\section{Diagrams with a cluster of four bodies without circles}\label{sec:k4}

In this section we treat diagrams $64$ and $66$. In both, the bodies of $K=\{3,4,5,6\}$ are pairwise joined by $z$-strokes and none is $z$-circled. In diagram $64$ the $w$-strokes form the binary $12$, and in diagram $66$ the triangle $\{1,2,6\}$. In both, $1$ and $2$ are the $w$-circled bodies. As in Section~\ref{sec:ak50}, $m\in(\R_{>0})^6$, $C\subseteq Y_m$ is an irreducible curve, $P$ is a singular place, the results are stated for the labelling of Table~\ref{tab:ten}, and $U\not\equiv0$ is not assumed.

\subsection{Leading systems}

\begin{lemma}\label{lem:d64type}
At a singular place with diagram $64$:
\begin{enumerate}[label=\textup{(\alph*)}]
\item on the six pairs of $K$, $\zeta=3x-2$ and $\omega=2-x$ with a common $x$, $0<x<1$;
\item $\zeta_{12}=2$, $\omega_{12}=-2$, $x_{12}=0$;
\item $\omega_{jk}=-2$ and $\zeta_{jk}=a_k<2$ for $j\in\{1,2\}$, $k\in K$;
\item $b_1=b_2=-2$, $a_1,a_2\ge2$, and for $k\in K$, $b_k>-2$ and $3x-2\le a_k<2$, with equality for at least three $k\in K$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) By (E3) and the remark after Lemma~\ref{lem:rules}, $\zeta$, $\omega$ and $x$ are equal on the six pairs, and (O3), (O4) give $\zeta=3x-2$, $\omega=2-x$, $0\le x<1$. If $x=0$, then $\zeta_{kl}=-2$, so $k$ or $l$ would be $z$-circled. (b) is (F1) with the colours exchanged. (c), (d) The bodies $1$, $2$ are the $w$-circled ones, so $\omega_{jk}=-2$ by (O6). Since $jk$ is not a $w$-stroke, $\zeta_{jk}<2$ by (O6) and (O1). In the $z$-equation of $j$ the term $Z_{12}$ has exponent $2$ and the terms $Z_{jk}$ have exponent $3-\frac12\zeta_{jk}>2$, so $a_j\ge2$, and then $a_k\le\zeta_{jk}<2\le a_j$ gives $\zeta_{jk}=a_k$. Let $a_*=\min_Ka_k$. If $a_*<3x-2$, all bodies of $K$ have exponent $a_*$ and a common leading coefficient $c\ne0$, since $\zeta_{kl}=3x-2>a_*$ on $K$, and the coefficient of exponent $a_*$ in $\sum_km_kz_k=0$ is $c\sum_{k\in K}m_k\ne0$. So $a_k\ge3x-2$, and two bodies with $a_k,a_l>3x-2$ would give $\zeta_{kl}>3x-2$.
\end{proof}

\begin{lemma}\label{lem:d64centre}
At a singular place with diagram $64$, $a_k=3x-2$ for all $k\in K$.
\end{lemma}

\begin{proof}
Suppose $a_{k_0}>3x-2$ for some $k_0\in K$. By Lemma~\ref{lem:d64type}(d) it is the only such body. For $j\in\{1,2\}$ and $k\in K$, (F4) applies to the pair $(k,j)$, since $a_k<a_j$, $b_j<b_k$ and $a_j+b_k>a_k+b_j$. In the torque $T_{\{1,2\}}$ the term of the pair $jk$ therefore has exponent $1-\frac12a_k$ and leading coefficient $m_jm_k\lc{p_{jk}}{}$. The least exponent, $1-\frac12a_{k_0}$, is attained only by the pairs $(j,k_0)$, and its coefficient is $m_{k_0}(m_1\lc{p_{1k_0}}{}+m_2\lc{p_{2k_0}}{})$. This is not zero by Lemma~\ref{lem:binary} with the colours exchanged and the binary $\{1,2\}$ in place of $\{5,6\}$.
\end{proof}

For four masses $\mu=(\mu_1,\dots,\mu_4)$, let $E_4(\mu)$ be the system
\[
\sum_{l\ne k}\mu_lP_{kl}^3(\xi_k-\xi_l)=0,\qquad\sum_{l\ne k}\mu_lP_{kl}^3(\eta_k-\eta_l)=0\quad(k=1,\dots,4),\qquad P_{kl}^2(\xi_k-\xi_l)(\eta_k-\eta_l)=1,
\]
in $\xi,\eta\in\C^4$ and $P_{kl}$. Its solutions are complex equilibria of four bodies in both colours.

