\documentclass[11pt]{amsart} \usepackage[utf8]{inputenc} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs} \usepackage{graphicx} \usepackage{xcolor} \usepackage[margin=0.95in]{geometry} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \hypersetup{pdftitle={A two-sided criterion for the Riemann hypothesis from weighted prime sums, verified up to 2e29},pdfauthor={Pedro Caceres}} \newtheorem{theorem}{Theorem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \setlength{\emergencystretch}{2.5em} \newcommand{\re}{\operatorname{Re}} \newcommand{\im}{\operatorname{Im}} \title[A two-sided criterion for RH verified up to $2\cdot10^{29}$]{A two-sided criterion for the Riemann hypothesis\\ from weighted prime sums, verified up to $2\cdot10^{29}$} \author{Pedro Caceres} \address{Universidad Europea de Valencia, Valencia, Spain} \email{pedrojesus.caceres@universidadeuropea.es} \subjclass[2020]{Primary 11M26; Secondary 11N05, 11M06, 42A75} \keywords{Riemann hypothesis, explicit formula, screw function, Riesz means, almost periodic functions, M\"obius function, Beurling primes} \date{October 2026 (v2)} \begin{document} \begin{abstract} For $N>1$ let $C(N)=2\sum_{n1$ put \[ C(N)=2\sum_{n1$. So RH is equivalent to confining a sum of size $8\sqrt N$ over the prime powers below $N$ to a window of fixed width, from both sides. This is as hard as RH, and we do not prove it. \emph{What the paper does.} It studies this one-variable inequality from three sides and records exactly where each side stops. \begin{enumerate} \item \emph{The margin} (Section~\ref{sec:margin}). The margin is a Fej\'er sum over the zeros (Proposition~\ref{prop:fejer}). The window width is $4(\log\xi)''(\frac12)$, computable without the zeros (Corollary~\ref{cor:sandwich}). The two-sided inequality holds for all real $11$ and $L=\log N$. Without any hypothesis, \[ M(N)=\sum_\rho F_L(\gamma_\rho),\qquad F_L(z)=\frac{4\sin^2(Lz/2)}{z^2}=\Big|\int_0^Le^{izu}du\Big|^2\ (z\in\mathbb R), \] the sum over all nontrivial zeros with multiplicity, absolutely convergent. More generally, for a window $\varphi\ge0$ on $[0,L]$ with autocorrelation $A(u)=\int_0^{L-u}\varphi(v)\varphi(v+u)dv$ and $t\in\mathbb R$, let $C^\varphi_N(t)=2\sum_{n0}\frac{1-\cos(\gamma\log N)}{\gamma^2}\in\Big[0,\ 8\sum_{\gamma>0}\gamma^{-2}\Big], \end{equation} so then $M(N)\ge0$ for every $N>1$. \end{theorem} \begin{proof} By Proposition~\ref{prop:fejer}, $M(e^L)=\sum_\rho F_L(\gamma_\rho)$. Under RH every term is $\ge0$, and pairing $\pm\gamma$ gives \eqref{eq:wave}. Conversely, suppose $M(e^L)\ge0$ for $L\ge L_0$. For $\re s>\frac12$, since $|F_L(\gamma)|\le(2+2e^{L/2})|\gamma|^{-2}$ and $\sum|\gamma_\rho|^{-2}<\infty$, Fubini gives \[ \int_0^\infty M(e^L)e^{-sL}dL=\sum_\rho\frac2{s(s^2+\gamma_\rho^2)}=:\mathcal G(s), \] meromorphic on $\mathbb C$ with poles at $0$ and at $s=\pm i\gamma_\rho$. A pole $s$ fixes $\gamma^2=-s^2$, so only $\pm\gamma$ meet there, with equal nonzero residues, and no cancellation occurs. An off-line zero $\frac12+\delta+i\gamma_0$, $\delta>0$, has $\gamma_0\ne0$ because $\zeta$ has no real zeros in $(0,1)$, and gives a pole at $\delta+i\gamma_0$ off the real axis; the zeros on the line give poles on the imaginary axis. So $\mathcal G$ is analytic on $(0,\infty)$. By Landau's theorem \cite{In,MV}, the Laplace integral of a function that is $\ge0$ on $[L_0,\infty)$ has a singularity at the real point of its abscissa of convergence; it differs from $\mathcal G$ by an entire function, so the abscissa is $\le0$, and $\mathcal G$ has no pole with $\re s>0$. \end{proof} Status: proved. It is Landau's oscillation method for a Riesz mean of $\psi(x)-x$ weighted by $n^{-1/2}$, and the theorem is Suzuki's: with $M=2\Psi(\log N)$, RH holds if and only if $\Psi\ge0$, and if and only if $\Psi=O(1)$ \cite[Theorems 1.6, 1.7]{Su1}; the criterion with the main term alone is \cite[Theorem 1]{Su2}. Our proof is independent. The constants in \eqref{eq:wave} are values of $\xi$ at the centre: Hadamard's product gives $(\log\xi)''(s)=-\sum_\rho(s-\rho)^{-2}$, and at $s=\frac12$ \begin{equation}\label{eq:xi2} (\log\xi)''(\tfrac12)=\sum_\rho\frac1{\gamma_\rho^2}=-8+\frac{\pi^2}4+2G+\Big(\frac{\zeta''}\zeta-\Big(\frac{\zeta'}\zeta\Big)^2\Big)(\tfrac12)=0.0462100, \end{equation} unconditionally; under RH it is $2\sum_{\gamma>0}\gamma^{-2}$, and with the $4520$ zeros below $5000$ and the tail the two agree to $6\cdot10^{-10}$. \begin{corollary}[the sandwich]\label{cor:sandwich} The following are equivalent: (1) RH; (2) $M(N)\ge0$ for all large $N$; (3) $M(N)\le B$ for some $B$ and all large $N$; (4) for every $N>1$, \begin{equation}\label{eq:sandwich} R(N)-4(\log\xi)''(\tfrac12)\;\le\;C(N)\;\le\;R(N),\qquad4(\log\xi)''(\tfrac12)=0.18484 . \end{equation} \end{corollary} \begin{proof} (1)$\Rightarrow$(4) is \eqref{eq:wave} with \eqref{eq:xi2}; (4)$\Rightarrow$(2),(3) are immediate; (2)$\Rightarrow$(1) is Theorem~\ref{thm:landau}; for (3)$\Rightarrow$(1) apply the same argument to $B-M(e^L)\ge0$, whose Laplace integral has the same poles off the real axis. \end{proof} Status: proved; the equivalences (1)--(3) and the bounds are in \cite{Su1}, and the evaluation \eqref{eq:xi2} of $\sum_\rho\gamma_\rho^{-2}$ is classical \cite{Vo}. Writing the window this way makes \eqref{eq:sandwich} checkable without the zeros, which is what the verification below uses. The two sides are not redundant in general: a real zero $\beta\in(\frac12,1)$ would push $M$ upwards only, and only the upper side sees it. For $\zeta$, which has no such zeros, either side alone is equivalent to RH; for the Beurling system and the fake curve of \cite{P20} the lower side never fails and the upper side fails for all large $N$. \subsection{What is known unconditionally} \begin{proposition}[unconditional bound, crude form]\label{prop:uncond} For every $N>1$, $M(N)\ge-8(3+\sqrt N)S\ge-2.29\cdot10^{-11}(3+\sqrt N)$, with $S=\sum_{\im\rho>T_0}(\im\rho)^{-2}\le2.854\cdot10^{-12}$. \end{proposition} \begin{proof} In Proposition~\ref{prop:fejer} the zeros on the line contribute $\ge0$. By Platt and Trudgian \cite{PT} every zero with $0<\im\rho\le T_0$ is on the line, so an off-line zero has $\gamma_\rho=\gamma_0+id$ with $|\gamma_0|>T_0$, $|d|<\frac12$. With $F_L(z)=(2-e^{iLz}-e^{-iLz})/z^2$, $|F_L(z)|\le(2+N^{|d|}+N^{-|d|})/|z|^2\le(3+\sqrt N)/\gamma_0^2$, and summing over the off-line ordinates, which come in quadruples $\pm\gamma_0\pm id$, gives at most $8(3+\sqrt N)S$ (a conservative constant). Stieltjes integration and Trudgian's bound for $N(u)$ \cite{Tr} give $S\le\int_{T_0}^\infty2N(u)u^{-3}du\le2.854\cdot10^{-12}$. \end{proof} The factor $\sqrt N$ is the price of a hypothetical zero near $\re s=1$. A zero-density estimate says that such zeros are rare, and it improves the bound by many orders of magnitude. Let $N(\sigma,T)$ be the number of zeros $\beta+i\gamma$ with $\beta>\sigma$, $0<\gamma1$ \begin{equation}\label{eq:EN} M(N)\ge-E(N),\qquad E(N)=4\Bigl[(3+N^{\sigma_1-\frac12})S_{1/2}+\sum_{j=1}^{J}\bigl(N^{\sigma_{j+1}-\frac12}-N^{\sigma_j-\frac12}\bigr)S_j\Bigr]. \end{equation} One may take $S_{1/2}=1.4822\cdot10^{-12}$, $S_j=S_{1/2}$ to four digits for $\sigma_j\le0.75$, $S(0.80)=4.71\cdot10^{-13}$, $S(0.85)=8.63\cdot10^{-15}$, $S(0.90)=1.57\cdot10^{-16}$, $S(0.95)=2.85\cdot10^{-18}$, $S(0.99)=1.14\cdot10^{-19}$, and $E(N)$ takes the values of Table~\ref{tab:EN}. In particular $|\min(M(N),0)|<10^{-3}$ for $N<5\cdot10^{24}$ and $E(N)<0.043$ for $N<2.3\cdot10^{29}$. \end{theorem} \begin{proof} As in Proposition~\ref{prop:uncond}, $M(N)\ge-\sum'|F_L(\gamma_\rho)|$, the sum over the off-line zeros, all of which have $|\im\rho|>T_0$, and $|F_L(\gamma_0+id)|\le(3+N^{|d|})/\gamma_0^2$. A zero $\rho=\beta+i\gamma$ with $\beta>\frac12$, $\gamma>T_0$, brings the quadruple $\rho,\bar\rho,1-\rho,1-\bar\rho$, so \[ \sum{}'|F_L|\le4\sum_{\beta>\frac12,\ \gamma>T_0}\frac{3+N^{\beta-1/2}}{\gamma^2}. \] Write $\Sigma(\sigma)=\sum_{\beta>\sigma,\gamma>T_0}\gamma^{-2}$. Since $N^{\beta-1/2}\le N^{\sigma_{j+1}-1/2}$ when $\sigma_j<\beta\le\sigma_{j+1}$, Abel summation over the bins gives \[ \sum_{\beta>\frac12,\gamma>T_0}\frac{N^{\beta-1/2}}{\gamma^2}\le N^{\sigma_1-\frac12}\Sigma(\tfrac12)+\sum_{j=1}^J\bigl(N^{\sigma_{j+1}-\frac12}-N^{\sigma_j-\frac12}\bigr)\Sigma(\sigma_j). \] It remains to show $\Sigma(\frac12)\le S_{1/2}$ and $\Sigma(\sigma_j)\le S_j$. Since $N(\sigma,T_0)=0$, Stieltjes integration gives $\Sigma(\sigma)=2\int_{T_0}^\infty N(\sigma,t)t^{-3}dt$. By the symmetry $\beta\leftrightarrow1-\beta$, at most half of the zeros up to height $t$ lie to the right of the line, so $N(\sigma,t)\le\frac12N(t)$; and $N(t)\le\widetilde N(t)$ for $t\ge T_0$, since $N(t)=\frac t{2\pi}\log\frac t{2\pi e}+\frac78+\theta(0.112\log t+0.278\log\log t+2.51+0.2/t)$ with $|\theta|\le1$ \cite{Tr}, and the term $-\frac t{2\pi}$ dominates. This gives $\Sigma(\frac12)\le S_{1/2}$. For $\sigma_j\ge0.6$, \eqref{eq:kln} applies since $T_0>H_0$, and $\Sigma(\sigma_j)\le S_j$. The values used come from adaptive quadrature in $u=\log t$ (\texttt{ebound.py}); they are rigorous upper bounds, because each integrand $g(u)$ is positive and decreasing for $t\ge T_0$ (both $N_{\rm KLN}(\sigma,t)t^{-2}$ and $\widetilde N(t)t^{-2}$ decrease), so a left Riemann sum of step $10^{-4}$ on $[\log T_0,\log T_0+120]$ plus an explicit tail bounds the integral from above, and every quadrature value exceeds that bound (\texttt{ebound\_rig.py}; the ratio is at most $0.9984$). For instance $S_{1/2}=(\log\frac{T_0}{2\pi}+1)/(2\pi T_0)=1.47970\cdot10^{-12}$ exactly, the Riemann sum gives $1.47977\cdot10^{-12}$ and the value used is $1.48222\cdot10^{-12}$. \end{proof} \begin{corollary}[unconditional two-sided bound]\label{cor:twosided} For every $N>1$, \[ -E(N)\le M(N)\le4(\log\xi)''(\tfrac12)+E(N)=0.18484+E(N). \] \end{corollary} \begin{proof} The lower side is Theorem~\ref{thm:explicit}. For the upper side split $M=\sum_\rho F_L(\gamma_\rho)$ into zeros on and off the line. A pair $\pm\gamma$ on the line contributes $4(1-\cos\gamma L)/\gamma^2\le8/\gamma^2$, and the off-line zeros contribute at most $E(N)$ in absolute value, as in the proof of Theorem~\ref{thm:explicit}. Identity \eqref{eq:xi2} holds without RH, with $\gamma_\rho=(\rho-\frac12)/i$ complex for an off-line zero. A quadruple $\pm\gamma_0\pm id$ contributes to it $4(\gamma_0^2-d^2)/(\gamma_0^2+d^2)^2>0$, since $|d|<\frac12<\gamma_0$. Hence $\sum_{\text{on the line}}\gamma^{-2}\le(\log\xi)''(\frac12)$, and the zeros on the line contribute at most $4(\log\xi)''(\frac12)$. \end{proof} So without any hypothesis \eqref{eq:sandwich} holds up to $E(N)$ on both sides; RH removes $E(N)$. \begin{table}[ht] \centering \caption{The off-line contribution: crude bound of Proposition~\ref{prop:uncond} against $E(N)$ of Theorem~\ref{thm:explicit}.} \label{tab:EN} \begin{tabular}{lcccccc} \toprule $N$ & $10^{10}$ & $10^{15}$ & $10^{20}$ & $10^{25}$ & $10^{27}$ & $10^{30}$\\ \midrule $8(3+\sqrt N)S$ & $2.3\cdot10^{-6}$ & $7.2\cdot10^{-4}$ & $0.23$ & $72$ & $720$ & $2.3\cdot10^4$\\ $E(N)$ & $1.0\cdot10^{-8}$ & $4.7\cdot10^{-7}$ & $2.4\cdot10^{-5}$ & $1.3\cdot10^{-3}$ & $6.3\cdot10^{-3}$ & $0.072$\\ \bottomrule \end{tabular} \end{table} The crude bound is below the least margin observed ($0.0427$, Proposition~\ref{prop:verified}) as long as $N<3.5\cdot10^{18}$; the explicit bound \eqref{eq:EN} stays below it up to $2.3\cdot10^{29}$. Beyond that it grows like $N^{0.35}$ (the bin $0.80<\beta\le0.85$ dominates) and says nothing. This is the familiar fact that verifying RH to a height $H$, together with zero-density estimates, gives bounds of RH quality only in a range polynomial in $H$ \cite{Jn}; it comes from the zeros, not from the primes. Theorem~\ref{thm:explicit} is not a step towards RH: it holds whether or not RH is true, and it is silent for $N>10^{30}$. \subsection{Verification} \begin{proposition}[the sandwich verified]\label{prop:verified} \begin{enumerate} \item For every real $N$ with $10$ for $10^{15}\le N\le2.15\cdot10^{29}$; $M(N)\le0.1377$ for $10^8\le N\le10^{27}$ and $M(N)<0.18484$ for $10^8\le N\le4.28\cdot10^{29}$; so \eqref{eq:sandwich} holds on $[10^8,2.15\cdot10^{29}]$. Here the $4520$ zeros below $5000$ are certified: with ball arithmetic, $N(5000)=4520$ and each ordinate used lies within $9.1\cdot10^{-12}$ of the true one. \end{enumerate} Since $0.14<0.18484$, parts (1) and (3) give the two-sided inequality \eqref{eq:sandwich} for every real $N$ with $10$ for $N\ge2$, so $m$ is convex there and its minimum on $[\log q,\log q']$ is bounded below by the tangent lines at the two ends; its maximum is at an end. \emph{Arithmetic}: Python's \texttt{decimal} at $34$ digits, whose $+,-,\times,/,\sqrt{\ },\log,\exp$ are correctly rounded; $\pi,\gamma_E,G$ entered to $48$ digits; a margin of $10^{-15}$ is subtracted from every lower bound and added to every upper bound. The computation ran over all $455\,062\,595$ prime powers below $10^{10}$, from a segmented sieve; $8$ intervals below $N=100$ needed subdivision. The least lower bound for $N\ge10^3$ is $0.042997$, on $[34\,186\,367,34\,186\,499]$, where the minimum of $M$ below $10^{10}$ is attained; the largest upper bound is $0.137816$ at $N=319\,439$. (2) Same convexity argument, over all $37\,607\,992\,088$ prime powers below $10^{12}$ (the count equals $\pi(10^{12})=37\,607\,912\,018$ plus the $80\,070$ powers $p^k$, $k\ge2$, which checks the sieve). On $[\log q,\log q']$, $\min m\ge\max\bigl(m(a)-(b-a)\max(0,-m'(a)),\,m(b)-(b-a)\max(0,m'(b))\bigr)$. The arithmetic is IEEE double precision. The operations $+,-,\times,/,\sqrt{\ }$ are correctly rounded; the library logarithm was checked at run time to be within $0.5$ ulp of the correctly rounded value on $2\cdot10^4$ test points, and an error of $2$ ulp is assumed in the bounds. The sums $A$ and $B$ are accumulated in segments of length $5\cdot10^7$, each total exactly rounded (\texttt{math.fsum}) and the totals added exactly in rational arithmetic. Every lower bound is lowered, and every upper bound raised, by an explicit bound for all rounding errors (the error in $A(10^{12})=1\,999\,997.3576058$ is at most $2.1\cdot10^{-9}$, and the error of one evaluation of $m$ is of order $10^{-7}$). The least lower bound is $0.042686$, near $N=105\,606\,345\,001$; the largest upper bound on $[10^{10},10^{12}]$ is $0.131211$, near $N=742\,864\,907\,663$. The same program reproduces part (1) on $[10^3,10^{10}]$ ($0.042997$, $0.137816$, $455\,062\,595$ prime powers). The minimum near $1.056\cdot10^{11}$ is also visible from the zeros: there the sum over the $4520$ zeros below $5000$ in part (3) is $0.041711$, and the difference $0.00098$ is the mean contribution $4\sum_{\gamma>5000}\gamma^{-2}=0.00098$ of the omitted zeros. (3) Unconditionally $M=\sum_\rho F_L(\gamma_\rho)$; the zeros up to $3\cdot10^{12}$ are on the line and contribute $\ge0$, so dropping those not in the list and bounding the rest as in Proposition~\ref{prop:uncond}, $M(N)\ge4\sum_{\gamma\in\text{list}}(1-\cos\gamma L)/\gamma^2-8(3+\sqrt N)S$. The sum is Lipschitz with constant $4\sum1/\gamma=14.132$; on a grid of $805\,906$ points covering $[\log10^8,\log10^{15}]$, the least value of the sum on a cell, minus the Lipschitz loss and minus $8(3+\sqrt N)S$ at the right end of the cell (at most $7.2\cdot10^{-4}$), is $0.04156$; it is attained near $N=1.06\cdot10^{11}$, where the sum is $0.041711$, the Lipschitz loss $1.41\cdot10^{-4}$ and the tail term $7\cdot10^{-6}$. For $10^{15}\le N\le10^{30}$ the same computation with $E(N)$ of Theorem~\ref{thm:explicit} in place of $8(3+\sqrt N)S$, on a grid of step $2\cdot10^{-5}$ in $L$ ($1\,726\,941$ points), gives the least lower bound $0.048104$ up to $10^{27}$, $0.047549$ up to $10^{28}$ and $0.034188$ up to $10^{29}$; the first cell where $W-E\le0$ is at $N=2.159\cdot10^{29}$, where $E=0.0418$ and the sum is near its least value on the range, $0.038908$ at $N=2.18\cdot10^{29}$ (Figure~\ref{fig:explicit}). For the upper side, the zeros on the line above $5000$ contribute at most $8\sum_{\gamma>5000}\gamma^{-2}$, and by Stieltjes integration with $N(5000)=4520$ and Trudgian's bound for $N(t)$ this is at most $8\cdot2.446\cdot10^{-4}=1.957\cdot10^{-3}$. (As a check, $2(\sum_{\rm list}4\gamma^{-2}+4\cdot2.446\cdot10^{-4})=0.184841$, against $4(\log\xi)''(\frac12)=0.184840$.) So $M\le W+1.957\cdot10^{-3}+E(N)$. The largest upper bound on the cells, the maximum of $W$ plus the Lipschitz gain, is $0.137298$ up to $10^{20}$, $0.137626$ up to $10^{27}$ and $0.161553$ up to $10^{29}$, and it first reaches $0.18484$ at $N=4.29\cdot10^{29}$; the maximum of $W$ on $[10^8,10^{30}]$ is $0.135632$. An error $\delta$ in each ordinate changes the sum by at most $4\delta L\sum\gamma^{-2}\le3.2\,\delta$ for $N\le10^{30}$. \end{proof} Status: (1) is computer-assisted, with correctly rounded arithmetic, and was checked in three independent ways (an extended-precision scan of $A$ and $B$ to $10^{10}$, $45$-digit values below $2\cdot10^6$, and an independent sieve on three windows). (2) is computer-assisted in double precision with rigorous error bounds; it relies on the stated accuracy of the library logarithm, which was tested but not proved. (3) uses the zeros below $5000$, certified with Arb \cite{Ar} through python-flint: \texttt{acb.zeta\_zeros} returns balls guaranteed to contain the $n$-th zeros, isolated by sign changes of Hardy's $Z$ with the count checked by Turing's method. The $4520$-th zero lies in $[4999.32968137297,4999.32968137298]$ and the $4521$-st above $5000.2343169313$, so the list is complete, and it is accurate to $9.1\cdot10^{-12}$, which changes $W$ by less than $3\cdot10^{-11}$. The grid evaluation of $W$ in double precision has error below $10^{-10}$ (each term $4(1-\cos\gamma L)/\gamma^2$ to a few units in the last place of $\gamma L\le3.5\cdot10^5$). Neither reaches the statement for all large $N$: the primes cost one evaluation per prime power, and the zeros stop near $2\cdot10^{29}$ by Theorem~\ref{thm:explicit}. \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{CE_explicit.png} \caption{(a) The bound on how far off-line zeros could move $M(N)$: the crude bound of Proposition~\ref{prop:uncond} and $E(N)$ of Theorem~\ref{thm:explicit}, which uses the zero density of \cite{KLN}. (b) For $10^{15}\le N\le10^{30}$: the contribution $W$ of the $4520$ zeros below $5000$ (grey), the bound $E(N)$ (red) the certified lower bound $W-E$ for $M$ (blue), positive up to $2.16\cdot10^{29}$ (dashed), and the upper bound $W+1.96\cdot10^{-3}+E$ (green), below $0.18484$ up to $4.29\cdot10^{29}$.} \label{fig:explicit} \end{figure} \emph{The wave.} Formula \eqref{eq:wave} says what the primes must produce: an almost periodic function of $\log N$ with mean $4\sum_{\gamma>0}\gamma^{-2}=0.0924$, frequencies at the ordinates and amplitudes $4/\gamma^2$. From the $5\,762\,859$ prime powers below $10^8$, at $6000$ values of $\log N$, $M$ agrees with the right side of \eqref{eq:wave} truncated at the zeros below $5000$ to $5.4\cdot10^{-5}$; its spectrum in $\log N$, from the primes alone, has peaks at $14.132$, $21.023$, $25.014$, $30.425,\dots$ with heights agreeing with $4/\gamma^2$ to $10^{-5}$ (Figure~\ref{fig:wave}). Over $4\cdot10^5$ values of $\log N$ in $[\log2,\log10^8]$ the mean is $0.092367$ (predicted $2(\log\xi)''(\frac12)=0.092420$) and the standard deviation $0.017219$ (predicted $\sqrt{8\sum\gamma^{-4}}=0.017245$ for independent phases). \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{CE20_wave.png} \caption{Top: the margin $M(N)$ computed from the primes below $10^8$ (black) and the right side of \eqref{eq:wave} from the $4520$ zeros below $5000$ (dashed); mean $0.0924$, least value $0.043$. Bottom: its spectrum in the variable $\log N$, from the primes, against $4/\gamma^2$ at the zeros.}\label{fig:wave} \end{figure} A least-squares fit of $m(L)=M(e^L)$ on the first $40$ ordinates, from the prime powers below $10^7$, gives cosine amplitudes equal to $-4/\gamma_k^2$ to within $0.4\%$ and sine amplitudes below $10^{-5}$, as \eqref{eq:wave} predicts under RH. The decay $4/\gamma^2$ is the regularity of $m$: it is continuous, and its slope jumps by $-2\Lambda(q)/\sqrt q$ at $L=\log q$ (Proposition~\ref{prop:minima}). \emph{Where the difficulty sits.} By blocks at $N=0.999\cdot10^8$: $R(N)=79853.63$, the primes give $79665.63$, the prime squares $130.68$, the higher powers $57.21$, and $M=0.114$. The squares are controlled without RH; the whole difficulty is in the sum over primes, which has to be known to $O(1)$ out of $8\sqrt N$. Without RH the best bound for that sum is of size $\sqrt N\exp(-c(\log N)^{3/5}(\log\log N)^{-1/5})$, and closing the gap to $O(1)$ is RH. For the Davenport--Heilbronn function the same construction gives its own window $w_f=0.78456$; its margin computed from its coefficients up to $2\cdot10^7$ stays inside, and the model with its zeros predicts the first exit near $N\approx2.3\cdot10^8$ \cite{P20}: in one variable the criterion is the simplest, but the slowest to detect an off-line zero. \section{The local minima and the slope}\label{sec:minima} The proof of Proposition~\ref{prop:verified} uses that the margin is convex between consecutive prime powers. This has a sharper consequence: the minima of $M$ are a discrete set, one at most between two prime powers, with an explicit location and value. The criterion of Theorem~\ref{thm:landau}, a statement about every real $N$, becomes a statement about countably many points given by sums over primes. We keep the notation of the proof of Proposition~\ref{prop:verified}: $m(L)=M(e^L)$, $r(L)=R(e^L)$, and for consecutive prime powers $qL_c=0.2824$ ($N>1.326$), $r''(L)=2e^{L/2}-2e^{-L/2}/(e^{2L}-1)>0$; so $m$ is strictly convex on every interval between consecutive prime powers. \item At a prime power $q=p^k$ the derivative of $m$ jumps down: $m'(\log q^+)-m'(\log q^-)=-2\Lambda(q)/\sqrt q<0$. Hence no prime power is a local minimum of $m$. \item For $N\ge2$, the local minima of $M$ are exactly the points $N^*_q=e^{L^*_q}$ with $r'(L^*_q)=2A_q$ and $q100$. \item On $11$, that is for $L>0.2824$. Between prime powers $m''=r''$. (2) $C(e^L)=2\sum_n\Lambda(n)n^{-1/2}(L-\log n)_+$ gains the slope $2\Lambda(q)/\sqrt q$ at $L=\log q$, and $r$ is smooth. A local minimum at $\log q$ would need $m'(\log q^-)\le0\le m'(\log q^+)$, which the jump forbids. (3) By (1) and (2), for $L\ge\log2$ every local minimum is an interior critical point of a strictly convex piece, so it is the unique solution of $m'=r'-2A_q=0$ in that piece, if it lies there. With $r'(L)=4e^{L/2}-c_0+\tau_1(L)$, $\tau_1=-\tau'$, the equation is $4e^{L/2}=2A_q+c_0-\tau_1(L)$, so $e^{L^*/2}=u_q-\tau_1(L^*)/4$. Inserting $e^{L^*/2}=u_q$ and $L^*=2\log u_q$ in $m=r-2A_qL+2B_q$ gives $8u_q-2\log u_q\,(c_0+2A_q)+K_0+2B_q=2B_q+K_0-8u_q(\log u_q-1)$; the difference with the exact value comes from $\tau$ and $\tau_1$, of order $N^{-5/2}$, and its size is the computed one. (4) On $(1,2)$ the sum over primes is empty, $r''<0$ for $N<1.326$ and $r''>0$ after, $r'$ vanishes once on $(1.326,2)$, at $N=1.5904$, and $r(0)=0$ with $r(L)\ge0.0148L$ near $0$ (Proposition~\ref{prop:verified}). \end{proof} Status: proved; the bounds for $\epsilon_q$ and the values in (4) are numerical, from the exact solution of $r'(L)=2A_q$ by Newton's method. Part (1) is the convexity already used in Proposition~\ref{prop:verified}; parts (2) and (3) are new to this paper as far as we