\section{What the arguments cannot see, and what they must use}\label{ce:sec:limits}
\subsection{What the energy programme cannot see: a planted zero}\label{ce:subsec:nogo}
Remark~\ref{ce:rem:programmestop} says that every route of Section~\ref{ce: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{ce: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{ce: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{ce: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{ce: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) $A_w(x)-A(x)=\sum_{q<x}\Lambda(q)\phi'(q)=\int_{2^-}^x\phi'(t)\,d\psi(t)$. Write $d\psi=dt+d(\psi(t)-t)$ and integrate the second part by parts:
\[
A_w(x)-A(x)=\phi(x)+\phi'(x)\bigl(\psi(x)-x\bigr)-\int_2^x\bigl(\psi(t)-t\bigr)\phi''(t)\,dt .
\]
Here $\phi(2)=0$, and for $x\ge x_0$, $\phi=f$. With $\psi(t)-t\ll te^{-c\sqrt{\log t}}$ and $\phi''(t)\ll\gamma_0t^{\delta-2}$, the last two terms are $\ll x^\delta e^{-c'\sqrt{\log x}}=o(x^\delta)$.

(iv) The same computation without the factor $q^{-1/2}$ gives $\sum_{q\le x}\Lambda(q)\varepsilon(q)=\int_2^x\sqrt t\,\phi'(t)\,d\psi(t)\ll x^{1/2+\delta}$, which is smaller than the classical error term.

(v) Write $\lfloor x/q\rfloor=x/q-\{x/q\}$. The defect is $D(x)=x\sum_{q\le x}\Lambda(q)\varepsilon(q)/q-\sum_{q\le x}\Lambda(q)\varepsilon(q)\{x/q\}$. The second sum is $\ll\sum_{q\le x}\Lambda(q)q^{\delta-1/2}\ll x^{1/2+\delta}$. The series $S=\sum_q\Lambda(q)\varepsilon(q)/q$ converges absolutely, and its tail beyond $x$ is $\ll x^{\delta-1/2}$. The bump changes $S$ by $\lambda\sum_q\Lambda(q)h'(q)q^{-1/2}\ne0$ for a suitable $h$, so one value of $\lambda$ gives $S=0$, and then $D(x)\ll x^{1/2+\delta}$.

(vi) By (iii), $f=\alpha_w-\alpha+o(x^\delta)$, so $\int_1^Xf^2\frac{dx}x\le3\int\alpha_w^2\frac{dx}x+3\int\alpha^2\frac{dx}x+o(X^{2\delta})$. Since $\sin^2$ has mean $\frac12$, $\int_1^Xf^2\frac{dx}x\sim\frac{4}{\gamma_0^2}\int^{\log X}e^{2\delta u}\sin^2(\gamma_0u)\,du\sim\frac{X^{2\delta}}{\delta\gamma_0^2}$.
\end{proof}

\emph{What the theorem says.} Every structural fact the programme uses holds for both systems: positivity of the weights, the energy identity, the recurrence of the minima (Proposition~\ref{ce:prop:betarec} is also pure bookkeeping), the prime number theorem with its classical error, and the arithmetic identity $\Lambda*1=\log$ up to $O(x^{1/2+\delta})$. Yet at most one of the two systems satisfies the energy bound of Proposition~\ref{ce:prop:energy}. So a proof of RH along these lines must use the exact values of $\Lambda$ at relative precision $x^{\delta-1/2}$, for every $\delta>0$. 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{ce: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{ce:sec:minima} and \ref{ce:sec:energy} uses it beyond the prime number theorem.


\emph{The numbers.} Figure~\ref{ce: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{ce:rem:nogo}
Theorem~\ref{ce: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{ce:eq:planted} cannot prove RH.
\end{remark}

\subsection{Using \texorpdfstring{$\Lambda=\mu*\log$}{Lambda = mu * log} exactly}\label{ce:subsec:mobius}
Theorem~\ref{ce: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{ce: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{ce: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{ce: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{ce: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{ce: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{ce: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,
\[
m_1(x)=\sum_{d<x}\frac{\mu(d)}d,\quad m_2(x)=\sum_{d<x}\frac{\mu(d)\log d}d,\quad m_{1/2}(x)=\sum_{d<x}\frac{\mu(d)}{\sqrt d},\quad S(y)=\sum_{m<y}\frac{\log m}{\sqrt m},
\]
and $S_0(y)=2\sqrt y\log y-4\sqrt y-\zeta'(\frac12)$. Euler--Maclaurin gives $S(y)-S_0(y)\to0$; the constant is $-\zeta'(\frac12)$ because $\sum_{m\le y}m^{-s}\log m=\frac{y^{1-s}\log y}{1-s}-\frac{y^{1-s}}{(1-s)^2}-\zeta'(s)+o(1)$ for $0<s<1$.