\begin{lemma}\label{lem:d64lead}
At a singular place with diagram $64$, put $\xi_k=\lc{z_k}{3x-2}$, $\eta_{kl}=\lc{w_{kl}}{2-x}$ and $P_{kl}=\lc{p_{kl}}{-x}$ for $k,l\in K$. Then $\eta_{kl}=\eta_k-\eta_l$ for some $\eta\in\C^K$, all $\xi_k-\xi_l$, $\eta_{kl}$ and $P_{kl}$ are nonzero, and $(\xi,\eta,P)$ solves $E_4(m_K)$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:d64centre}, $\zeta_{kl}=3x-2$ exactly, so $\xi_k-\xi_l=\lc{z_{kl}}{3x-2}\ne0$. Also $\omega_{kl}=2-x$ exactly, so $\eta_{kl}\ne0$, and $\eta_{kl}+\eta_{lj}=\eta_{kj}$. The distance equations at exponent $0$ give $P_{kl}^2(\xi_k-\xi_l)\eta_{kl}=1$. In the $z$-equation of $k\in K$, the terms $Z_{kl}$, $l\in K$, have exponent $-2$ and coefficient $P_{kl}^3(\xi_k-\xi_l)$, the terms $Z_{kj}$, $j=1,2$, have exponent $4-\frac32x>-2$, and $a_k>-2$. At exponent $-2$ this gives $\Phi^\xi_k:=\sum_lm_lP_{kl}^3(\xi_k-\xi_l)=0$. In the $w$-equation of $k$, the terms $W_{kl}$, $l\in K$, have exponent $2-4x$ and coefficient $P_{kl}^3\eta_{kl}$, and the terms $W_{kj}$ have exponent $(2-4x)+(2-\frac12x)$. So $\lc{w_k}{2-4x}=\Phi^\eta_k:=\sum_lm_lP_{kl}^3\eta_{kl}$. If $b_k>2-4x$ for all $k\in K$, then $\Phi^\eta_k=0$. Otherwise some $b_k\le2-4x<2-x=\omega_{kl}$, so all bodies of $K$ have the same exponent $b_k$ and the same leading coefficient $c'\ne0$. If $b_k<2-4x$, the $w$-equation at exponent $b_k$ gives $c'=0$. If $b_k=2-4x$, then $\Phi^\eta_l=c'$ for all $l\in K$, while $\sum_{k\in K}m_k\Phi^\eta_k=0$ by antisymmetry, so $c'\sum_{k\in K}m_k=0$. Both are impossible.
\end{proof}

The leading system uses both colours, each at its own order: $z$ at exponent $-2$, $w$ at exponent $2-4x$, the distances at $0$. A count of equations and unknowns suggests that $E_4$ is overdetermined by one equation. By the two momentum identities and the torque identity, at most five of its eight force equations are independent, so it has at most eleven independent equations. Its $14$ unknowns modulo the symmetries $(\xi,\eta,P)\mapsto(\lambda s\xi,\lambda s^{-1}\eta,\lambda^{-1}P)$ and the translations of $\xi$ and $\eta$ leave ten. We will show in Proposition~\ref{prop:E4} that $E_4$ has no solution at generic masses.

\begin{lemma}\label{lem:d66type}
At a singular place with diagram $66$, let $S=\{3,4,5\}$. Then \textup{(a)} of Lemma~\ref{lem:d64type} holds. On $12$, $16$, $26$, $\omega=-2$, $\zeta=2$ and $x=0$. For $j\in\{1,2\}$ and $k\in S$, $\omega_{jk}=-2$ and $\zeta_{jk}=a_k=3x-2$. Moreover $b_1=b_2=-2$, $b_k>-2$ for $k\in K$, and $a_1,a_2,a_6\ge2$. Hence $\lc{z_6}{3x-2}=0$ and $\sum_{k\in S}m_k\lc{z_k}{3x-2}=0$.
\end{lemma}

\begin{proof}
(a) as before. On the $w$-triangle $\{1,2,6\}$, (F1) with the colours exchanged gives $x=0$, $\omega=-2$, $\zeta=2$. As in Lemma~\ref{lem:d64type}, $\omega_{jk}=-2$, $\zeta_{jk}<2$ and $a_j\ge2$ for $j=1,2$, $k\in S$, and $\zeta_{jk}=a_k$. By (F3), $a_6\ge2$. For $k\in S$, $\zeta_{k6}=3x-2$ and $a_k<2\le a_6$, so $a_k=3x-2$. At exponent $3x-2$ only the bodies of $S$ appear in $\sum_km_kz_k=0$.
\end{proof}

