itself give a bound at each total; conversely our lower bounds do not give cancellation of $C_\lambda(N)$. Cantarini, Gambini and Zaccagnini \cite{CantariniEtAl} study weighted averages and analytic properties of related convolutions; their Theorem~1 is an explicit-formula result under RH and simplicity of the nontrivial zeta zeros, not a pointwise sign-count theorem. The stronger cancellation problem originates with Corr\'adi and K\'atai \cite{CorradiKatai}. The relationship of Liouville convolution with hypothetical Siegel zeros also predates the present argument; see De~Koninck, Germ\'an and K\'atai \cite{DeKoninckGermanKatai} for a result under existence assumptions on such zeros and with its own ranges and coprimality restrictions. The local attributions below distinguish standard or inherited tools from the quantitative way they are combined: Mangerel's disagreement lemma, Pollack's convolution and reciprocal-prime method, the Montgomery--Vaughan inequality, and divisor scaling. The contribution asserted here is the counting mechanism, its losses and its uniform domains, not those general principles in isolation. The comparisons concern the stated hypotheses and conclusions, not a ranking of the methods by a single cutoff or exponent. \subsection{Reading order} Section~\ref{sec:prelim} states the analytic inputs and uniform costs. To reconstruct the inverse, read Sections~\ref{sec:propagation}--\ref{sec:spectral}, returning to the unit-to-prefix lemma in Section~\ref{sec:reconstruction} when it is called at the end of the spectral argument. Sections~\ref{sec:means}--\ref{sec:allproof} complete the signed count for Liouville and its uniform class together; Sections~\ref{sec:primetheorem}--\ref{sec:zeros} give the separate prime route. Section~\ref{sec:boundary} and Appendix~\ref{sec:ledger} record the boundaries and parameter budgets. No result about a random multiplicative model or normality is an input. \section{Notation, classical inputs, and uniform costs}\label{sec:prelim} For an integer $N\ge3$, set \begin{equation}\label{eq:notation} X=N-1,\qquad L=\log N,\qquad \ell=\log(3L),\qquad \nu=\frac{\phi(N)}N,\qquad H_x=\sum_{1\le n\le x}\frac1n. \end{equation} A sum or product indexed by $p,q,r$ is over primes whenever this is indicated or stated locally. Dirichlet characters are extended by zero off the units. The principal primitive character has conductor $1$; it is allowed in the generic inverse theorem, but not in the structured branch used for the lower bounds. For the means $A_f$ and $C_f$ defined in Section~\ref{sec:intro}, expanding the two sign indicators records the identity announced there: \begin{equation}\label{eq:indicator} 4N^f_{s,t}(N)=X+(s+t)A_f(X)+stC_f(N). \end{equation} Equivalently, \begin{equation}\label{eq:split} N^f_{+,-}=N^f_{-,+}=\frac{P_-(f;N)}2,\qquad N^f_{\pm,\pm}=\frac{P_+(f;N)\pm A_f(X)}2. \end{equation} In the branch where both products have a logarithmic-scale lower bound, these identities and the first moment separate all four counts. The structured branch will instead count each rare signed pair directly. No estimate for $C_f(N)$ is assumed. \subsection{Analytic inputs} We use the following precise forms of classical results. They are external theorems, rather than consequences of the finite diagnostics accompanying this manuscript. \begin{inputthm}[Prime counts and the Liouville mean]\label{in:mean} Chebyshev's bounds and Mertens' estimates are used in their usual effective forms. For an absolute $c>0$, \begin{equation}\label{eq:liouvillemean} A_\lambda(x)\ll x\exp(-c\sqrt{\log x})\qquad(x\ge3). \end{equation} In particular $\sup_{x\ge1}\abs{\sum_{n\le x}\lambda(n)/n}<\infty$. The first-moment input is recalled, in a stronger form, in \cite[Section~6.1]{MangerelShusterman}. \end{inputthm} \begin{inputthm}[Burgess at parameter two]\label{in:burgess} If $\xi$ is primitive and nonprincipal with conductor $Q$, then for every fixed $\eta>0$, \begin{equation}\label{eq:burgess} \left|\sum_{n\le t}\xi(n)\right|\ll_\eta t^{1/2}Q^{3/16+\eta}. \end{equation} No cube-free condition is required at this parameter; see \cite[equation~(1)]{HeathBrown}. Periodicity extends the bound, if needed, to lengths exceeding the conductor. \end{inputthm} \begin{inputthm}[Brun--Titchmarsh]\label{in:bt} For $x>q$ and $(a,q)=1$, \begin{equation}\label{eq:bt} \pi(x;q,a)\le\frac{2x}{\phi(q)\log(x/q)}. \end{equation} We need the classical Montgomery--Vaughan constant $2$ \cite{MontgomeryVaughan}, in the form recalled in \cite[Introduction]{Maynard}. \end{inputthm} \begin{inputthm}[A quadratic convolution and Siegel]\label{in:pollack} For each fixed $\eta_1>0$, there is $\eta_2>0$ such that, uniformly for a primitive nonprincipal quadratic character $\chi$ of conductor $Q$, \begin{equation}\label{eq:pollack} B_\chi(t):=\sum_{n\le t}(1*\chi)(n) =tL(1,\chi)+O_{\eta_1}(t^{1-\eta_2}) \quad(t\ge Q^{1/4+\eta_1}). \end{equation} This is \cite[Proposition~3.1]{Pollack}; even conductors are permitted. We also use Siegel's theorem \begin{equation}\label{eq:siegel} L(1,\chi)\gg_\zeta Q^{-\zeta}\qquad(\zeta>0\text{ fixed}), \end{equation} with a positive ineffective constant, as used in \cite[proof of Theorem~1.3]{Pollack}. \end{inputthm} \begin{inputthm}[Nonvanishing and Landau--Page]\label{in:page} For real primitive nonprincipal $\chi$, $L(1,\chi)>0$. There is an absolute effective $c_0>0$ such that, among primitive real characters of conductors at most $T$, at most one has a real zero $\beta>1-c_0/\log T$, for sufficiently large $T$. The latter form is recalled in \cite[Section~4.2]{GranvilleMangerel}. \end{inputthm} Only \eqref{eq:siegel} introduces ineffectivity into the all-total argument. The prime logarithmic argument uses \cref{in:mean,in:burgess,in:page}, but neither \eqref{eq:pollack} nor \eqref{eq:siegel} nor Brun--Titchmarsh. Its elementary Gauss-sum truncation is supplied in \cref{sec:zeros}. \subsection{Costs uniform in the factorization of the total} Define \[ P_N=\prod_{p\mid N}(1+p^{-1}),\qquad T_N=\sum_{p\mid N}\frac{\log p}{p+1}. \] \begin{lemma}\label{lem:uniformcosts} Uniformly in $N$, \begin{equation}\label{eq:uniformcosts} \omega(N)\ll\frac{\log(3N)}{\log\log(3N)},\qquad 4^{\omega(N)}=N^{o(1)},\qquad \nu^{-1},P_N,T_N\ll\ell. \end{equation} Moreover, if \begin{equation}\label{eq:unitprefix} u_N(t)=\#\{1\le n\le t:(n,N)=1\},\qquad J_N=\max_{1\le t\le X}\frac{t}{u_N(t)}, \end{equation} where the maximum is over integers, then $J_N\ll\log(3N)$. \end{lemma} \begin{proof} If $k=\omega(N)$, the product of the $k$ distinct prime divisors is at least $(k+1)!$. This gives the first two assertions. Split the prime factors at $L=\log N$. For the smaller ones, Mertens' estimates give the asserted bounds for the products and for $\sum\log p/p$. For the larger ones, use $\sum_{p\mid N}\log p\le L$. Their contribution to $\sum1/p$ is $O(1)$, and to $\sum\log p/p$ is at most $1$. This proves the last three bounds. For $t\ge C\log N$ with a sufficiently large absolute $C$, Chebyshev's lower bound gives $\pi(t)\ge2\omega(N)$, and therefore $u_N(t)\ge\pi(t)/2\gg t/\log N$. For smaller $t$, the unit $1$ gives $u_N(t)\ge1$. \end{proof} The factor $4^{\omega(N)}$ will only be absorbed by a fixed power saving in $N$. It is not treated as bounded, nor absorbed by a fixed power of $\log N$. This distinction matters when the prime support of $N$ is unrestricted. \section{Reconstruction from the units}\label{sec:reconstruction} Agreement on $(\Z/N\Z)^\times$ initially says nothing about integers divisible by the primes of $N$. We first show how complete multiplicativity reconstructs all prefixes, once those prime values have been matched. The fixed-scale passage from few integer disagreements to small reciprocal prime disagreement has a direct antecedent in \cite[Lemma~3.6]{MangerelRigidity}. The point here is the initial restriction to units, the prescribed quotient values at primes dividing $N$, and simultaneous control of every prefix with a cost uniform in the factorization of $N$. The argument is given in full rather than invoking that lemma outside its hypotheses. In the application, $h=fg$ is the quotient of the source and the character extension $g$ from \eqref{eq:extension}; its values at primes dividing $N$ are fixed to be $1$. The lemma has two outputs: agreement on \emph{every} prefix, and control of the reciprocal mass of prime disagreements. The first completes the inverse; the second is consumed by the signed inventory and by the value-and-derivative calculation near $s=1$. \begin{lemma}[Rectangular product energy]\label{lem:energy} Let $A,H\ge1$ be integers and let $d_{A,H}(n)$ count the pairs $a\le A$, $b\le H$ with $ab=n$. Then \begin{align} \sum_n d_{A,H}(n)^2 &=\sum_{\substack{u,v\le\min(A,H)\\(u,v)=1}} \floor{\frac A{\max(u,v)}}\floor{\frac H{\max(u,v)}}\label{eq:energyexact}\\ &\le AH\bigl(2H_{\min(A,H)}-1\bigr).