know. \begin{corollary}[a discrete criterion]\label{cor:discrete} RH holds if and only if $M(N^*_q)\ge0$ for every prime power $q$ with $q0$ on $(1,2)$ by Proposition~\ref{prop:verified}). Let $L_0=\log N_0$. If $m'(L_0^-)>0$, follow $m$ to the left: on the convex piece containing $L_0$ it decreases towards the left end unless it meets a critical point, which would be a local minimum with value $L_0$; in the second case $M\le M(N_0)$ for all $N\ge N_0$, so $M$ is bounded above, and Corollary~\ref{cor:sandwich}(3) gives RH, hence $M\ge0$, a contradiction. One of the two cases occurs, since $m'(L_0^-)\ge m'(L_0^+)$ and they cannot be both $\le0$ and $\ge0$ unless $L_0$ is itself a critical point, which is then a local minimum with negative value. \end{proof} So RH is equivalent to countably many inequalities between two explicit sums over the prime powers below $q$, at the prime powers $q$ for which the stationary point $u_q^2$ falls before the next prime power. The reduction removes the need to look at every real $N$; it does not make the inequality easier. Both sides of \eqref{eq:discrete} are of size $4\sqrt q\log q$, and their difference is $M(N^*_q)\in[0,0.185]$ under RH. \begin{proposition}[no margin]\label{prop:nomargin} If RH holds, then $\liminf_{N\to\infty}M(N)=0$, and the local minima of Proposition~\ref{prop:minima} accumulate at $0$: $\liminf M(N^*_q)=0$ over the prime powers $q$ with $q0}(1-\cos\gamma L)/\gamma^2$ by \eqref{eq:wave}, an absolutely convergent trigonometric series, so $m$ is uniformly almost periodic with mean $2(\log\xi)''(\frac12)=0.0924$. Given $\varepsilon\in(0,0.09)$, choose $T$ with $8\sum_{\gamma>T}\gamma^{-2}<\varepsilon/2$. By Dirichlet's simultaneous approximation theorem there are arbitrarily large $L_0$ with $\|\gamma L_0/2\pi\|$ so small for every $0<\gamma\le T$ that $4\sum_{\gamma\le T}(1-\cos\gamma L_0)/\gamma^2<\varepsilon/2$; then $m(L_0)<\varepsilon$. Since $m$ is almost periodic with mean $0.0924>\varepsilon$, the set $\{m>\varepsilon\}$ is relatively dense, so there are $a\varepsilon$. The minimum of $m$ on $[a,b]$ is $<\varepsilon$, is not at $a$ or $b$, and is not at a prime power by Proposition~\ref{prop:minima}(2); so it is a local minimum $N^*_q$ with $M(N^*_q)<\varepsilon$, and $N^*_q\ge e^a$ is arbitrarily large. \end{proof} Status: proved, conditional on RH, and needs no linear independence of the ordinates. So no proof of $M\ge0$ can rest on a uniform positive lower bound: under RH the margin is positive without margin, as the least eigenvalues of the windows are in Section~\ref{sec:windows}. \emph{The minima from the primes.} Below $10^7$ there are $665\,134$ prime powers and $10\,341$ of the intervals between them, $1.6\%$, contain a local minimum. The least is $M(N^*_q)=0.055041$ at $N^*_q=3099.71$, between $q=3089$ and $q'=3109$; the largest is $0.137816$ at $N^*_q=319\,433.67$, next to the maximum of $M$ found in Proposition~\ref{prop:verified}; the mean is $0.091059$, against the mean $0.0924$ of $M$. The first ones are $M=0.0718$, $0.0754$, $0.0862$, $0.0652$ at $N^*_q=2.448$, $3.598$, $4.298$, $5.934$. The minimum $0.042997$ of $M$ below $10^{10}$, near $N=3.4\cdot10^7$, is necessarily one of these points. Both sides of \eqref{eq:discrete} are of size $4\sqrt q\log q$, and their difference is at most $0.185$. In this subsection we remove the common smooth part of the two sums $A$ and $B$ and look at what remains. The criterion becomes a condition on the fluctuation of one weighted Chebyshev sum, $B$, at the points where the fluctuation of the other, $A$, vanishes. The identities below are exact and elementary, and they contain no information beyond Corollary~\ref{cor:discrete}. What they do is separate the two roles that the primes play at a minimum, where it lies and how deep it is, and show the size of the problem. For real $x\ge1$ let $A(x)=\sum_{n\le x}\Lambda(n)n^{-1/2}$ and $B(x)=\sum_{n\le x}\Lambda(n)n^{-1/2}\log n$, so that $A_q=A(q)$ and $B_q=B(q)$, and define their fluctuations \begin{equation}\label{eq:alphabeta} \alpha(x)=A(x)-\Big(2\sqrt x-\frac{c_0}2\Big),\qquad \beta(x)=B(x)-\big(2\sqrt x\log x-4\sqrt x\big). \end{equation} The smooth parts are the integrals $\int_0^x t^{-1/2}\,dt=2\sqrt x$ and $\int_0^x t^{-1/2}\log t\,dt=2\sqrt x\log x-4\sqrt x$, that is, the two sums with $d\psi(t)$ replaced by $dt$; the constant $-c_0/2$ in $\alpha$ is the one that the closed form \eqref{eq:rL} imposes. By the prime number theorem and partial summation, $\alpha(x)=o(\sqrt x)$ and $\beta(x)=o(\sqrt x\log x)$; under RH, $\psi(x)-x=O(\sqrt x\log^2x)$ gives $\alpha(x)=O(\log^2x)$ and $\beta(x)=O(\log^3x)$. Between two consecutive prime powers $A$ and $B$ are constant, so $\alpha$ decreases, with $d\alpha/d\log x=-\sqrt x$, and at a prime power $q$ it jumps up by $\Lambda(q)/\sqrt q$. It is a sawtooth (Figure~\ref{fig:floors}(a)). \begin{proposition}[the two sums give the slope and the value of the margin]\label{prop:slope} Let $N\ge2$ not be a prime power, $L=\log N$, and $\tau_1=-\tau'$, that is, $\tau_1(L)=2\sum_{k\ge1}e^{-(2k+\frac12)L}/(2k+\frac12)$. Then \begin{align} \alpha(N)&=\frac{\tau_1(L)-m'(L)}2,\label{eq:slope}\\ \beta(N)-L\,\alpha(N)&=\frac{m(L)-K_0+\tau(L)}2.\label{eq:value} \end{align} Equivalently, $M(N)=K_0+2\big(\beta(N)-\alpha(N)\log N\big)-\tau(\log N)$ and $\beta(N)=\frac12\big(m-Lm'-K_0+\tau+L\tau_1\big)$. \end{proposition} \begin{proof} Let $q100$. \end{corollary} \begin{proof} By Proposition~\ref{prop:minima}(3) the local minima are the solutions of $m'(L)=0$ inside the intervals $(\log q,\log q')$, and by \eqref{eq:slope} these are the points where $\alpha=\frac12\tau_1$. On $L\ge\log2$ the function $\alpha-\frac12\tau_1$ is strictly decreasing between prime powers, since its derivative in $L$ is $-e^{L/2}+\sum_{k\ge1}e^{-(2k+\frac12)L}<0$; so there is at most one such point in each interval, and $\alpha$ crosses the level from above. At that point \eqref{eq:value} gives $m=K_0-\tau+2\beta-2L\alpha=K_0-\tau+2\beta-L\tau_1$, which is \eqref{eq:floorbeta}. The equivalence with RH is Corollary~\ref{cor:discrete}. The bound for the correction is the computed maximum of $\frac12(\tau+L\tau_1)$ over the minima with $N^*_q>100$. \end{proof} In words: a minimum of the margin lies where the staircase $A$ comes down to its smooth curve $2\sqrt x-\frac{c_0}2$, inside an interval free of prime powers. Its depth is twice the excess of the staircase $B$ over its smooth curve at that point, minus $7.40$. The large term $\alpha\log N$ in \eqref{eq:value} is switched off exactly at the minima, which is why the value of a minimum depends on $\beta$ alone. One function of the primes locates the minima and the other measures them. \emph{The numbers.} On the $10\,341$ local minima below $10^7$ the identities hold to the precision of the computation: $|\alpha(N^*_q)-\frac12\tau_1|\le5.5\cdot10^{-12}$, and \eqref{eq:floorbeta} holds to $2.2\cdot10^{-10}$. The values of $\beta$ at the minima lie in $[3.72819,3.79680]$, with mean $3.74621$, against the threshold $3.70067$ of \eqref{eq:betacrit} (Figure~\ref{fig:floors}(c)). The least value is at $N^*_q=3099.71$, the least minimum of $M$ below $10^7$, and its excess over the threshold, $0.02752$, is half of $M(N^*_q)=0.05504$. The least value of $M$ below $10^{10}$, $0.042997$, corresponds to $\beta=3.72217$. \emph{The minima one after another.