\begin{proposition}[the slope through Möbius]\label{ce:prop:mobiusrep}
For every $x>1$ 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_{d<x}\frac{\mu(d)}{\sqrt d}\,(S-S_0)\Big(\frac xd\Big).
\end{align*}
Moreover $(S-S_0)(y)=-B_1(\{y\})\frac{\log y}{\sqrt y}+O\big(\frac{\log ey}{y^{3/2}}\big)$ with $B_1(t)=t-\frac12$, so that
\begin{equation}\label{ce:eq:sawtooth}
\mathfrak T_3(x)=-\frac1{\sqrt x}\sum_{d<x}\mu(d)\,B_1\Big(\Big\{\frac xd\Big\}\Big)\log\frac xd+\sum_{d<x}\frac{\mu(d)}{\sqrt d}\,O\Big(\frac{\log(ex/d)}{(x/d)^{3/2}}\Big).
\end{equation}
\end{proposition}

\begin{proof}
By \eqref{ce:eq:mulog}, $A(x)=\sum_{dm<x}\mu(d)\log m\,(dm)^{-1/2}=\sum_{d<x}\mu(d)d^{-1/2}S(x/d)$. Write $S=S_0+(S-S_0)$. The $S_0$ part is
\[
\sum_{d<x}\frac{\mu(d)}{\sqrt d}\Big[2\sqrt{\tfrac xd}\bigl(\log x-\log d\bigr)-4\sqrt{\tfrac xd}-\zeta'(\tfrac12)\Big]=2\sqrt x\,(m_1\log x-m_2-2m_1)-\zeta'(\tfrac12)\,m_{1/2}.
\]
Subtracting $2\sqrt x$ and adding $\frac{c_0}2=\frac{\zeta'}\zeta(\frac12)$ gives $\alpha=\mathfrak T_1+\mathfrak T_2+\mathfrak T_3$. For \eqref{ce:eq:sawtooth}, apply Euler--Maclaurin to $f(t)=t^{-1/2}\log t$ with the first Bernoulli function: $S(y)=\int_1^yf+C-B_1(\{y\})f(y)-\int_y^\infty B_1(\{t\})f'(t)dt$. The last integral is $O(|f'(y)|)$ after one more integration by parts, and $C$ is fixed by $S-S_0\to0$.
\end{proof}

Status: proved; the identity is checked to $4.4\cdot10^{-9}$ at $400$ points in $[10^3,10^7]$ (Figure~\ref{ce:fig:mobius}(c)). The three pieces behave very differently. $\mathfrak T_1$ and $\mathfrak T_2$ are each large: their root mean squares on $[10^3,10^7]$ are $0.694$ and $0.680$, against $0.218$ for $\alpha$. But they are almost exact negatives of each other: their correlation is $-0.9999999$, and $\mathfrak T_1+\mathfrak T_2$ has root mean square $0.0140$. So $\alpha$ is essentially $\mathfrak T_3$, and $\mathfrak T_3$ is essentially the \emph{Möbius-weighted sawtooth} in \eqref{ce:eq:sawtooth}: the difference has root mean square $0.012$. This difference in turn nearly cancels $\mathfrak T_1+\mathfrak T_2$: $\alpha$ minus the sawtooth has root mean square $0.0020$, against $0.249$ for $\alpha$ at the same $50$ points. To within $1\%$,
\[
\alpha(x)\approx-\frac1{\sqrt x}\sum_{d<x}\mu(d)\Big(\Big\{\frac xd\Big\}-\frac12\Big)\log\frac xd .
\]
The slope of the margin is the fractional parts of $x/d$, weighted by $\mu(d)\log(x/d)$. This is the one place where the additive structure of $\mathbb N$, through $\lfloor\cdot\rfloor$, enters the margin. The $1\%$ agreement is not a coincidence. By Remark~\ref{ce:rem:mobius}, the sawtooth sum is $-(\psi(x)-x)$ up to Mertens-type sums. Under RH, $(\psi(x)-x)/\sqrt x$ has the coefficients $-1/\rho=\frac i\gamma+O(\gamma^{-2})$, the same as $\alpha$ up to $O(\gamma^{-2})$.

\emph{The spectral content of the pieces.} Under RH the coefficient of $e^{i\gamma u}$ in $\alpha(e^u)$ is $\frac{im_\gamma}\gamma$ (Proposition~\ref{ce:prop:variogram}). It is rigid: it depends only on $\gamma$ and the multiplicity. The pieces of Proposition~\ref{ce:prop:mobiusrep} are not rigid.