\label{eq:energybound} \end{align} \end{lemma} \begin{proof} In $ab=cd$, write $a=gu$, $c=gv$, with $(u,v)=1$. Then $b=tv$ and $d=tu$. The possible $g$ and $t$ give \eqref{eq:energyexact}. The contribution of $\max(u,v)=1$ is $AH$; for each larger maximum $j$, there are at most $2j$ pairs. Summing $2AH/j$ proves \eqref{eq:energybound}. \end{proof} \begin{lemma}[Uniform reconstruction]\label{lem:reconstruct} Let $h:\N\to\{-1,1\}$ be completely multiplicative, with $h(p)=1$ for every prime $p\mid N$. Put \[ D_U=\#\{n\le X:(n,N)=1,\ h(n)=-1\},\qquad \delta=2J_N^2\sqrt{\frac{D_UH_X}{X}}. \] If $\delta<1/2$, every unit prefix and every complete prefix of $[1,X]$ has negative proportion at most $\delta$. If $\delta<3/32$, then in addition \begin{equation}\label{eq:reconstruct-primes} \sum_{\substack{q\le X\ \mathrm{prime}\\h(q)=-1}}\frac1q\le3\delta. \end{equation} \end{lemma} \begin{proof} Fix an integer $H\le X$ and put $A=\floor{X/H}$, so $AH\ge X/2$. Choose $a$ and $b$ independently and uniformly among the units in $[1,A]$ and $[1,H]$. If $h(ab)=-1$, the product lies among the $D_U$ exceptional unit positions. By Cauchy--Schwarz and \cref{lem:energy}, \[ \Pr(h(ab)=-1) \le\frac{\sqrt{2D_UAHH_X}}{u_N(A)u_N(H)} \le2J_N^2\sqrt{D_UH_X/X}=\delta. \] If $t_H$ is the negative proportion in the unit prefix of length $H$, independence and multiplicativity give \[ \Pr(h(ab)=-1)=t_A(1-t_H)+(1-t_A)t_H \ge\min\{t_H,1-t_H\}. \] Every $t_H$ is therefore at most $\delta$ or at least $1-\delta$. As units enter a prefix one at a time, a change of majority would have to pass through equal positive and negative counts. This is impossible when $\delta<1/2$. Since $h(1)=1$, the majority is positive at every prefix. Every integer $n\le t$ has a unique factorization $n=am$, where $a$ is supported on the prime divisors of $N$ and $(m,N)=1$. Our hypothesis gives $h(n)=h(m)$, and hence \begin{align*} \#\{n\le t:h(n)=-1\} &\le\delta\sum_{\substack{a\le t\\p\mid a\Rightarrow p\mid N}} u_N(\floor{t/a}) =\delta t. \end{align*} For the prime assertion, let $S_t=\sum_{q\le t,h(q)=-1}1/q$. The integers $qa\le t$ for which $q$ is negative and $a$ has no negative prime factor are disjoint as $q$ varies. They are all negative. If $S_t\le1$, their number is at least \begin{equation}\label{eq:singleprimecount} (1-S_t)\sum_{\substack{q\le t\ \mathrm{prime}\\h(q)=-1}}\floor{t/q} \ge\frac t2 S_t(1-S_t). \end{equation} Indeed, for each $q$, a union bound removes at most an $S_t$ proportion of $a\le\floor{t/q}$. At a first crossing of $1/4$, the value of $S_t$ lies in $[1/4,3/4]$, since the increment is at most $1/2$. Formula~\eqref{eq:singleprimecount} would give a negative proportion at least $3/32$, contradicting $\delta<3/32$. Thus $S_X<1/4$. Let $e$ be the negative proportion in the complete prefix $[1,X]$. The number of negative primes up to $X$ is at most $eX$. Retaining the floors in \eqref{eq:singleprimecount} gives \[ e\ge(1-S_X)(S_X-e). \] Since $S_X<1/4$, this implies $S_X\le3e\le3\delta$. \end{proof} \begin{remark} The hypothesis at primes dividing $N$ cannot be dropped. For $N=2^k$, the function $h(n)=(-1)^{v_2(n)}$ equals $1$ on all units but not on all integers. In our application the quotient of the source and its character extension is deliberately made equal to $1$ at those primes. \end{remark} \section{From reflection to polylogarithmic dilations}\label{sec:propagation} For a completely multiplicative sign function $f$ and $\kappa\in\{-1,1\}$, define the reflection error \begin{equation}\label{eq:reflectionerror} \mathscr E_\kappa(f;N)=\#\{1\le x0,\\ 0,&r_b(x)=0, \end{cases} \qquad \widehat D_b=\#\{1\le xN/2$, put $y=N-x$. When $y$ and $2y$ are outside $\mc E$, \[ f(x)=\kappa f(y),\qquad f(2x-N)=\kappa f(2y)=\kappa f(2)f(y)=f(2)f(x). \] For $2x0$, its quotient is strictly larger than $r$. Away from the states where $\vartheta$, $r_b(u)$, or $v=r_b(\vartheta)$ is zero, direct multiplication of signs gives \begin{equation}\label{eq:descentidentity} \Lambda_b(u)=\Lambda_r(r_b(u))\Lambda_b(\vartheta)w(ru)w(v), \qquad r_r(r_b(u))=N-v. \end{equation} No inverse of $r$ modulo $N$ is used. A state with quotient $r$ has at most $2^{r-1}$ earlier origins: the previous quotients form a strictly increasing subsequence of $1,\ldots,r-1$, and each such subsequence determines at most one origin through $u_{j-1}=(N-u_j)/r_j$. At a fixed quotient $r$, at most $2b+1$ states are singular. The equality $\vartheta=0$ gives at most one; $r_b(u)=0$ gives at most $b$; and the congruence $bru\equiv0\pmod N$ within $N/(r+1)R$, distinct members of $\mc A$ are distinct modulo $d_k$. Additive orthogonality therefore gives \[ \sum_{x\bmod N}|W_k(x)|^2=\frac N{M_A}. \] Averaging the approximate invariance in \eqref{eq:cocycle} and applying Cauchy--Schwarz yields \begin{equation}\label{eq:fourier-defect} |S_b(k)|\le\sqrt{D_bN/M_A}+2\overline D\qquad(d_k>R). \end{equation} For $d_k\le R$, retain the trivial bound $D_b$. Let $I=[1,h]$ with $h=\floor{X/b}$. At frequencies of reduced denominator $d$, the sum of the absolute Fourier coefficients of $\1_I$ is at most $dH_d$, by the geometric-sum bound. Thus all denominators at most $R$ cost at most $R^2H_R$. The normalized sum over all nonzero frequencies is at most $H_N$. Fourier inversion in the empty interval from \eqref{eq:cocycle}, followed by \eqref{eq:fourier-defect}, now gives \[ \frac{D_bh}N\le\frac{D_bR^2H_R}N +H_N\bigl(\sqrt{D_bN/M_A}+2\overline