} Let $N^*_10}\frac{m_\gamma}{\gamma}\sin(\gamma u), \] where $\gamma$ runs over the distinct ordinates of the zeros (both signs in the first sum) and $m_\gamma$ is the multiplicity. Consequently, for every $L>0$, \begin{align} V(L)&:=\bigl\langle|\alpha(e^{u+L})-\alpha(e^u)|^2\bigr\rangle=4\sum_{\gamma>0}\frac{m_\gamma^2\,(1-\cos\gamma L)}{\gamma^2}\ \ge\ M(e^L),\label{eq:variogram}\\ \langle\alpha^2\rangle&=2\sum_{\gamma>0}\frac{m_\gamma^2}{\gamma^2}\ \ge\ (\log\xi)''(\tfrac12)=0.046210\ldots,\label{eq:energy} \end{align} with equality in both, for all $L$, if and only if all zeros are simple. In that case $M(e^L)=V(L)$ is a mean of squares, and the sandwich $0\le M\le4(\log\xi)''(\frac12)$ is the elementary inequality \[ 0\le\langle|f(u+L)-f(u)|^2\rangle\le2\langle|f(u+L)|^2\rangle+2\langle|f(u)|^2\rangle=4\langle|f|^2\rangle . \] \end{proposition} \begin{proof} Let $\phi(u)=e^{-u/2}(\psi(e^u)-e^u)$. Under RH, $\phi\in B^2$ with Fourier series $-\sum_\rho e^{i\gamma_\rho u}/\rho$ \cite{RuSa,ANS}. This is the explicit formula $\psi(x)=x-\sum_\rho x^\rho/\rho-\log2\pi-\frac12\log(1-x^{-2})$, with the series converging in $B^2$ because $\sum_\rho|\rho|^{-2}<\infty$. Partial summation with $A(x)=\sum_{n\frac12$, the Laplace transform of $\alpha(e^u)$ is \[ \mathcal L[\alpha](s)=-\frac{\zeta'}{\zeta}\bigl(s+\tfrac12\bigr)\frac1s-\frac2{s-\frac12}+\frac{c_0}{2s}. \] By the Abelian theorem \cite{Wi}, the mean equals $\lim_{\sigma\to0^+}\sigma\mathcal L[\alpha](\sigma)=-\frac{\zeta'}{\zeta}(\frac12)+\frac{c_0}2$, which is $0$ by the identity $c_0=2\frac{\zeta'}{\zeta}(\frac12)$ of Section~\ref{sec:margin}. Translation and difference preserve $B^2$. So $\alpha(e^{u+L})-\alpha(e^u)$ has coefficients $\frac{im_\gamma}{\gamma}(e^{i\gamma L}-1)$, and Parseval gives \eqref{eq:variogram}, because $|e^{i\gamma L}-1|^2=2(1-\cos\gamma L)$ and each $\gamma>0$ appears with its negative. Parseval for $\alpha$ itself gives \eqref{eq:energy}. For the comparisons, \eqref{eq:wave} counts zeros with multiplicity, so $M(e^L)=4\sum_{\gamma>0}m_\gamma(1-\cos\gamma L)/\gamma^2$. Similarly $(\log\xi)''(\frac12)=-\sum_\rho(\frac12-\rho)^{-2}=2\sum_{\gamma>0}m_\gamma/\gamma^2$. Since $m^2\ge m$, with equality iff $m=1$, both inequalities and the equality cases follow, because $1-\cos\gamma L$ does not vanish identically in $L$. \end{proof} \begin{proposition}[RH as finite energy]\label{prop:energy} The following are equivalent: \begin{enumerate} \item[(i)] the Riemann hypothesis; \item[(ii)] $\alpha$ has bounded mean energy in $\log x$: $\frac1U\int_0^U\alpha(e^u)^2\,du=O(1)$; \item[(iii)] $\int_0^U\alpha(e^u)^2\,du=e^{o(U)}$, that is, $\int_1^X\alpha(x)^2\frac{dx}x=X^{o(1)}$. \end{enumerate} \end{proposition} \begin{proof} (i)$\Rightarrow$(ii) is Proposition~\ref{prop:variogram}. In substance it is Cram\'er's theorem that, under RH, $\int_2^X(\psi(x)-x)^2x^{-2}dx\sim c\log X$ \cite{In}. (ii)$\Rightarrow$(iii) is trivial. (iii)$\Rightarrow$(i): let $I(U)=\int_0^U\alpha(e^u)^2du=e^{o(U)}$ and $\sigma>0$. Integration by parts against $dI$ shows that $\int_0^\infty\alpha(e^u)^2e^{-2\sigma'u}du<\infty$ for every $\sigma'>0$. By Cauchy--Schwarz with $0<\sigma'<\sigma$, \[ \int_0^\infty|\alpha(e^u)|e^{-\sigma u}du\le\Big(\int_0^\infty\alpha^2e^{-2\sigma'u}du\Big)^{1/2}\Big(\int_0^\infty e^{-2(\sigma-\sigma')u}du\Big)^{1/2}<\infty . \] So $\mathcal L[\alpha](s)$ converges absolutely in $\Re s>0$ and is holomorphic there. For $\Re s>\frac12$ it equals $-\frac{\zeta'}{\zeta}(s+\frac12)\frac1s-\frac2{s-\frac12}+\frac{c_0}{2s}$, by partial summation from $\sum_n\Lambda(n)n^{-1/2-s}=-\frac{\zeta'}\zeta(s+\frac12)$. By analytic continuation, $-\frac{\zeta'}\zeta(w)/(w-\frac12)-2/(w-1)$ is holomorphic in $\Re w>\frac12$. The pole of $-\frac{\zeta'}\zeta$ at $w=1$ is cancelled, since its residue $1$ divided by $w-\frac12=\frac12$ gives $\frac2{w-1}$. So $\zeta$ has no zeros in $\Re w>\frac12$. \end{proof} \begin{remark}[what Proposition~\ref{prop:energy} changes, and what it does not]\label{rem:energy} Proposition~\ref{prop:energy} replaces the positivity $M\ge0$ by an energy bound for the slope, $\int_1^X\alpha(x)^2\frac{dx}x=X^{o(1)}$. Positivity is then automatic: under (ii), Proposition~\ref{prop:variogram} applies and $M$ is a mean of squares. The bound in (iii) is weak: it allows sub-exponential growth in $\log X$. Yet it is equivalent to RH. Unconditionally, $\alpha(x)\ll x^{\Theta-1/2+\varepsilon}$, where $\Theta$ is the supremum of the real parts of the zeros. This gives only $\int_1^X\alpha^2\frac{dx}x\ll X^{2\Theta-1+\varepsilon}$, and the gap between $X^{2\Theta-1}$ and $X^{o(1)}$ is exactly the gap between the known zero-free regions and RH. By the Paley--Wiener theorem, (ii) says that $\mathcal L[\alpha]$ belongs to the Hardy space $H^2$ of every half-plane $\Re s>\sigma>0$. Its holomorphy there is the content of RH. Its $L^2$ size on vertical lines is a mean-square bound for $\frac{\zeta'}{\zeta}$ to the right of the critical line, which is known under RH \cite{MV}. So an unconditional proof of (iii) would have to control $\frac{\zeta'}\zeta$ in mean square on every line $\Re w=\frac12+\sigma$ and show that it extends holomorphically. No method we know does the first without the second. We do not claim progress on it. \end{remark} Numerically, with the prime powers below $10^8$, $M(e^L)$ and the variogram of $\alpha$ agree to three digits at $L=0.2,0.6,\dots,3$, and the empirical variance of $\alpha(e^u)$ is $0.04625$, against $(\log\xi)''(\frac12)=0.04621$. \subsection{The energy programme: dyadic blocks and averages in \texorpdfstring{$L$}{L}}\label{subsec:programme} Proposition~\ref{prop:energy} reduces RH to an energy bound for $\alpha$. This subsection follows the two natural ways of attacking that bound: in dyadic blocks, using the recurrence of the minima, and by averaging the margin in $L$. Each route gives exact statements and a clear account of where it stops. Neither route proves RH. \emph{An exact identity: the energy of the slope is a C-transform of the margin.} Write $m(L)=M(e^L)$, so that $m'=\tau_1-2\alpha$ by Proposition~\ref{prop:slope}. \begin{theorem}[energy identity]\label{thm:energyidentity} For $10}m_\gamma^2/\gamma^2$. This is $\ge2\sum_{\gamma>0}m_\gamma/\gamma^2=(\log\xi)''(\frac12)$, with equality iff every $m_\gamma=1$. \end{proof} So the energy of the slope is a computable test of ``RH and simple zeros'' together, with an excess $2\sum(m_\gamma^2-m_\gamma)/\gamma^2$ for multiple zeros. The empirical value is $0.04625$ up to $10^8$, and $0.0465$ from the block energies to $10^{10}$ below. \emph{Dyadic blocks.} Let $E_k=\int_{k\log2}^{(k+1)\log2}\alpha(e^u)^2du$ be the energy in the block $[2^k,2^{k+1})$; by Proposition~\ref{prop:energy}, RH is equivalent to $E_k=2^{o(k)}$. We computed $E_k$ and $W_k=\int m'^2$ for every block below $10^{10}$, exactly between consecutive prime powers, from all $455\,062\,595$ prime powers (Figure~\ref{fig:energy}(a)). The block energies do not grow: the mean of $E_k/\log2$ over $6\le k\le32$ is $0.0465$, against $0.0462$, and $W_k/\log2$ has mean $0.1858$, against the rate $0.1848$ of Corollary~\ref{cor:rate}. Theorem~\ref{thm:energyidentity} holds block by block to $4\cdot10^{-11}$ for $k\ge15$. Why this does not give a proof: on a block, $\alpha(x)=\alpha(2^k)+D_k(x)$, where $D_k$ depends only on the prime powers of the block. Its energy is the energy of the normalized error $\bigl(\psi(x)-\psi(2^k)-(x-2^k)\bigr)/2^{k/2}$ of the prime number theorem on an interval as long as its position. A bound $2^{o(k)}$ for it is already RH. The recurrence of Proposition~\ref{prop:betarec} moves $\alpha$ exactly from one prime power to the next and reproduces $D_k$, so it gives nothing more. Finally, no computation can see an off-line zero. Such a zero has $\gamma>T_0$, so its term in $\alpha$ is at most $10^{-12}x^{\beta-1/2}$, which reaches the size $0.2$ of $\alpha$ only beyond $x=e^{54}$ even for $\beta=1$. \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{CE_energy.png} \caption{(a) Energy per dyadic block of the slope, $E_k/\log2$, and of the margin's derivative, $W_k/(4\log2)$, from all prime powers below $10^{10}$, against $(\log\xi)''(\frac12)$. (b) The same block energy, below $2^{26}$, for the true weights and for the weights \eqref{eq:planted} with a planted zero: (A) $\delta=0.15$, $\gamma_0=10$; (B) $\delta=0.25$, $\gamma_0=1000$ (Theorem~\ref{thm:nogo}).