\begin{lemma}\label{lem:d66lead}
At a singular place with diagram $66$, put $\xi_k=\lc{z_k}{3x-2}$ ($\xi_6=0$), $\eta_{kl}=\lc{w_{kl}}{2-x}=\eta_k-\eta_l$ and $P_{kl}=\lc{p_{kl}}{-x}$ for $k,l\in K$. Then the differences $\xi_k-\xi_l$, the $\eta_{kl}$ and the $P_{kl}$ are nonzero, and:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $P_{kl}^2(\xi_k-\xi_l)\eta_{kl}=1$ for the six pairs of $K$;
\item $\Phi^\xi_k=0$ for all $k\in K$;
\item $\Phi^\eta_k=c$ for $k\in S$, with $c\in\C$ independent of $k$, possibly $0$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) as in Lemma~\ref{lem:d64lead}. (ii) For $k\in S$ as there. For $k=6$ the terms $Z_{6j}$, $j=1,2$, have exponent $2$ and $a_6\ge2$, so the coefficient of exponent $-2$ is $\Phi^\xi_6=0$. (iii) For $k\in S$, $\lc{w_k}{2-4x}=\Phi^\eta_k$. If $b_k>2-4x$ for all $k\in S$, then $c=0$. Otherwise, as in Lemma~\ref{lem:d64lead}, all bodies of $K$ have exponent $b_k=2-4x$ and a common coefficient $c'$, and $\Phi^\eta_k=c'$ for $k\in S$. The $w$-equation of body $6$ at exponent $2-4x$ contains subleading terms of $W_{61}$, $W_{62}$ and is not used.
\end{proof}

Renumber $K$ as $1,2,3,4$ with the centre $6$ last, and let $\mu=(m_3,m_4,m_5,m_6)$. We call $C_4(\mu)$ the system of (i), (ii), (iii), $\xi_4=0$ and $\sum_{k\le3}\mu_k\xi_k=0$, in the unknowns $\xi,\eta,P,c$: a cluster of four bodies with body $4$ at its centre, in equilibrium in the colour $z$ and with the same force $c$ on the three other bodies in the colour $w$. If $c=0$, then $\Phi^\eta_4=0$ as well, by antisymmetry, and the solution solves $E_4(\mu)$.

\subsection{The four-body equilibria}\label{ss:E4}
The system $E_4$ is the leading system of the fully edged four-body diagram of \cite[\S5.5]{AK12}. In that section Albouy and Kaloshin call its mass condition complicated and do not discuss it. Aquino-Michaels \cite[Lemmas~5.3--5.5 and Theorem~5.6]{AM26} proves that the fully balanced four-body core of diagrams $63$, $65$ and $67$, which is the system $E_4$, is confined to a proper algebraic subset of masses. His reduction is of the same kind as Lemma~\ref{lem:E4} below: a rank-one equilibrium stress in the non-collinear case, where he then uses the Cayley--Menger relation, and sign classes in the collinear case. For diagrams $64$ and $66$, whose cores he describes as not fully balanced, he confines the collinear case and leaves the non-collinear case open \cite[Rem.~5.7]{AM26}, and he asks for centred next-scale models of them \cite[Problem~8.2]{AM26}. Lemmas~\ref{lem:d64lead} and~\ref{lem:d66lead} give their leading systems. We give the reduction in the form we use.

\begin{lemma}\label{lem:E4}
For each of the $3^6$ symmetric patterns $\varepsilon=(\varepsilon_{kl})$ of cube roots of unity, the powers of $\mathrm{j}=e^{2\pi i/3}$, and each of the $2^6$ antisymmetric patterns $\sigma=(\sigma_{kl})$ of signs, let $A_k=\sum_{l\ne k}\varepsilon_{kl}\mu_lq_l$ and
\[
\mathcal F_\varepsilon=\Big\{q_j^2A_k-q_k^2A_j\ (j<k),\ \sum_k\mu_kq_k^3\Big\},\qquad \mathcal G_\sigma=\Big\{\sum_{l\ne k}\sigma_{kl}\mu_l\prod_{j\ne k,l}(\xi_k-\xi_j)^2\ (k=1,\dots,4)\Big\},
\]
seven cubic forms in $q\in\mathbb P^3$ and four quartic forms in $(\xi_2:\xi_3:\xi_4)\in\mathbb P^2$ with $\xi_1=0$. If $\mu_1\mu_2\mu_3\mu_4\ne0$ and $E_4(\mu)$ has a solution, then for some pattern the forms $\mathcal F_\varepsilon$ or $\mathcal G_\sigma$ have a common zero.
\end{lemma}

\begin{proof}
Let $L$ be the matrix with $L_{kl}=\mu_lP_{kl}^3$ for $k\ne l$ and zero row sums. The equations of $E_4$ say $L\xi=L\eta=0$, and $L\mathbf1=0$. The span $V$ of $\mathbf1,\xi,\eta$ has dimension at least $2$, since $\xi$ is not constant.