D\bigr). \] Since $h\ge N/(2b)$ and $R^2H_R\le N/(4b)$, we have $h-R^2H_R\ge N/(4b)$. Moving the small-denominator term to the left and writing $z=D_b/N$ gives \[ z\le4bH_N\sqrt{z/M_A}+8bH_N\overline D/N. \] Apply $u\sqrt z\le(z+u^2)/2$ with $u=4bH_N/\sqrt{M_A}$, and move $z/2$ to the left. The resulting coefficients are $u^2=16b^2H_N^2/M_A$ and $16bH_N\overline D/N$, exactly as in \eqref{eq:averaging}. \end{proof} The low-denominator frequencies are paid for explicitly. Omitting them would restrict the argument to totals with a roughness condition and would not suffice for \cref{thm:all}. \subsection{A fixed box of generators} Fix a block $\mc Q_0=\{q_1,\ldots,q_d\}$ of distinct primes and let $B_0=\prod_{q\in\mc Q_0}q$. Suppose $(N,B_0)=1$. For \[ R=\floor{N^{1/4}},\quad q_{\max}=\max\mc Q_0,\quad J=\floor{\frac{\log R}{d\log q_{\max}}}, \] the box \[ \mc A=\{q_1^{j_1}\cdots q_d^{j_d}:0\le j_i\le J\} \] is contained in $[1,R]$, consists of units, and has $M_A=(J+1)^d\gg_{B_0}L^d$ elements. Applying \cref{lem:descent} only to the fixed $q_i$ and using \eqref{eq:cocycle}, we have \[ \overline D\ll_{B_0}L(E+1). \] Consequently, for unit $b$ in any fixed polylogarithmic range and sufficiently large $N$, \begin{equation}\label{eq:boxbound} \frac{D_b}N\ll_{B_0}\frac{b^2}{L^{d-2}}+\frac{bL^2(E+1)}N. \end{equation} The hypothesis of \cref{lem:averaging} holds in these ranges because $R^2H_R\ll N^{1/2}L$. We retain two budgets, rather than obscuring their different uses in one asymptotic symbol: \begin{center}\small \begin{tabular}{lcccc} \toprule Use & $d$ & dilation cutoff $Y$ & reflection error & $\max_{b\le Y,(b,N)=1}D_b/N$\\ \midrule All-total counting & $1024$ & $L^{256}$ & $E\le N/L^{512}$ & $\ll_{B_0}L^{-254}$\\ Prime logarithmic scale & $128$ & $L^{32}$ & $E\le N/L^{96}$ & $\ll_{B_0}L^{-62}$\\ \bottomrule \end{tabular} \end{center} For the second row, for example, the two exponents in \eqref{eq:boxbound} are $64-126=-62$ and $32+2-96=-62$. The number of generators is fixed, and their descent costs can be enormous. These choices are not optimized and do not give practical numerical thresholds. \section{The short-product spectral test}\label{sec:spectral} The unit group permits characters of proper conductor, which must be kept when $N$ is composite. There are two tasks: confine the Fourier mass to a small family, then use short integer products to select one real character with a coefficient close to $+1$. The second task is not a consequence of small dilation defects alone. \subsection{Prime labels and high moments} Let $G=(\Z/N\Z)^\times$, and view $F(x)=f(x)$ on the representatives $1\le x1/2\}. \] The labels are the actual prime signs $f(q)$, not invariably $-1$. \begin{lemma}[Prime-product moment]\label{lem:moment} If $Y^j0$ after reducing the constant in \eqref{eq:liouvillemean}. \end{lemma} \begin{proof} The convolution identity is checked at prime powers. The quotient of the two local Dirichlet factors is $(1+z)/(1-z)$ at a prime of $E$ and $1$ elsewhere. Positivity justifies the infinite product. For $0\le z\le2^{-\theta}$, \[ \log\frac{1+z}{1-z}\le\frac{2z}{1-z^2}\le C_\theta z, \] proving \eqref{eq:sourceweight}. In $A_f(x)=\sum_{d\le x}a_E(d)A_\lambda(x/d)$, split at $d=\sqrt x$. The first part is $O_{\theta,M}(x\exp(-c\sqrt{\log x}/\sqrt2))$. The second has modulus at most \[ x\sum_{d>\sqrt x}\frac{a_E(d)}d \le\exp(C_\theta M)x^{(1+\theta)/2}. \] For fixed $\theta<1$, this is absorbed by the first estimate in \eqref{eq:uniformmean} after reducing $c_1$. For the harmonic estimate, convolve the truncated harmonic sums and use the bounded Liouville harmonic sums and $\sum a_E(d)/d\le\exp(C_\theta M)$. \end{proof} \subsection{Harmonic masks on the units} Fix from now on \begin{equation}\label{eq:alpha} \alpha=\frac{13}{50},\qquad Z=\floor{N^\alpha},\qquad R=\floor{X/Z}. \end{equation} \paragraph{Compatibility of the two cutoffs.} The cutoff must be large enough for Pollack's mean theorem after truncating the filter, but small enough that the Brun--Titchmarsh loss leaves a positive signed population. The inventory and the convolution therefore use the same $\alpha$. With the convolution cutoff $d\le N^{1/400}$ and Pollack parameter $\eta_1=1/200$, sufficient simultaneous constraints are \[ \frac{103}{400}<\alpha< \frac{\exp(7/16)-1}{2\exp(7/16)-1}\approx0.2616392277. \] The left inequality ensures $\alpha-1/400>1/4+1/200$. The right is exactly \[ c_\alpha:=\frac12-\log\frac{1-\alpha}{1-2\alpha}>\frac1{16}. \] Our choice $\alpha=0.26$ lies strictly between the endpoints. It gives $c_\alpha=\tfrac12-\log(37/24)=0.0671359177\ldots$, with margin $0.0046359177\ldots$ above $1/16$. These decimal values are explanatory; only the strict inequalities are used. The remaining tail saving is $(1-\alpha)(1-\theta)>0$ for every fixed $\theta<1$, not a condition $\alpha+\theta<1$. Suppose that \cref{prop:structured} applies, and let $\chi$ and $S$ be as there. Write $\rho=\chi(-1)$. \begin{lemma}[Unit means and sign masks]\label{lem:mask} Uniformly in $f\in\FF_{\theta,M}$ and in the factorization of $N$, \begin{align} \sum_{\substack{n\le t\\(n,N)=1}}\frac1n &=\nu\log t+O(P_N(1+T_N)),\label{eq:unit-harmonic}\\ \left|\sum_{\substack{n\le t\\(n,N)=1}}\frac{f(n)}n\right| &\ll_{\theta,M}P_N.