}\label{fig:energy} \end{figure} \emph{Averages in $L$.} Averaging the margin in $L$ is a second-order Riesz mean: \[ M_2(N):=\int_0^{\log N}M(e^L)\,dL=R_2(N)-\sum_{n0}\gamma^{-2}\log N-4\sum_{\gamma>0}\sin(\gamma\log N)/\gamma^3=2\langle\alpha^2\rangle\log N+O(1)$, which is the identity $\langle M\rangle=2\langle\alpha^2\rangle$. \begin{proposition}[averaging in $L$]\label{prop:average} The following are equivalent: \begin{enumerate} \item[(i)] RH; \item[(ii)] for every $\varepsilon>0$, $M_2(N)\ge-C_\varepsilon N^\varepsilon$; \item[(iii)] $|M_2(N)|=N^{o(1)}$. \end{enumerate} \end{proposition} \begin{proof} (i)$\Rightarrow$(ii),(iii): under RH, $0\le M\le0.18484$ by Corollary~\ref{cor:sandwich}, so $0\le M_2(N)\le0.18484\log N$. (iii)$\Rightarrow$(i): integrating by parts, $\int_0^\infty M(e^L)e^{-sL}dL=s\int_0^\infty M_2(e^L)e^{-sL}dL$ converges for $\Re s>0$, so the function $\mathcal G$ of the proof of Theorem~\ref{thm:landau} is holomorphic in $\Re s>0$. But each zero $\rho=\frac12+\delta+it$ with $\delta>0$ gives $\mathcal G$ a pole at $s=\delta+it$. (ii)$\Rightarrow$(i): apply Landau's theorem \cite{Wi} to $M_2(e^L)+C_\varepsilon e^{\varepsilon L}\ge0$. Its Laplace transform is $\mathcal G(s)/s+C_\varepsilon/(s-\varepsilon)$, which is holomorphic at every real point $s>\varepsilon$, because zeros of $\zeta$ are not real and $\zeta(\sigma)\ne0$ for $\frac12<\sigma<1$. So the integral converges in $\Re s>\varepsilon$, and $\mathcal G$ has no pole there. \end{proof} Proposition~\ref{prop:average} is weaker than Theorem~\ref{thm:landau} in its hypothesis, since it needs only a one-sided bound on an average and allows $N^\varepsilon$. It is equivalent all the same, and it is the same statement as Proposition~\ref{prop:energy}, because $M_2=2\langle\alpha^2\rangle\log N+O(1)$ under RH. The least eigenvalue of a window (Section~\ref{sec:windows}) is an infimum of such energies over test functions, so it enters through the same door. \begin{remark}[where both routes stop]\label{rem:programmestop} Every statement in this subsection that is equivalent to RH is equivalent through a Landau-type argument. The analytic content is always the same: the Laplace transform of $m$, of $\alpha$, or of the kernel's remainder is $-\frac{\zeta'}\zeta$ shifted by $\frac12$, divided by a polynomial in $s$, plus explicit terms. Each bound in the programme (energy, average, one-sided bound) is equivalent to the holomorphy of $\frac{\zeta'}\zeta$ in $\Re w>\frac12$. The two routes make this concrete. The block computation shows that ``local'' in $\log x$ means a prime number theorem on intervals as long as their position. Averaging shows that averaging changes the order of the Riesz mean but not the abscissa. Theorem~\ref{thm:energyidentity} is the one exact link between the energy and the primes. To be useful it would need an independent lower bound for the continuum term or an upper bound for the prime term, and each of these is again a statement about $\frac{\zeta'}\zeta$ to the right of the line. We record the programme as a reformulation with exact identities and verified numerics. It does not prove RH. \end{remark} \section{What the arguments cannot see, and what they must use}\label{sec:limits} \subsection{What the energy programme cannot see: a planted zero}\label{subsec:nogo} Remark~\ref{rem:programmestop} says that every route of Section~\ref{subsec:programme} ends at the holomorphy of $\frac{\zeta'}\zeta$. This subsection makes that statement precise in the opposite direction. We build a second system of weights on the prime powers. It satisfies every structural ingredient the programme uses: positivity, the energy identity, the prime number theorem with the classical error term, the local recurrence, and Chebyshev's identity up to $O(x^{1/2+\delta})$. Yet its slope carries an oscillation of an off-line zero. So any proof of the energy bound must use the exact values of $\Lambda$ at a precision that these ingredients do not give. \emph{The construction.} Fix $\delta\in(0,\frac12)$, $\gamma_0>0$, a switch-on point $x_0$ and a smooth function $\chi$ with $\chi=0$ on $(-\infty,0]$, $\chi=1$ on $[1,\infty)$. Put \[ f(x)=\frac2{\gamma_0}\,x^\delta\sin(\gamma_0\log x)\,\chi\Big(\log\frac x{x_0}\Big),\qquad \phi=f+\lambda h, \] where $h$ is a fixed smooth bump with compact support in $(1,x_0)$ and $\lambda$ is a real constant chosen below. Define new weights on the prime powers, \begin{equation}\label{eq:planted} w(q)=\frac{\Lambda(q)}{\sqrt q}\,\bigl(1+\varepsilon(q)\bigr),\qquad \varepsilon(q)=\sqrt q\,\phi'(q), \end{equation} and let $A_w$, $\alpha_w=A_w-2\sqrt x+\frac{c_0}2$, $C_w$, $M_w=R-C_w$ and $m_w$ be built from $w$ exactly as $A$, $\alpha$, $C$, $M$ and $m$ are built from $\Lambda(n)n^{-1/2}$. \begin{theorem}[a planted zero]\label{thm:nogo} With the notation above, for $x_0$ large enough: \begin{enumerate} \item[(i)] \emph{positivity, and the weights are asymptotically the true ones:} $|\varepsilon(q)|\le C_\phi\,q^{\delta-\frac12}<1$, so $w(q)>0$ and $w(q)\sqrt q/\Lambda(q)\to1$; \item[(ii)] \emph{the energy identity:} Theorem~\ref{thm:energyidentity} holds verbatim for $m_w$, with $\Lambda(q)q^{-1/2}$ replaced by $w(q)$; \item[(iii)] \emph{the planted oscillation:} unconditionally, $\alpha_w(x)=\alpha(x)+f(x)+o(x^\delta)$; \item[(iv)] \emph{the prime number theorem is unchanged at the classical scale:} $\sum_{q\le x}\Lambda(q)(1+\varepsilon(q))=x+O\big(x\,e^{-c\sqrt{\log x}}\big)$; \item[(v)] \emph{Chebyshev's identity up to $x^{1/2+\delta}$:} for a suitable $\lambda$, $\sum_{q\le x}\Lambda(q)(1+\varepsilon(q))\lfloor x/q\rfloor=\log\lfloor x\rfloor!+O(x^{1/2+\delta})$; \item[(vi)] \emph{one of the two systems violates the energy bound:} for all large $X$, \[ \max\Big(\int_1^X\alpha^2\frac{dx}x,\ \int_1^X\alpha_w^2\frac{dx}x\Big)\ \ge\ \frac{X^{2\delta}}{7\,\delta\,\gamma_0^2}. \] \end{enumerate} \end{theorem} \begin{proof} (i) $|f'(x)|\le\frac2{\gamma_0}(\delta+\gamma_0+\|\chi'\|_\infty)x^{\delta-1}$ and $h'$ is bounded with compact support, so $|\varepsilon(q)|\le C_\phi q^{\delta-1/2}$. On the support of $h$ the bound is $|\lambda|\sqrt q\|h'\|_\infty$, which is $<1$ for the small $\lambda$ needed in (v). For $q\ge x_0$, $C_\phi x_0^{\delta-1/2}<1$ once $x_0$ is large. (ii) The proof of Theorem~\ref{thm:energyidentity} uses only that $m_w=r-2A_wL+2B_w$ is smooth with $m_w''=r''$ between prime powers and that $m_w'$ jumps by $-2w(q)$ at $q$. Nothing else about the weights enters. (iii) $A_w(x)-A(x)=\sum_{q0$. That is the strength of the statement $\psi(x)=x+O(x^{1/2+\delta})$ itself, the zero-free strip of width $\frac12-\delta$. Beurling systems give a stronger form of the same obstruction. Their weights come from a genuine Euler product, the integers satisfy $N(x)=cx+O(x^\theta)$ with $\theta<1$, and yet the error term of the prime number theorem can be as large as $x\exp(-c\sqrt{\log x})$ \cite{DMV}. The weights \eqref{eq:planted} do not come from an Euler product. Their point is that they stay within $q^{\delta-1/2}$ of the true ones and violate $\Lambda*1=\log$ only at the scale $x^{1/2+\delta}$. The information that rules them out is the exact identity $\sum_{n\le x}\Lambda(n)\lfloor x/n\rfloor=\log\lfloor x\rfloor!$, equivalently $\zeta(s)=\sum n^{-s}$ with its analytic continuation. No step of Sections~\ref{sec:minima} and \ref{sec:energy} uses it beyond the prime number theorem. \emph{The numbers.