If $\dim V=3$, then $\rank L\le1$, and $L\ne0$. The matrix $L\,\mathrm{diag}(\mu)^{-1}$ is symmetric of rank one, so $P_{kl}^3=u_ku_l$ for $k\ne l$ with all $u_k\ne0$. Then $L\mathbf1=0$, $L\xi=0$ and $L\eta=0$ give $\sum\alpha_k=\sum\alpha_k\xi_k=\sum\alpha_k\eta_k=0$ with $\alpha_k=\mu_ku_k$. For $d_{kl}=(\xi_k-\xi_l)(\eta_k-\eta_l)=P_{kl}^{-2}$ this gives $\sum_l\alpha_ld_{kl}=\sum_l\alpha_l\xi_l\eta_l=:-\beta$ for each $k$. With $q_k^3=u_k$ we have $P_{kl}=\varepsilon_{kl}q_kq_l$ for cube roots of unity $\varepsilon_{kl}$, and $d_{kl}=\varepsilon_{kl}q_k^{-2}q_l^{-2}$. So $A_k=-\beta q_k^2$ for all $k$, and $\sum_k\mu_kq_k^3=\sum_k\alpha_k=0$: $q$ is a common zero of $\mathcal F_\varepsilon$.

If $\dim V=2$, then $\eta=\lambda\xi+\theta\mathbf1$ with $\lambda\ne0$, so $P_{kl}(\xi_k-\xi_l)=\sigma_{kl}\kappa$ with $\kappa^2=1/\lambda$ and $\sigma$ antisymmetric. The $\xi$-equations become $\kappa^3\sum_{l\ne k}\sigma_{kl}\mu_l(\xi_k-\xi_l)^{-2}=0$. After a translation $\xi_1=0$ and multiplication by $\prod_{j\ne k}(\xi_k-\xi_j)^2$, the point $(\xi_2:\xi_3:\xi_4)$ is a common zero of $\mathcal G_\sigma$.
\end{proof}

\begin{lemma}\cert\label{lem:E4cert}
At $\mu=(1,1,1,1)$ no $\mathcal F_\varepsilon$ has a common zero in $\mathbb P^3$, and no $\mathcal G_\sigma$ has a common zero in $\mathbb P^2$.
\end{lemma}