\label{eq:signed-harmonic} \end{align} For $\tfrac12N^{1-\alpha}\le t\le N$, each sign has $(\nu/2+o(\nu))t$ representatives among the units up to $t$. Furthermore, for a prime $q\le N^\alpha$ with $q\nmid N$ and $\kappa=\pm1$, \begin{equation}\label{eq:mask} \sum_{\substack{at$, since then $\log t\le\log d$. The coefficient of $\log t$ is $\nu$. For the signed sum, inclusion--exclusion has weights $\mu(d)f(d)/d$, and \cref{lem:sourcemean} proves \eqref{eq:signed-harmonic}. For the unweighted signed sum on units, the same expansion, split at $d=\sqrt t$, gives \begin{equation}\label{eq:unitmeanerror} \sum_{\substack{n\le t\\(n,N)=1}}f(n) \ll_{\theta,M}tP_N\exp(-c_2\sqrt{\log t})+2^{\omega(N)}\sqrt t =o(\nu t) \end{equation} in the specified range. Combine this with $u_N(t)=\nu t+O(2^{\omega(N)})$ to separate the signs. A unit $a$ at which $f(a)\ne\chi(a)$ has a prime divisor appearing in \eqref{eq:strongprime}. Thus the harmonic mass of mismatches below $q$ is at most $H_{q-1}S$. With $\log(N/(aq))\gg_\alpha L$, replacing the $\chi$ mask in \eqref{eq:mask} by the $f$ mask costs $O(S)$. The $f$ mask is $(1+\kappa f(a))/2$ on units. Apply partial summation to \eqref{eq:unit-harmonic} and \eqref{eq:signed-harmonic} with weight $1/\log(N/(qt))$. Its endpoint size and total variation on $[1,q]$ are $O_\alpha(1/L)$. The main integral is \[ \frac\nu2\int_1^q\frac{\dd t}{t\log(N/(qt))} =\frac\nu2\log\!\left(\frac{\log(N/q)}{\log(N/q^2)}\right). \] Changing the endpoint from $q$ to $q-1$ costs $O(1/L)$. Since $P_N(1+T_N)\ll\ell^2$, the remaining errors are as stated. \end{proof} \subsection{Unite candidates before removing a common exceptional set} A useful prime has $q\nmid N$, $f(q)=-1$ and $\chi(q)=1$. To see what it creates, suppose that $qmT_q$. As $r$ is an integer, $r>X/q$. Write $N-qm=ar$. Then $aT_q$ with $f(r)=1$ and $\chi(r)=-1$. Since $T_q\ge R$, every such failed survivor, whatever its marking prime, belongs to the same set \[ V_R=\{1\le nR,\text{ divides }N-n\}. \] The involution $n\mapsto N-n$ gives $|V_R|\le W_E(N,R)$. Also $|\mc A_q^\sigma\cap\mc A_r^\sigma|\le\floor{X/(qr)}$, since an intersection consists of multiples of $qr$. For finite sets $A_i,V$, Bonferroni followed by one deletion gives \[ \left|\left(\bigcup_i A_i\right)\setminus V\right| \ge\sum_i|A_i|-\sum_{i0. \end{equation} Both patterns with product $\rho$ have at least $\phi(N)/3$ representatives. \end{lemma} \begin{proof} For $q\in\PP$, we have $T_q\ge\tfrac12N^{1-\alpha}$ for large $N$, and $1/q\le S$. Thus $q\to\infty$ uniformly over nonempty families in this branch. The sign count in \cref{lem:mask}, Brun--Titchmarsh, and \eqref{eq:mask} give \[ B_q^\sigma\ge\frac Nq\left\{ \nu c_\alpha-O_{\theta,M}\left(S+\frac{\ell^2}L\right)-o(\nu)\right\}, \] where \begin{equation}\label{eq:margin} c_\alpha=\frac12-\log\frac{1-\alpha}{1-2\alpha} =\frac12-\log\frac{37}{24}>\frac1{16}. \end{equation} Indeed, the progression is reduced, $N/a>q$, and Brun--Titchmarsh contributes the factor $2N/(q-1)$ times the masked harmonic sum. Replacing $q/(q-1)$ by $1+o(1)$ is legitimate since $q\to\infty$. The expression inside the logarithm is largest at $q=N^\alpha$. All errors are $o(\nu)$ by \eqref{eq:strongprime} and \eqref{eq:uniformcosts}. Hence $B_q^\sigma\ge\nu N/(32q)$. The pair intersections in \eqref{eq:inventory} have total at most $NU^2/2$. As $U\le S=o(\nu)$, they leave at least $\nu NU/64$. The common deleted population satisfies the direct bound \begin{equation}\label{eq:tailW} W_E(N,R)\le X\sum_{\substack{r\in E(f)\\r>R}}\frac1r \le MXR^{\theta-1}\ll_\theta M N^{1-\beta_\theta}. \end{equation} This proves \eqref{eq:signedsupply}. On the unit group the nonprincipal real character has exactly $\phi(N)/2$ occurrences of each sign, and its reflected product is $\rho$. Removing at most two unit disagreement sets, whose density is bounded in \eqref{eq:strongunit}, leaves at least $\phi(N)/3$ occurrences of each of those two patterns. \end{proof} For $f=\lambda$, $E(f)$ is empty and the tail in \eqref{eq:tailW} is exactly zero. For a general source it has a power saving for every fixed $\theta<1$. No restriction $\alpha+\theta<1$ is needed after the common deletion. \Needspace{9\baselineskip} \section{A filtered convolution supplies disagreement