} Figure~\ref{fig:energy}(b) plants two zeros into the true prime powers below $2^{26}$. The bump $h=\sin^2$ in $\log x$ on $[30,300]$ fixes $S=0$ on the computed range. In (A), $\delta=0.15$, $\gamma_0=10$, $x_0=10^3$: $\max|\varepsilon(q)|=0.112$, the Chebyshev defect satisfies $|D(x)|\le0.31\,x^{0.65}$ for $10^6\le x\le6\cdot10^7$, and the block energy grows like $2^{2\delta k}$. In (B), $\delta=0.25$, $\gamma_0=1000$, $x_0=10^5$: $\max|\varepsilon(q)|=0.077$, and the block energies agree with those of $\zeta$ within $0.002$ up to $k=16$. Case (B) is the realistic one; a true off-line zero has $\gamma_0>T_0$ and is invisible below $e^{54}$. \begin{remark}[status]\label{rem:nogo} Theorem~\ref{thm:nogo} is proved and unconditional. It does not say that RH is false, or that the energy programme is useless. It says exactly which input is missing: an argument that uses $\Lambda=\mu*\log$ exactly, not only through the prime number theorem. The same test applies to any other argument in this paper: an argument that remains valid for the weights \eqref{eq:planted} cannot prove RH. \end{remark} \subsection{Using \texorpdfstring{$\Lambda=\mu*\log$}{Lambda = mu * log} exactly}\label{subsec:mobius} Theorem~\ref{thm:nogo} says that a proof of the energy bound must use the identity $\Lambda=\mu*\log$ exactly, not only through the prime number theorem. This subsection carries that identity through the slope $\alpha$ and asks what it gives. \emph{The identity.} For every $n\ge1$, \begin{equation}\label{eq:mulog} \Lambda(n)=\sum_{d\mid n}\mu(d)\log\frac nd,\qquad\text{equivalently}\qquad\sum_{d\mid n}\Lambda(d)=\log n,\qquad -\frac{\zeta'}{\zeta}(s)=\frac1{\zeta(s)}\cdot\bigl(-\zeta'(s)\bigr). \end{equation} The divisor sum is a massive cancellation (Figure~\ref{fig:mobius}(a)). For $n$ with two distinct prime factors, the positive and negative terms are each as large as $\log n$, and they cancel exactly to $0$. Summed up to $x$, \eqref{eq:mulog} is Chebyshev's identity $\sum_{m\le x}\psi(x/m)=\log\lfloor x\rfloor!$. Its Möbius inverse $\psi(x)=\sum_{d\le x}\mu(d)\log\lfloor x/d\rfloor!$ adds terms of size $x\log x$ that cancel to $x$ (Figure~\ref{fig:mobius}(b): at $x=10^4$ the first term is $82\,109$ and the sum is $10\,013.4$). The cleanest exact form for what follows is \begin{equation}\label{eq:psifloor} \psi(x)+\log x=\sum_{d\le x}\mu(d)\Big\lfloor\frac xd\Big\rfloor\log\frac xd , \end{equation} which follows from \eqref{eq:mulog} by summing $\sum_{d\mid n}\mu(d)\log(x/d)=[n=1]\log x+\Lambda(n)$ over $n\le x$. It holds to $10^{-11}$ in floating point at $x=10^4$, $12\,345.6$ and $99\,999.5$. \emph{The slope through Möbius.} Write, for $x>1$ not an integer, \[ m_1(x)=\sum_{d1$ that is not an integer, exactly, \begin{align*} \alpha(x)&=\mathfrak T_1(x)+\mathfrak T_2(x)+\mathfrak T_3(x),\\ \mathfrak T_1&=2\sqrt x\,\bigl(m_1\log x-m_2-2m_1-1\bigr),\qquad \mathfrak T_2=-\zeta'(\tfrac12)\Big(m_{1/2}-\frac1{\zeta(\frac12)}\Big),\\ \mathfrak T_3&=\sum_{d0}|\rho\zeta'(\rho)|^{-2}$.} \label{fig:mobius} \end{figure} \emph{What the factorization does to the energy.} The identity \eqref{eq:mulog} factors $-\frac{\zeta'}\zeta$ into $\frac1\zeta$ times $-\zeta'$. The second factor is harmless: it is holomorphic in $\Re s>\frac12$ except at $s=1$, and of polynomial growth there. The first factor carries all the zeros, with residues $\frac1{\zeta'(\rho)}$. The rigidity of $\alpha$ is the identity $\zeta'(\rho)\cdot\frac1{\zeta'(\rho)}=1$, and it reappears only when the three pieces are added. The energy shows the difference. \begin{proposition}[the Möbius energy]\label{prop:mobiusenergy} Let $M(x)=\sum_{n\le x}\mu(n)$. \begin{enumerate} \item[(i)] Unconditionally: if $\int_1^XM(x)^2x^{-2}dx=X^{o(1)}$, then RH holds. \item[(ii)] (Ng \cite{Ng}) If RH holds and $J_{-1}(T)=\sum_{0<\gamma\le T}|\zeta'(\rho)|^{-2}\ll T$, then $e^{-u/2}M(e^u)$ has a limiting distribution with mean $0$ and mean square \[ \big\langle(M(x)/\sqrt x)^2\big\rangle=2\sum_{\gamma>0}\frac1{|\rho\,\zeta'(\rho)|^2}. \] \item[(iii)] Numerically, the sum over the first $96$ zeros is $0.0282$. The Gonek--Hejhal conjecture $J_{-1}(T)\sim\frac{6T}{\pi^3}$ \cite{Go} estimates the tail beyond $\gamma=229.3$ at $0.0017$, for a total of $0.0299$. The block energies of $M(x)/\sqrt x$ for $k=10,\dots,25$ have mean $0.0296$ (Figure~\ref{fig:mobius}(d)). \end{enumerate} \end{proposition} \begin{proof} (i) For $\Re s>1$, $\int_1^\infty M(x)x^{-s-1}dx=\frac1{s\zeta(s)}$. In the variable $u=\log x$, $\int_0^\infty M(e^u)e^{-u/2}e^{-wu}du=\frac1{(w+\frac12)\zeta(w+\frac12)}$. If the energy is $e^{o(U)}$, the Cauchy--Schwarz argument of Proposition~\ref{prop:energy} makes this Laplace integral converge absolutely in $\Re w>0$. So $\frac1\zeta$ is holomorphic in $\Re s>\frac12$. (ii) is proved in \cite{Ng}, and (iii) is a computation. \end{proof} The comparison is the point of this subsection. Under RH, the energy of $\alpha$ is $2\sum m_\gamma^2/\gamma^2$. It is finite, equal to $(\log\xi)''(\frac12)$ for simple zeros, and needs nothing more. The Möbius energy needs control of $\zeta'(\rho)$ from below, which is not known even under RH. Through \eqref{eq:mulog}, the energy problem for $\alpha$ becomes an energy problem for $M$ together with a harmless smooth factor. That problem is no easier: it is strictly less rigid. \emph{Three levels of exactness.} Which arithmetic facts exclude the systems of Theorem~\ref{thm:nogo}? \begin{itemize} \item[(0)] \emph{The planted weights} \eqref{eq:planted} have an Euler product only formally. Defining $\zeta_w(s)=\exp\big(\sum_n\Lambda(n)(1+\varepsilon(n))n^{-s}/\log n\big)$, the local factor at $p$ is $\exp\big(\sum_k(1+\varepsilon(p^k))p^{-ks}/k\big)$. So $\mu_w(p)=-(1+\varepsilon(p))$ and $\mu_w(p^2)=\frac12\big((1+\varepsilon(p))^2-(1+\varepsilon(p^2))\big)$. For the system (A) of Figure~\ref{fig:energy}(b), $\max|\mu_w(p)|=1.112$ on $10^31$, and $\max|\mu_w(p^2)|=0.113$. The local facts $\mu(p)=-1$ and $\mu(p^2)=0$ already exclude them. \item[(1)] \emph{Beurling systems} have $\mu_B(p)=-1$ and $\mu_B(p^k)=0$ for $k\ge2$, because they come from a genuine Euler product over generalized primes. Diamond, Montgomery and Vorhauer \cite{DMV} construct such systems with integer counting function $N_B(x)=\kappa x+O(x^\theta)$, $\frac12<\theta<1$, whose primes satisfy $\pi_B(x)=\operatorname{li}(x)+\Omega\big(x\exp(-c\sqrt{\log x})\big)$. So the multiplicative structure, even with integers known to precision $x^\theta$ for any $\theta>\frac12$, does not give the energy bound. In Proposition~\ref{prop:mobiusrep} for such a system, the sawtooth $B_1(\{x/d\})$ is replaced by the error of $N_B$ at $x/d$, which can be as large as $(x/d)^\theta$ instead of $O(1)$. \item[(2)] \emph{The integers $\mathbb N$}: $N(x)=\lfloor x\rfloor$, with error $O(1)$. This is the only information left, and Proposition~\ref{prop:mobiusrep} shows where it enters: the sawtooth in \eqref{eq:sawtooth} and the two constants $\zeta(\frac12)$, $\zeta'(\frac12)$. \end{itemize} \begin{remark}[status: what an exact argument must do]\label{rem:mobius} By Proposition~\ref{prop:mobiusrep} and Theorem~\ref{thm:nogo}, a proof along the lines of this paper must show that the Möbius-weighted sawtooth \[ \mathcal S(x)=\sum_{d\le x}\mu(d)\,B_1\Big(\Big\{\frac xd\Big\}\Big)\log\frac xd \] is $x^{1/2+o(1)}$ in mean square on $[X,2X]$, and it must use the cancellation between $\mu(d)$ and the fractional parts of $x/d$. The identity \eqref{eq:psifloor} shows that nothing is gained for free. Inserting $\lfloor y\rfloor=y-\frac12-B_1(\{y\})$ gives, exactly, \[ \mathcal S(x)=x\sum_{d\le x}\frac{\mu(d)}d\log\frac xd-\frac12\sum_{d\le x}\mu(d)\log\frac xd-\psi(x)-\log x , \] so $\mathcal S$ is $-\bigl(\psi(x)-x\bigr)$ up to Mertens-type sums that are $O(x^{1/2+\varepsilon})$ precisely when RH holds. The identity $\Lambda=\mu*\log$ is reversible, and rewriting $\psi$ through it changes the form of the problem but not its content. The form is still informative. $\mathcal S$ is a type~I sum: Möbius against the explicit, bounded, additive