\begin{proof}
\emph{The forms $\mathcal F_\varepsilon$.} At equal masses $A=Eq$, where $E=(\varepsilon_{kl})$ has zero diagonal. At a common zero $q$ of $\mathcal F_\varepsilon$ the vectors $A$ and $q\circ q=(q_k^2)_k\ne0$ are proportional, so $A=-\beta\,q\circ q$ for some $\beta\in\C$, and $\sum_kq_k^3=0$. If $\beta=0$, then $Eq=0$. If $\beta\ne0$, then $p=-\beta q\ne0$ satisfies $Ep=p\circ p$ and $\sum_kp_k^3=0$. Write $\varepsilon_{kl}=\mathrm j^{e_{kl}}$. The substitution $q_l\mapsto\mathrm j^{c_l}q_l$ maps the common zeros of $\mathcal F_\varepsilon$ to those of $\mathcal F_{\varepsilon'}$ with $e'_{kl}=e_{kl}-c_k-c_l$. These substitutions act freely on the $729$ patterns, so they have $9$ orbits. The triple $(e_{12}+e_{34},e_{13}+e_{24},e_{14}+e_{23})$ modulo $(1,1,1)$ is constant on the orbits and takes $9$ values, so it separates them. Relabelling the bodies permutes its entries, and complex conjugation, which maps common zeros to common zeros, negates them. So it suffices to treat three patterns, with the triples $(0,0,0)$, $(0,0,1)$ and $(0,1,2)$:
\begin{enumerate}[label=\textup{(\Roman*)}]
\item all $e_{kl}=0$. Then $E=J-I$ is invertible. If $Ep=p\circ p$, then $p_k^2+p_k=S:=\sum_lp_l$ for all $k$, so each $p_k$ is a root $r_1$ or $r_2$ of $r^2+r-S$, and $r_1+r_2=-1$. If all $p_k$ equal $r$, then $r(r-3)=0$ and $r\ne0$, so $\sum p_k^3=108$. If three of them equal $r_1$ and one equals $r_2$, then $r_1r_2=-S$ gives $r_1^2-r_1+1=0$ and $\sum p_k^3=3r_1^3+r_2^3=-3\mp3\sqrt3\,i$. If two equal $r_1$ and two equal $r_2$, then $S=-2$, $r_1r_2=2$ and $\sum p_k^3=2(r_1^3+r_2^3)=10$.
\item $e_{14}=1$ and all other $e_{kl}=0$. Then $E$ is invertible, and $Ep=p\circ p$ reads $p_2+p_3+\mathrm jp_4=p_1^2$, $p_1+p_3+p_4=p_2^2$, $p_1+p_2+p_4=p_3^2$, $\mathrm jp_1+p_2+p_3=p_4^2$. Subtracting gives $(p_2-p_3)(p_2+p_3+1)=0$ and $(p_1-p_4)(p_1+p_4+\mathrm j)=0$. If $p_2=p_3=a$ and $p_1=p_4=b$, then $a^3+b^3=0$, and $b\ne0$ since $p\ne0$, so $a=-\zeta b$ with $\zeta^3=1$. The second and the first equation give $b=(2-\zeta)\zeta$ and $b=\mathrm j-2\zeta$, so $\zeta^2-4\zeta+\mathrm j=0$, which fails for $\zeta=1,\mathrm j,\mathrm j^2$. If $p_2=p_3=a$ and $p_1+p_4=-\mathrm j$, then $a^2-a+\mathrm j=0$ and $p_1p_4=\mathrm j^2-2a$, so $\sum p_k^3=2a^3+2-6\mathrm ja$, which equals $2a(1-4\mathrm j)+2-2\mathrm j$ because $a^2=a-\mathrm j$, and vanishes only for $a=(\mathrm j-1)/(1-4\mathrm j)$. For this value $a^2-a+\mathrm j=(21-3\mathrm j)/(1-4\mathrm j)^2\ne0$. The case $p_2+p_3=-1$, $p_1=p_4=b$ is the same with $b=(\mathrm j-1)/(\mathrm j^2-4)$ and $b^2-\mathrm jb+1=(21-3\mathrm j)/(\mathrm j^2-4)^2\ne0$. If $p_2+p_3=-1$ and $p_1+p_4=-\mathrm j$, then $p_2p_3=-\mathrm j^2$, $p_1p_4=-\mathrm j$ and $\sum p_k^3=4+6\mathrm j\ne0$.
\item $e_{12}=e_{34}=0$, $e_{13}=e_{24}=2$, $e_{14}=e_{23}=1$. Then $E$ has the eigenvectors $\chi_0=(1,1,1,1)$, $\chi_1=(1,1,-1,-1)$, $\chi_2=(1,-1,1,-1)$, $\chi_3=(1,-1,-1,1)$, with eigenvalues $0$, $2$, $\lambda_2=-2-2\mathrm j$, $\lambda_3=2\mathrm j$. If $Eq=0$, then $q\in\C\chi_0$ and $\sum q_k^3=4q_1^3$, so $q=0$. Write $p=\frac14\sum_is_i\chi_i$. Since $\chi_i\circ\chi_k=\chi_{i\oplus k}$ for the group law $\oplus$ of $\{0,1,2,3\}\cong(\Z/2)^2$, the equation $Ep=p\circ p$ reads
\[
s_0^2+s_1^2+s_2^2+s_3^2=0,\quad s_1(4-s_0)=s_2s_3,\quad s_2(2\lambda_2-s_0)=s_1s_3,\quad s_3(2\lambda_3-s_0)=s_1s_2,
\]
and then $\sum p_k^3=\frac18(3s_1s_2s_3-s_0^3)$. If $s_1s_2s_3\ne0$, the last three equations give $s_1^2=(2\lambda_2-s_0)(2\lambda_3-s_0)$, $s_2^2=(4-s_0)(2\lambda_3-s_0)$ and $s_3^2=(4-s_0)(2\lambda_2-s_0)$. Since $4+2\lambda_2+2\lambda_3=0$ and $8\lambda_2+8\lambda_3+4\lambda_2\lambda_3=0$, the first equation becomes $4s_0^2=0$, and then $\sum p_k^3=\frac38s_1s_2s_3\ne0$. If $s_1s_2s_3=0$, then $\sum p_k^3=-\frac18s_0^3$, so a solution has $s_0=0$. Then the last three equations show that two of $s_1,s_2,s_3$ vanish, and the first that the third vanishes too, so $p=0$.
\end{enumerate}
In each case $Eq=0$ and $\sum_kq_k^3=0$ only for $q=0$, and $\sum p_k^3\ne0$ unless $p=0$, so $\mathcal F_\varepsilon$ has no common zero.