\begin{proposition}[the pieces carry $1/\zeta'(\rho)$]\label{ce:prop:mobiusspec}
Assume RH, that the zeros are simple, and that the Perron expansions of $m_1$, $m_2$ and $m_{1/2}$ over the zeros converge in $B^2$ (as they do for $M(x)$ under Ng's hypothesis below). Then the coefficients of $e^{i\gamma u}$, $\gamma$ the ordinate of $\rho=\frac12+i\gamma$, are
\[
c_1(\gamma)=\frac{4(1-i\gamma)}{(i\gamma-\frac12)^2\,\zeta'(\rho)},\qquad c_2(\gamma)=-\frac{\zeta'(\frac12)}{i\gamma\,\zeta'(\rho)},\qquad c_3(\gamma)=\frac i\gamma-c_1(\gamma)-c_2(\gamma),
\]
for $\mathfrak T_1$, $\mathfrak T_2$ and $\mathfrak T_3$. In particular, as $\gamma\to\infty$,
\[
c_1(\gamma)+c_2(\gamma)=\frac{\bigl(4+\zeta'(\frac12)\bigr)\,i}{\gamma\,\zeta'(\rho)}\big(1+O(\gamma^{-1})\big),\qquad 4+\zeta'(\tfrac12)=0.0774 .
\]
\end{proposition}

\begin{proof}
Perron's formula for $m_{1/2}$ is $\frac1{2\pi i}\int\frac{x^w}{w\,\zeta(\frac12+w)}dw$. The zero $\rho$ gives the residue $\frac{x^{i\gamma}}{i\gamma\zeta'(\rho)}$, so $\mathfrak T_2$ has $c_2$. For $m_1$ the residue is $\frac{x^{w_0}}{w_0\zeta'(\rho)}$, $w_0=\rho-1$. For $m_2$, the Dirichlet series is $-(1/\zeta)'(1+w)$, which has a double pole with principal part $\frac1{\zeta'(\rho)(w-w_0)^2}$; the residue of $\frac{x^w}w\cdot\frac1{\zeta'(\rho)(w-w_0)^2}$ is $\frac{x^{w_0}}{\zeta'(\rho)}\big(\frac{\log x}{w_0}-\frac1{w_0^2}\big)$. Hence $m_1\log x-m_2-2m_1$ contributes $\frac{x^{w_0}}{\zeta'(\rho)}\big(\frac1{w_0^2}-\frac2{w_0}\big)$. Multiplying by $2\sqrt x$ gives $c_1$, since $1-2w_0=2-2i\gamma$. The sum of the three coefficients is the coefficient of $\alpha$, which gives $c_3$. The asymptotic form follows from $c_1=\frac{4i}{\gamma\zeta'(\rho)}(1+O(\gamma^{-1}))$.
\end{proof}

Status: the residues are a computation; the convergence in $B^2$ is an assumption of the same type as Ng's. A least-squares fit of the first fourteen frequencies to the computed pieces on $\log x\in[\log10^3,\log10^7]$ gives amplitudes $2|c|$ that agree with the predictions to the accuracy of the fit. At $\gamma_1$ the predicted and fitted values are $0.6998$ and $0.7058$ for $\mathfrak T_2$, $0.7145$ and $0.7206$ for $\mathfrak T_1$, and $0.0147$ and $0.0148$ for the residual $\mathfrak T_1+\mathfrak T_2$, which is $50$ times smaller than either piece.

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{fig/CE_mobius.png}
\caption{$\Lambda=\mu*\log$. (a) The divisor sum \eqref{ce:eq:mulog} for $n\le60$: positive and negative terms, and their sum $\Lambda(n)$. (b) Partial sums of $\psi(x)=\sum_{d\le x}\mu(d)\log\lfloor x/d\rfloor!$ at $x=10^4$. (c) The exact decomposition of Proposition~\ref{ce:prop:mobiusrep}: $\mathfrak T_1$ and $\mathfrak T_2$ are mirror images, and $\alpha$ follows the Möbius-weighted sawtooth \eqref{ce:eq:sawtooth} (dots). (d) Energy per dyadic block of $\alpha$ (to $10^{10}$) and of the Mertens function $M(x)/\sqrt x$ (to $2^{26}$), with their predicted means $(\log\xi)''(\frac12)$ and $2\sum_{\gamma>0}|\rho\zeta'(\rho)|^{-2}$.}
\label{ce:fig:mobius}
\end{figure}

\emph{What the factorization does to the energy.} The identity \eqref{ce: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{ce: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{ce: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{ce: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{ce: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{ce:thm:nogo}?
\begin{itemize}
\item[(0)] \emph{The planted weights} \eqref{ce: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{ce:fig:energy}(b), $\max|\mu_w(p)|=1.112$ on $10^3<p<2^{13}$, where $55\%$ of the primes have $|\mu_w(p)|>1$, 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{ce: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{ce:prop:mobiusrep} shows where it enters: the sawtooth in \eqref{ce:eq:sawtooth} and the two constants $\zeta(\frac12)$, $\zeta'(\frac12)$.
\end{itemize}

\begin{remark}[status: what an exact argument must do]\label{ce:rem:mobius}
By Proposition~\ref{ce:prop:mobiusrep} and Theorem~\ref{ce: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{ce: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}