function $B_1(\{x/d\})$. Expanding $B_1$ in its Fourier series turns $\mathcal S$ into exponential sums $\sum_d\mu(d)e(hx/d)$. The best unconditional bounds for $\psi(x)-x$, and so for $\mathcal S$, save only a factor $\exp\big(-c(\log x)^{3/5}(\log\log x)^{-1/5}\big)$ over the trivial bound. That is the Vinogradov--Korobov zero-free region, seen from the margin. A saving of $x^{-1/2+\varepsilon}$ would be RH. We do not claim any progress on closing this gap. \end{remark} \section{Windows and Weil positivity: a summary}\label{sec:windows} The margin is Weil's form $Q$ for one test function. The companion papers \cite{P20,P21,P22} study $Q$ on whole spaces of test functions. We summarize the results, with their status, and refer to those papers for proofs and data. \emph{Dirichlet polynomials} \cite{P20}. Because $\Lambda\ge0$, the prime part of $Q$ on Gaussian-smoothed Dirichlet polynomials is unbounded below. So positivity is a cancellation between the archimedean place and the coherent sum of all primes, not a diagonal dominance (proved, elementary). With ball arithmetic, $Q$ is positive definite on these polynomials for every length $N\le179$ (computer-assisted, for smoothing width $0.005$). No translation-invariant square $\|Xk\|^2$ with $X$ a Fourier multiplier fits below $Q$ on the whole line; in particular, no Euler-product or mollifier multiplier does (proved). \emph{The margin as $W-E$} \cite{P21}. Following Weil's proof for curves, write $Q=W-E$ with the pole $W=2\widehat g(\frac i2)\widehat g(-\frac i2)$. Then $W$ is a hyperbolic plane, like the two fibres of $C\times C$, and $Q\ge0$ on a space implies that $E$ has at most one positive direction there (proved; not conversely). The archimedean place alone has $(t_0/\pi+o(1))\log N$ positive directions, $t_0=6.2898$ (proved, of Szeg\H{o} type). Numerically, the primes remove them only all together and only with their powers. \emph{Windows} \cite{P22}. In a window of length $L=\log N$ the form is one symbol. A positive completion of that symbol gives an exact sum of squares, and a coercive form has the completion $\pi/|E_L|^2$ with $E_L$ the de Branges function of the window (proved, classical in substance). For even $k$ of bounded variation supported in $[-\frac L2,\frac L2]$, with $K=\widehat k$, the explicit formula applied to $k*\tilde k$ gives \[ Q(k)=\sum_\rho K(\gamma_\rho)^2 , \] so under RH the least eigenvalue of a window is the minimum of $2\sum_{\gamma>0}K(\gamma)^2/\|k\|^2$ (proved). Connes observed that the minimizer reproduces the zeros \cite{Co26}. In high precision, it does so to $10^{-2}$ up to $1.3$ to $1.9$ times $T^*=2\pi N$, the height where the density $\frac1{2\pi}\log\frac t{2\pi}$ of the zeros equals the Nyquist density $L/2\pi$ of the window (numerical; Figure~\ref{fig:edge}). For the Davenport--Heilbronn function, the agreement stops at the corresponding scale $2\pi N/5$. The zeros below $T^*$ carry a fraction between $10^{-25}$ and $10^{-8}$ of the least eigenvalue. \begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{CE22_borde.png} \caption{The minimizer of Weil's form in the windows $N=4.5,\,7.4,\,13,\,20.1$ \cite{P22}. Left: $\log_{10}$ of the distance from each zero of $\zeta$ to the nearest zero of the minimizer's transform. Right: the same data with the height in units of $T^*=2\pi N$ and the error in units of $\lambda_{\min}$; dashed $T^*$, dotted $eT^*$.}\label{fig:edge} \end{figure} \section{Companion papers}\label{sec:companions} \emph{The Caceres prime function} \cite{CPF}. The weight $a(n)=\sum_{p^k\|n}\log p\,p^{-k/2}$ has Dirichlet series $\zeta(s)\mathcal K(s)$. Its kernel $\mathcal K$ is $-\frac{\zeta'}\zeta(s+\frac12)+\frac{\zeta'}\zeta(2s+\frac12)-\frac1{12s}$ plus a function holomorphic in $\Re s>-\frac18$, so it sees each zero twice, at $i\gamma$ and at $i\gamma/2$. In the language of this paper, the kernel's margin is $M(N)-2M(\sqrt N)$ under RH, and its slope is $\alpha(x)-\alpha(\sqrt x)$ without hypothesis; its energy is $2(\log\xi)''(\frac12)$, less $2\sum\gamma^{-2}$ over the pairs of ordinates with $\gamma'=2\gamma$. Since $a=1*b$ with $b$ fixed by the primes, the kernel carries no information beyond $\Lambda$, and Theorem~\ref{thm:nogo} applies to it verbatim. \emph{The $6k\pm1$ race} \cite{SixK}. The same construction for the character $\chi_{-3}$ gives a margin $M_\chi$ whose positivity is equivalent to GRH for $L(s,\chi_{-3})$. It holds with $0.102\le M_\chi\le0.355$ for $2\le N\le2^{27}$. In the margin's weighting, the prime powers $\equiv2\pmod3$ lead those $\equiv1\pmod 3$ by exactly $2\frac{L'}L(\frac12,\chi_{-3})\log N-\frac12\psi'(\frac34)+M_\chi(N)$, up to exponentially small terms. So under GRH the weighted $6k\pm1$ race never reverses. \section{Discussion}\label{sec:discussion} Every statement in this paper that is equivalent to RH is equivalent through a Landau-type argument, and each reduces to the holomorphy of $\frac{\zeta'}\zeta$, or of $\frac1\zeta$, to the right of the critical line. The unconditional results are of two kinds. Some are exact identities between the primes and the margin: the Fej\'er formula, the description of the minima, the energy identity and the M\"obius representation. The others are statements of what cannot work: no uniform positive lower bound for $M$ under RH (Proposition~\ref{prop:nomargin}), and no argument that survives planted weights (Theorem~\ref{thm:nogo}). The second kind gives a test that any proposed argument can be run against: if it also works for the weights \eqref{eq:planted}, or for a Beurling system \cite{DMV}, it cannot prove RH. What remains is an estimate that uses $N(x)=\lfloor x\rfloor$ exactly, through the sawtooth of Proposition~\ref{prop:mobiusrep}. The best unconditional estimates of that kind save $\exp(-c(\log x)^{3/5}(\log\log x)^{-1/5})$, the Vinogradov--Korobov region. A saving of $x^{-1/2+\varepsilon}$ would be RH. We do not prove the Riemann hypothesis. \section*{Data and code} \sloppy The verification of Proposition~\ref{prop:verified} is \texttt{certificado\_segmentado.py} (Python's \texttt{decimal} only). The minima and their recurrence (Section~\ref{sec:minima}, Figure~\ref{fig:floors}) come from \texttt{minimos.py}, \texttt{recurrencia.py}, \texttt{fig\_fondos.py} and \texttt{fig\_fondos\_rec.py} in \texttt{anc/minimos}. The block energies to $10^{10}$ and the energy identity come from \texttt{blocks.py} (segmented sieve, extended precision); the planted systems from \texttt{fake.py}; Figure~\ref{fig:energy} from \texttt{fig\_energy.py}, all in \texttt{anc/energy}. The M\"obius representation and Figure~\ref{fig:mobius} come from \texttt{compute.py}, \texttt{saw.py} and \texttt{figure.py} in \texttt{anc/mobius}. The wave (Figure~\ref{fig:wave}) and the windows are computed by the ancillary files of \cite{P20,P22}. Theorem~\ref{thm:explicit}, the extended certificate and Figure~\ref{fig:explicit} come from \texttt{ebound.py}, \texttt{ebound\_rig.py}, \texttt{cert\_ext.py}, \texttt{cert\_up.py}, \texttt{certify\_zeros.py} and \texttt{fig\_ext.py}, and part (2) of Proposition~\ref{prop:verified} from \texttt{verify\_margin.py} in \texttt{anc/verification} in \texttt{anc/explicit}, with the ordinates in \texttt{zeros5000.npy}. All are included with this paper. \section*{Use of AI assistance} The cross-energy, the one-variable formulation and the research programme are the author's. Claude (Anthropic) was used as a research and writing aid: to derive and check proofs, to write and run the computations, to test statements against the controls, to search the literature, and to draft and edit text and figures. The author checked all results and is responsible for the content. {\footnotesize \begin{thebibliography}{GORZ} \bibitem[Arb]{Ar} F. 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