\emph{The forms $\mathcal G_\sigma$.} Let $(\xi_2:\xi_3:\xi_4)$ be a common zero, with $\xi_1=0$. Not all four $\xi_k$ coincide. If exactly two coincide, $\xi_a=\xi_b$, every term of the form for $a$ except the one with $l=b$ contains $(\xi_a-\xi_b)^2$, so this form equals $\sigma_{ab}\prod_{j\ne a,b}(\xi_a-\xi_j)^2\ne0$. The same holds if two pairs coincide. If $\xi_a=\xi_b=\xi_c\ne\xi_d$, the form for $d$ equals $(\xi_d-\xi_a)^4(\sigma_{da}+\sigma_{db}+\sigma_{dc})\ne0$. So the $\xi_k$ are distinct, and $\sum_{l\ne k}\sigma_{kl}(\xi_k-\xi_l)^{-2}=0$ for all $k$. These equations are invariant under relabelling and under $\sigma\mapsto-\sigma$. We read $\sigma$ as a tournament, with $k\to l$ when $\sigma_{kl}=1$. A tournament on four vertices is determined up to isomorphism by its score sequence, $(3,2,1,0)$, $(3,1,1,1)$, $(2,2,2,0)$ or $(2,2,1,1)$, and reversal exchanges the second and the third. So it suffices to treat the transitive tournament $\mathcal T_1$, the tournament $\mathcal T_2$ with a source and a $3$-cycle, and the strongly connected tournament $\mathcal T_3$, with $\sigma_{kl}=1$ for $k<l$ except $\sigma_{24}=-1$ in $\mathcal T_2$ and $\sigma_{14}=-1$ in $\mathcal T_3$. After a translation and a scaling, $\xi=(0,1,t,s)$. Let $e_k$ be the numerator of $\sum_{l\ne k}\sigma_{kl}(\xi_k-\xi_l)^{-2}$, and let $R_1=\Res_s(e_1,e_2)$ and $R_2=\Res_s(e_1,e_3)$, which vanish at the $t$ of a solution. In all three cases $R_2=t^4R_2'$ with
\begin{align*}
\mathcal T_1:\quad R_1&=5t^8-20t^7+26t^6-20t^5+19t^4-4t^3+2t^2+1,\\
\mathcal T_2:\quad R_1&=5t^8-20t^7+30t^6-36t^5+43t^4-28t^3+22t^2-8t+1,\\
\mathcal T_1,\mathcal T_2:\quad R_2'&=t^8+14t^6-28t^5+47t^4-52t^3+42t^2-20t+5,\\
\mathcal T_3:\quad R_1&=-3t^8+12t^7-18t^6+20t^5-17t^4+4t^3+2t^2+1,\\
\mathcal T_3:\quad R_2'&=t^8-10t^6+28t^5-37t^4+36t^3-26t^2+12t-3.
\end{align*}
Here $R_1(0)=1$, and $\Res_t(R_1,R_2')$ is $2^{16}3^45^47^2\cdot3881$, $2^{16}3^85^{10}$ and $-2^{16}5^4\cdot23\cdot433$ for $\mathcal T_1$, $\mathcal T_2$ and $\mathcal T_3$. Thus $R_1$ and $R_2$ have no common zero, so there is no solution.
\end{proof}

\begin{proposition}\label{prop:E4}
Let $B_E\subset\C^4$ be the union of $\{\mu_1\mu_2\mu_3\mu_4=0\}$ and of the set of $\mu$ for which some $\mathcal F_\varepsilon$ or some $\mathcal G_\sigma$ has a common projective zero. Then $B_E$ is a proper Zariski-closed subset of $\C^4$, defined over $\Q$, and $E_4(\mu)$ has no solution for $\mu\notin B_E$.
\end{proposition}

\begin{proof}
The second set is the projection of a closed subset of $\C^4\times\mathbb P^3$, resp.\ $\C^4\times\mathbb P^2$, so it is closed. It is a union over the patterns of sets defined over $\Q(\mathrm{j})$, since the forms have coefficients in $\Z[\mathrm{j}][\mu]$, and the union is stable under $\mathrm{j}\mapsto\mathrm{j}^2$, which permutes the patterns. Therefore $B_E$ is defined over $\Q$. By Lemma~\ref{lem:E4cert}, $(1,1,1,1)\notin B_E$. The last statement is Lemma~\ref{lem:E4}.
\end{proof}