mass}\label{sec:filter} It remains to lower-bound $U$ in \eqref{eq:useful}. Pollack's use of the positive convolution $1*\chi$, including its connection with reciprocal sums of prime residues, is an antecedent of this step \cite[Proposition~3.1, proof of Theorem~1.3 and the subsequent remark]{Pollack}. Here we additionally remove both the prime factors of $N$ and the possibly infinite set of positive primes of $f$. The point of \cref{prop:filteredmean} is to retain a uniform error and a positive Euler factor after this filtering, so that only useful disagreements are detected. Define \begin{equation}\label{eq:filtered} b_{\chi,N,E}(n)=(1*\chi)(n)\1_{(n,N)=1} \1_{p\mid n\Rightarrow p\notin E},\qquad B_{\chi,N,E}(x)=\sum_{n\le x}b_{\chi,N,E}(n). \end{equation} The local coefficients explain what this detects. For $\chi(q)=1$, $(1*\chi)(q^j)=j+1$; for $\chi(q)=-1$, the coefficient is $1$ when $j$ is even and $0$ when $j$ is odd. All ramified primes are removed because $Q\mid N$. After the filter, a contribution without a useful prime must therefore come from a square. A mean exceeding this square contribution forces a supply of useful primes; the following bound quantifies that observation before we estimate the infinite filter. \begin{lemma}[A positive upper bound]\label{lem:positiveupper} For $x=Z$ and $U\le1/2$, \begin{equation}\label{eq:positiveupper} B_{\chi,N,E}(Z)\le\floor{\sqrt Z}+4ZU. \end{equation} \end{lemma} \begin{proof} If a contributing integer has no useful prime divisor, all its prime divisors have character value $-1$. It contributes only when it is a square, with weight $1$. For a useful prime $q$, the local factors give $b_{\chi,N,E}(qm)\le2b_{\chi,N,E}(m)$. For $y\le Z$, the weight $b_{\chi,N,E}(n)$ is at most the number of divisors of $n$ supported on useful primes. Consequently, \[ B_{\chi,N,E}(y)\le y\prod_{q\in\PP}(1-q^{-1})^{-1} \le\frac y{1-U}\le2y. \] A positive union bound over the useful prime divisors now costs at most $\sum_{q\in\PP}2B_{\chi,N,E}(Z/q)\le4ZU$, proving the assertion. \end{proof} \begin{proposition}[Uniform infinite-prime filtering]\label{prop:filteredmean} Fix $0<\theta<1$ and $M\ge0$. There exists $c_\theta>0$ such that, uniformly for $f\in\FF_{\theta,M}$ and for a primitive nonprincipal quadratic character $\chi$ of conductor $Q\mid N$, \begin{equation}\label{eq:filteredmean} B_{\chi,N,E}(Z) =ZC_{N,E}(\chi)L(1,\chi)+O_{\theta,M}(ZN^{-c_\theta}), \end{equation} where the convergent product \begin{equation}\label{eq:localfactor} C_{N,E}(\chi)=\prod_{p\mid N\ \text{or}\ p\in E} (1-p^{-1})(1-\chi(p)p^{-1}) \ge\e^{-4M}\nu^2 \end{equation} is not discarded from the main term. \end{proposition} \begin{proof} At each removed prime, the filter polynomial is \[ (1-z)(1-\chi(p)z)=1-(1+\chi(p))z+\chi(p)z^2. \] Let $c_N$ be the filter on the primes of $N$, and $c_{E\setminus N}$ the filter on the primes of $E$ not dividing $N$. Their prime supports are disjoint. Coefficientwise, \[ b_{\chi,N,E}=c*(1*\chi),\qquad c=c_N*c_{E\setminus N}, \] and every individual coefficient is a finite sum. The weighted absolute norm is \begin{equation}\label{eq:filternorm} \sum_{d\ge1}\frac{|c(d)|}{d^\theta} \le4^{\omega(N)}\prod_{p\in E}(1+p^{-\theta})^2 \le4^{\omega(N)}\e^{2M}. \end{equation} This replaces an unusable dependence on the possibly infinite number of removed primes. It also implies absolute convergence of $\sum c(d)/d$, whose value is $C_{N,E}(\chi)$. At primes of $N$, the factor in \eqref{eq:localfactor} is at least $(1-1/p)^2$. At the remaining removed primes, use $\log(1-1/p)\ge-2/p$ and $\sum_{p\in E}1/p\le M$. This proves its lower bound. Apply \eqref{eq:pollack} with $\eta_1=1/200$, and decrease its $\eta_2>0$ if necessary so that $1-\eta_2\ge\theta$. Split the convolution at $d=N^{1/400}$. For the smaller $d$, \[ Z/d\ge\tfrac12N^{103/400}\ge Q^{51/200} \] for all sufficiently large $N$, since $Q\le N$ and $103/400-51/200=1/400>0$. This is the lower-end constraint displayed after \eqref{eq:alpha}; the harmless factor $1/2$ is absorbed by that strict exponent margin. Thus the domain of \eqref{eq:pollack} is respected. The sum of its errors is bounded by \[ Z^{1-\eta_2}\sum_d\frac{|c(d)|}{d^{1-\eta_2}} \ll_{\theta,M}Z\,4^{\omega(N)}N^{-\alpha\eta_2}. \] For the tail, use $B_\chi(t)\le t(1+\log^+t)$, $L(1,\chi)\ll1+\log Q$, and \[ \sum_{d>N^{1/400}}\frac{|c(d)|}d \le N^{-(1-\theta)/400}\sum_d\frac{|c(d)|}{d^\theta}. \] The actual tail and the omitted tail of the main term together cost \[ O_{\theta,M}\bigl(Z(1+L)4^{\omega(N)}N^{-(1-\theta)/400}\bigr). \] Since $4^{\omega(N)}=N^{o(1)}$ uniformly, both errors have a fixed power saving. For example one may take a sufficiently small positive \[ c_\theta\le\frac14\min\{\alpha\eta_2,(1-\theta)/400\}. \] This