We have also checked by computation that a second mass point lies outside $B_E$. Let $\mu_0=(3,5,7,11)$ and $p=2147483629$. For each of the $729$ patterns $\varepsilon$, the Macaulay matrix of $\mathcal F_\varepsilon$ in degree $7$, with entries in $\Z[\mathrm j]$ reduced by $\mathrm j\mapsto882503254$ (a root of $t^2+t+1$ modulo $p$), has full rank $120$ modulo $p$, and for each of the $64$ patterns $\sigma$ the Macaulay matrix of $\mathcal G_\sigma$ in degree $8$ has full rank $45$ modulo $p$. A nonzero maximal minor modulo $p$ is a nonzero element of $\Z[\mathrm j]$, since $\mathrm j\mapsto882503254$ defines a ring map $\Z[\mathrm j]\to\F_p$. Thus these Macaulay matrices have full rank over $\Q(\mathrm j)$, every monomial of degree $7$ (resp.\ $8$) lies in the ideal of the forms, and $\mu_0\notin B_E$. The ranks are computed with python-flint, and Gr\"obner bases over $\Q(\mathrm j)$ and over $\Q$ in Singular, of dimension zero, give the same conclusion.\cert{}

\subsection{The system with a centre}
Eliminate $\eta$ from $C_4(\mu)$ by $\eta_{kl}=1/(P_{kl}^2(\xi_k-\xi_l))$, and write the solutions of $\xi_4=0$, $\sum_{k\le3}\mu_k\xi_k=0$ as $\xi=(\mu_3s,\mu_3t,-\mu_1s-\mu_2t,0)$, which is possible when $\mu_3\ne0$. With $a_{kl}=P_{kl}^2(\xi_k-\xi_l)$, the equations become
\begin{align*}
&\textstyle\sum_{l\ne k}\mu_lP_{kl}^3(\xi_k-\xi_l)=0&&(k=1,\dots,4),\\
&\textstyle\sum_{l\ne k}\mu_lP_{kl}\prod_{j\ne k,l}(\xi_k-\xi_j)-c\prod_{l\ne k}(\xi_k-\xi_l)=0&&(k=1,2,3),\\
&a_{ln}a_{nk}+a_{kl}a_{nk}+a_{kl}a_{ln}=0&&(\text{the four triangles }\{k,l,n\}).
\end{align*}
The second line is (iii) multiplied by $\prod_{l\ne k}(\xi_k-\xi_l)$, and the third is the relation $\eta_{kl}+\eta_{ln}+\eta_{nk}=0$ multiplied by $a_{kl}a_{ln}a_{nk}$. The equations are bihomogeneous of bidegrees $(1,3)$, $(2,1)$ and $(2,4)$ for the grading $\deg s=\deg t=(1,0)$, $\deg P_{kl}=(0,1)$, $\deg c=(-1,1)$. Let $X$ be the quotient of $(\C^2\setminus0)\times(\C^7\setminus0)$, with coordinates $(s,t)$ and $(P_{12},\dots,P_{34},c)$, by the action $(\lambda_1,\lambda_2)\cdot(s,t,P,c)=(\lambda_1s,\lambda_1t,\lambda_2P,\lambda_1^{-1}\lambda_2c)$ of $(\C^*)^2$. It is a $\mathbb P^6$-bundle over $\mathbb P^1$, hence projective. Let $Z(\mu)\subseteq X$ be the zero set of the equations. A solution of $C_4(\mu)$ with $\mu_3\ne0$ gives a point of $Z(\mu)$.

\begin{lemma}\cert\label{lem:C4cert}
Let $\mu_0=(3,5,7,11)$. On each of the $14$ affine charts $\{s=1\text{ or }t=1\}\times\{\text{one of }P_{12},\dots,P_{34},c\text{ equal to }1\}$, which cover $X$, the equations at $\mu_0$ generate the unit ideal over $\F_{32003}$ and over $\F_{1073741789}$.
\end{lemma}

The first prime is used by Singular and the second by msolve, from two programs that build the equations and produce identical polynomials.

\begin{lemma}\label{lem:transfer}
Let $\mu\in\Z^4$. If the equations of $Z(\mu)$ generate the unit ideal over $\F_p$ on the $14$ charts, then $Z(\mu)=\emptyset$.
\end{lemma}

\begin{proof}
Let $I_p\subseteq\F_p[s,t,P,c]$ be the ideal of the equations at $\mu$ and $B=(s,t)\cdot(P,c)$. Every point of $\bar\F_p^9$ outside the zero set of $B$ maps to a point of $X(\bar\F_p)$ in one of the charts, so the zero set of $I_p$ lies in that of $B$, and $B^N\subseteq I_p$ for some $N$ by the Nullstellensatz. A monomial of bidegree $(a,b)$ has $(s,t)$-degree $a+\deg_c\ge a$ and $(P,c)$-degree $b$, so for $a,b\ge N$ it lies in $B^N$. Hence the Macaulay matrix of the equations in bidegree $(N,N)$, an integer matrix with finitely many columns, has full rank modulo $p$, hence over $\Q$, and every monomial of bidegree $(N,N)$ lies in the ideal of the equations at $\mu$. At a point of $X$ one of $s,t$, say $s$, is nonzero, and either some $P_{kl}\ne0$, in which case $s^NP_{kl}^N\ne0$, or $c\ne0$, in which case $s^{2N}c^N\ne0$, and both monomials have bidegree $(N,N)$. Therefore $Z(\mu)=\emptyset$.
\end{proof}