proves \eqref{eq:filteredmean}. \end{proof} Combining \cref{lem:positiveupper,prop:filteredmean} and decreasing $c_\theta$ to absorb $Z^{-1/2}$ yields \begin{equation}\label{eq:Ulower} U\ge\frac14\bracket{C_{N,E}(\chi)L(1,\chi)-C_{\theta,M}'N^{-c_\theta}}. \end{equation} In particular, \cref{lem:signedsupply} gives the useful hybrid estimate \begin{equation}\label{eq:hybrid} \begin{split} N^f_{-\sigma,\rho\sigma}(N)\ge{}& \frac{\nu N}{256}\bracket{C_{N,E}(\chi)L(1,\chi)-C_{\theta,M}'N^{-c_\theta}}\\ &{}-C_\theta M N^{1-\beta_\theta}. \end{split} \end{equation} The main term contains both the local factors of the total and the possibly very small value $L(1,\chi)$. The proof does not replace either by a positive absolute constant. \section{Completion at every total}\label{sec:allproof} \subsection{The fixed-block conclusion} Fix $\theta,M,\varepsilon$ and a block of $1024$ primes. Let $N$ avoid that block and be sufficiently large. If \[ E_*(f;N)\ge N/L^{512}, \] then \eqref{eq:split} and \cref{lem:sourcemean} give \begin{equation}\label{eq:unstructured} m_f(N)\ge\frac{N}{4L^{512}}. \end{equation} This exceeds a constant times $N^{1-\varepsilon}$. Otherwise \cref{prop:structured} applies. Both patterns of product $\rho$ have at least $\phi(N)/3$ representatives. For the other two, use \eqref{eq:hybrid}. Choose \[ 0<\zeta<\frac14\min\{c_\theta,\beta_\theta,\varepsilon\}. \] By \eqref{eq:siegel} and \eqref{eq:localfactor}, \[ C_{N,E}(\chi)L(1,\chi) \gg_{\theta,M,\zeta}\nu^2 Q^{-\zeta} \ge c_{\theta,M,\zeta}\nu^2N^{-\zeta}. \] The positive part of \eqref{eq:hybrid} therefore has size at least a constant times $\nu^3N^{1-\zeta}$, and dominates both power-saving error terms. Since $\nu^{-1}\ll\ell$, it is $\gg_{\theta,M,\varepsilon}N^{1-\varepsilon}$. We have proved \begin{equation}\label{eq:blockconclusion} m_f(N)\gg_{B_0,\theta,M,\varepsilon}N^{1-\varepsilon} \qquad((N,B_0)=1), \end{equation} uniformly in $f$ in the class. The ineffectivity enters at the application of Siegel, and nowhere in the block selection that follows. \subsection{Reserve a block inside a large divisor} Divisor lifting by simultaneous scaling is an elementary established observation; see Bloom's comment in the original discussion \cite{BloomMO} and \cite[Lemma~2.1]{Meng}. We use it together with a selection from several disjoint fixed prime blocks. It is this selection and its exponent bookkeeping, rather than scaling alone, that passes from a fixed-block estimate to every large total. The lift preserves a collection of distinct representations; it does not multiply its size by $N/D$. \begin{lemma}[Disjoint-block selection]\label{lem:blockselection} Let $\mc Q_1,\ldots,\mc Q_J$ be disjoint finite sets of primes, with products $B_i$. Every $N\ge2$ has a divisor $D$ and an index $i$ such that \begin{equation}\label{eq:blockselection} (D,B_i)=1,\qquad D\ge N^{1-1/J}. \end{equation} For every completely multiplicative sign function $f$ and every divisor $D\mid N$ with $D\ge2$, \begin{equation}\label{eq:lifting} m_f(N)\ge m_f(D). \end{equation} \end{lemma} \begin{proof} Put $A_i=\prod_{q\in\mc Q_i}q^{v_q(N)}$. These integers are pairwise coprime and $\prod_iA_i\mid N$, so some $A_i\le N^{1/J}$. Set $D=N/A_i$. For \eqref{eq:lifting}, multiply both summands in a representation of $D$ by $N/D$. This is injective and multiplies both signs by the same number $f(N/D)\in\{-1,1\}$. It permutes the four sign pairs, proving the minimum inequality. \end{proof} \begin{proof}[Proof of \cref{thm:robust,thm:all}] It is enough to consider $0<\varepsilon<1$. Fix an integer $J\ge2/\varepsilon$, then fix $J$ disjoint blocks of $1024$ primes. Apply \eqref{eq:blockconclusion} with exponent $\varepsilon/2$ to the divisor supplied by \cref{lem:blockselection}. There are only finitely many blocks, so take the minimum of their positive constants and the maximum of their thresholds. The divisor tends uniformly to infinity with $N$. Thus \[ m_f(N)\ge m_f(D)\gg_{\theta,M,\varepsilon}D^{1-\varepsilon/2} \ge N^{(1-1/J)(1-\varepsilon/2)}\ge N^{1-\varepsilon}. \] The function and its prime budget have not changed on passage to the divisor. This proves \cref{thm:robust}; taking $M=0$ proves \cref{thm:all}. \end{proof} The order of choices is fixed: first $\theta,M,\varepsilon$, then the finitely many blocks and the analytic exponents, and only then the threshold for $N$. No boundedness assumption on $\omega(N)$ or $\Omega(N)$ is used. The argument neither selects a subsequence of totals nor changes the values of Liouville. \section{The logarithmic scale at prime totals}\label{sec:primetheorem} We now return to $f=\lambda$, but use the less stringent reflection threshold and the generic inverse of \cref{thm:inverse}. This is a separate route: the signed inventory, filtered