\begin{proposition}\label{prop:C4}
Let $B_C\subset\C^4$ be the union of $\{\mu_3=0\}$ and of the set of $\mu$ with $Z(\mu)\ne\emptyset$. Then $B_C$ is a proper Zariski-closed subset of $\C^4$, defined over $\Q$, and $C_4(\mu)$ has no solution for $\mu\notin B_C$.
\end{proposition}

\begin{proof}
Since $X$ is projective, the projection to $\C^4$ of the closed set $\{(\mu,y):y\in Z(\mu)\}$ is closed. The equations have coefficients in $\Z[\mu]$, so this projection, and hence $B_C$, is defined over $\Q$. By Lemmas~\ref{lem:C4cert} and~\ref{lem:transfer}, $\mu_0\notin B_C$. A solution of $C_4(\mu)$ with $\mu_3\ne0$ gives a point of $Z(\mu)$.
\end{proof}

\subsection{The diagrams}\label{ss:k4diagrams}

\begin{theorem}\label{thm:d64}
If a singular place has diagram $64$ with $w$-binary $\{a,b\}$ and $z$-cluster $K=\{1,\dots,6\}\setminus\{a,b\}$, then $m_K\in B_E$.
\end{theorem}

\begin{proof}
By Lemmas~\ref{lem:d64centre} and~\ref{lem:d64lead}, $E_4(m_K)$ has a solution. Apply Proposition~\ref{prop:E4}.
\end{proof}

\begin{theorem}\label{thm:d66}
If a singular place has diagram $66$ with $z$-cluster $K$, centre $o\in K$ and $w$-triangle $\{a,b,o\}$, then $(m_{K\setminus\{o\}},m_o)\in B_C$, where $m_{K\setminus\{o\}}$ lists the masses in increasing order of the labels.
\end{theorem}

\begin{proof}
By Lemmas~\ref{lem:d66type} and~\ref{lem:d66lead}, $C_4(m_{K\setminus\{o\}},m_o)$ has a solution. Apply Proposition~\ref{prop:C4}.
\end{proof}

Both theorems hold for all labellings. By Lemma~\ref{lem:E4cert}, $(1,1,1,1)\notin B_E$. The computation of Lemma~\ref{lem:C4cert} gives the same result at $\mu=(1,1,1,1)$, so $(1,1,1,1)\notin B_C$ by Lemma~\ref{lem:transfer}.\cert{} Theorems~\ref{thm:d64} and~\ref{thm:d66} therefore exclude diagrams $64$ and $66$ at six equal masses. We leave open whether $E_4$ or $C_4$ has solutions at positive masses.

\begin{remark}\label{rem:BEsums}
The set $B_E$ contains every $\mu$ in which one mass is the sum of two others. Let $\mu_k=\mu_i+\mu_j$ with $i,j,k$ distinct, let $l$ be the fourth index, and take $\xi_i=\xi_j=\xi_k$ and $\xi_l=\xi_i+1$, translated so that $\xi_1=0$. Each term of the form of $\mathcal G_\sigma$ for one of the bodies $i,j,k$ contains the square of the difference of two coinciding points, so these three forms vanish, and the form for $l$ is $\sigma_{li}\mu_i+\sigma_{lj}\mu_j+\sigma_{lk}\mu_k$, which vanishes when $\sigma_{li}=\sigma_{lj}=-\sigma_{lk}$. Hence the polynomial $F$ of Section~\ref{sec:proof} vanishes at every $m\in(\R_{>0})^6$ in which one mass is the sum of two others. These common zeros have coinciding points and do not come from solutions of $E_4$. At $\mu=(1,1,1,2)$, $(5,2,11,3)$, $(1,2,3,4)$ and $(2,3,5,7)$ the construction applies to exactly $36$, $16$, $28$ and $28$ of the $64$ patterns $\sigma$. An exact computation over $\Q$ finds common zeros of $\mathcal G_\sigma$ for exactly these numbers of patterns, but none of them survives the saturation by the product of the differences $\xi_k-\xi_l$.\cert
\end{remark}