convolution, and Siegel lower bound are not needed. Fix a block of $128$ primes. Every sufficiently large prime $p$ avoids it. Write $L=\log p$ and $T_p=p/L^{96}$. Suppose first that $E_*(\lambda;p)L^{-11}$. We obtain \begin{equation}\label{eq:primeclose} \chi_p(-1)=\kappa,\qquad \max_{tL^{-11}$. In particular, \[ \chi_p(2)=\chi_p(3)=-1. \] The supplementary laws of quadratic reciprocity then force $p\equiv5$ or $19\pmod{24}$. Next, suppose $P_{\rho_p}(\lambda;p)1$, the Liouville Dirichlet series is $\zeta(2s)/\zeta(s)$. Estimate~\eqref{eq:liouvillemean} and partial summation justify its boundary value and derivative at $s=1$, giving \begin{equation}\label{eq:lambda-boundary} \sum_{n\ge1}\frac{\lambda(n)}n=0,\qquad -\sum_{n\ge1}\frac{\lambda(n)\log n}{n}=\zeta(2). \end{equation} Their tails after $p$ are $O(\exp(-c'\sqrt L))$, after reducing $c'>0$ to absorb powers of $L$. No Liouville series is evaluated to the left of $1$. For the character, the Gauss-sum identity \[ \chi_p(n)=\tau(\overline{\chi_p})^{-1} \sum_{a=1}^{p-1}\overline{\chi_p(a)}\e^{2\pi ian/p},\qquad |\tau(\overline{\chi_p})|=\sqrt p, \] and the geometric-sum bound yield $\sup_t|\sum_{n\le t}\chi_p(n)|\ll\sqrt p\log p$, after reducing by complete periods. Partial summation therefore bounds the corresponding two character tails by $O(p^{-1/2}L)$ and $O(p^{-1/2}L^2)$. Combining these estimates with $S_p\le3L^{-11}$ gives \begin{align*} L(1,\chi_p)&=O\bigl(LS_p+\e^{-c'\sqrt L}+p^{-1/2}L\bigr),\\ L'(1,\chi_p)&=\zeta(2)+O\bigl(L^2S_p+\e^{-c'\sqrt L}+p^{-1/2}L^2\bigr). \end{align*} Positivity of $L(1,\chi_p)$ follows from the positive Euler product for real $s>1$ and classical nonvanishing at $1$. This is not an ineffective quantitative lower bound. \end{proof} \begin{lemma}\label{lem:secondderivative} Uniformly for real $1-1/L\le s\le1$, \begin{equation}\label{eq:secondderivative} |L''(s,\chi_p)|\ll L^3. \end{equation} \end{lemma} \begin{proof} Up to $p$, the absolute sum is at most $\e\sum_{n\le p}(\log n)^2/n\ll L^3$. In the tail, periodicity bounds partial character sums by $p$. For large $p$ the weight $(\log t)^2t^{-s}$ is decreasing on $t\ge p$, and partial summation gives $O(p^{1-s}L^2)=O(L^2)$. The differentiated character series and these tail bounds are valid uniformly on the indicated compact interval in $s>0$. \end{proof} Let $a=L(1,\chi_p)>0$ and $t=2a/\zeta(2)$. By \cref{prop:value-slope}, $t\ll L^{-10}<1/L$. Taylor's theorem and \cref{lem:secondderivative} give \[ L(1-t,\chi_p) =a-t\bigl(\zeta(2)+O(L^{-9})\bigr)+O(t^2L^3) =-a+o(a)<0. \] The derivative stays positive on $[1-t,1]$. There is therefore a unique real zero in that interval, and \begin{equation}\label{eq:precisezero} 1-\beta_p=\frac{L(1,\chi_p)}{\zeta(2)} \bigl(1+O(L^{-7})\bigr)\ll L^{-10}. \end{equation} The uniqueness assertion is confined to this short interval. \begin{proof}[Completion of \cref{thm:prime}] If $p\in\XX$ lies in $[\exp((\log T)^{1/8}),T]$, then \[ 1-\beta_p\ll(\log p)^{-10}\le(\log T)^{-5/4} (\log p)^8,\qquad \log\log r>8\log\log p. \] The double logarithms grow geometrically, giving \eqref{eq:sparsecount}. This argument allows an infinite but very sparse family; it does not eliminate it. \end{proof} \section{Boundaries of the conclusions}\label{sec:boundary} \subsection{Why the reciprocal endpoint is not included} The single common deletion in \cref{prop:inventory} has error $O_\theta(MN^{1-\beta_\theta})$, with $\beta_\theta=(37/50)(1-\theta)>0$ for each fixed $\theta<1$. The saving disappears at $\theta=1$, and two separate observations prevent extending the theorem by an informal limiting argument. First, the required first moment is not uniform under a reciprocal budget, even a budget tending to zero. For a large integer $x$, flip the prime values of Liouville only in $(x/2,x]$, obtaining $f_x$. Exactly, \begin{equation}\label{eq:endpointmean} \sum_{n\le x}f_x(n)=A_\lambda(x)+2\bigl(\pi(x)-\pi(x/2)\bigr). \end{equation} Indeed, the only multiple of such a prime at most $x$ is the prime itself. The prime number theorem gives \[ \sum_{p:f_x(p)=1}\frac1p\sim\frac{\log2}{\log x}\longrightarrow0, \qquad \sum_{n\le x}f_x(n)\sim\frac{x}{\log x}. \] This contradicts a uniform version of \eqref{eq:uniformmean} at $\theta=1$. It is \emph{not} a counterexample to an all-pattern theorem at that endpoint. Second, an endpoint theorem would have an additional arithmetic consequence. \begin{proposition}[A necessary condition for uniform reciprocal robustness]\label{prop:endpoint} Suppose that for every $M>0$ there exists $N_M$ such that every completely multiplicative sign function with $\sum_{f(q)=1}1/q\le M$ possesses all four reflected patterns at every total $N\ge N_M$. Then \begin{equation}\label{eq:endpointnecessary} \sum_{\substack{q