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\section{Point certificates for the clusters of zero energy}\label{sec:points}

For a body $j$ write $Q_j=\{1,\dots,5\}\setminus\{j\}$. By Theorem~\ref{thm:Kgating}, the pieces $\pc{d14a}(j)$ and $\pc{d14b}(j,k)$ are attained at $m$ only if $\mathrm{K14}(m_{Q_j})$ has a solution, and $\pc{d15}(j,k)$ only if $\mathrm{K15}(m_{Q_j},k)$ has one. In \S\ref{sec:curves} the separation of \S\ref{sec:lex} succeeds at finitely many mass points, given exactly, only after the pieces of some clusters $Q_j$, usually one, are removed. At these points we show that $\mathrm{K14}(m_{Q_j})$ has no solution. By \S\ref{ss:certs}, the proof needs this only for the pieces $\pc{d14a}(j)$.

All these certificates have the same structure. We reduce the cluster system modulo a prime $\ell$ at an $\ell$-adic point over the given masses. A standard basis over $\mathbb F_\ell$ shows that neither the reduced system nor any of its initial systems on the cells of its tropical prevariety has a zero in the torus. A valuation argument lifts this to characteristic $0$. The computations over $\mathbb F_\ell$ and the enumeration of the cells are exact. Where a point of \S\ref{sec:curves} is a certified point, we identify the two exactly.

\subsection{The cluster systems in the torus}\label{ss:Ksystems}
Let $Q$ be a set of four bodies, $\mu=(\mu_i)_{i\in Q}$ complex masses, $M_Q=\sum_{i\in Q}\mu_i$, and $k\in Q$. In this section we write $h$ for the second index of a pair of bodies, since $\ell$ denotes a prime. Recall from \eqref{eq:K14} and \eqref{eq:K15} that $\mathrm{K14}(\mu)$ asks for $z,w\in\C^Q$ and $p_{ih}=p_{hi}$ with $p_{ih}^2z_{ih}w_{ih}=1$ and
\[
F^z_i=\sum_{h\in Q\setminus i}\mu_h\,z_{ih}\,p_{ih}^3=0,\qquad F^w_i=\sum_{h\in Q\setminus i}\mu_h\,w_{ih}\,p_{ih}^3=0\qquad(i\in Q),
\]
and that $\mathrm{K15}(\mu,k)$ asks for the same with $F^w_i=c+d\,[i=k]$, $d\ne0$. Whether they have solutions depends only on the ratios of the masses and on $k$, and is equivariant under relabelling (\S\ref{ss:Kgating}).

We number the bodies of $Q$ increasingly as $1,\dots,4$. The unknowns $\xi$ are $z_{ih}$, $w_{ih}$ and $p_{ih}$ for the five pairs $\{i,h\}\ne\{1,2\}$, and for $\mathrm{K15}$ also $c$ and $d$, which gives $N=15$, respectively $N=17$, unknowns. The generators are:
\begin{enumerate}[label=(\roman*)]
\item the seven circuit relations of the complete graph on $Q$ (four triangles and three four-cycles), for $z$ and for $w$, with $z_{12}=w_{12}=1$;
\item $p_{ih}^2z_{ih}w_{ih}-1$ for the five pairs;
\item the eight rows $F^z_i$ and $F^w_i-c-d\,[i=k]$, with $p_{12}=1$, and $c=d=0$ for $\mathrm{K14}$;
\item with $U=\sum_{i<h}\mu_i\mu_hp_{ih}$, $V_{zz}=\sum_{i<h}\mu_i\mu_hz_{ih}^2p_{ih}^3$ and $V_{ww}=\sum_{i<h}\mu_i\mu_hw_{ih}^2p_{ih}^3$: for $\mathrm{K14}$ the three polynomials $U$, $V_{zz}$, $V_{ww}$; for $\mathrm{K15}$ the five polynomials
\[
cM_Q+d\mu_k,\qquad U,\qquad \textstyle\sum_{h\in Q}\mu_hz_{kh},\qquad V_{zz},\qquad V_{ww}+c\sum_{h\in Q}\mu_hw_{kh}.
\]
\end{enumerate}
This gives $30$ generators for $\mathrm{K14}$ and $32$ for $\mathrm{K15}$. Each generator has the form $F_\nu=\sum_{a\in A_\nu}c_{\nu,a}(\mu)\,\xi^a$, where each $c_{\nu,a}$ is a nonzero element of $\Z[\mu]$. Each $c_{\nu,a}$ is an integer multiple of a monomial in $\mu$, except the coefficient $M_Q$ of $c$ in $cM_Q+d\mu_k$. We call $A_\nu$ the \emph{universal support} of $F_\nu$.

\begin{lemma}\label{lem:torus}
If $\mathrm{K14}(\mu)$ holds, the generators of $\mathrm{K14}$ have a common zero in $(\C^*)^{15}$. If $\mu_k\ne0$, $M_Q\ne0$ and $\mathrm{K15}(\mu,k)$ holds, the generators of $\mathrm{K15}$ with this $k$ have a common zero in $(\C^*)^{17}$.
\end{lemma}

\begin{proof}
Replacing $(z,w,p)$ by $(sz,tw,(st)^{-1/2}p)$ multiplies $F^z$ by $s(st)^{-3/2}$ and $F^w$ by $t(st)^{-3/2}$. Replacing $p$ by $-p$ changes the sign of both. For $\mathrm{K15}$ the factor of $F^w$ is absorbed by $(c,d)$. Since $z_{12}w_{12}=p_{12}^{-2}\ne0$, each solution has a representative with $z_{12}=w_{12}=p_{12}=1$. Conditions (i)--(iii) hold there. The circuit relations give positions $z_i,w_i$. From $p_{ih}^2z_{ih}w_{ih}=1$ and antisymmetry we obtain
\[
\textstyle\sum_i\mu_iw_iF^z_i=\sum_i\mu_iz_iF^w_i=U,\qquad\sum_i\mu_iz_iF^z_i=V_{zz},\qquad\sum_i\mu_iw_iF^w_i=V_{ww},\qquad\sum_i\mu_iF^w_i=0.
\]
For $\mathrm{K14}$ the left sides vanish. For $\mathrm{K15}$ the last identity gives $cM_Q+d\mu_k=0$. Then $0=U=\sum_i\mu_iz_i(c+d[i=k])=d\mu_k(z_k-z_G)$ with $z_G=\sum_i\mu_iz_i/M_Q$. So $\sum_h\mu_hz_{kh}=M_Q(z_k-z_G)=0$. Similarly, $V_{ww}=cM_Qw_G+d\mu_kw_k=-c\sum_h\mu_hw_{kh}$. Finally, $p_{ih}^2z_{ih}w_{ih}=1$ makes $z_{ih}$, $w_{ih}$, $p_{ih}$ nonzero, $d\ne0$, and $c=-d\mu_k/M_Q\ne0$.
\end{proof}

\emph{Cells.} For a finite set $A\subset\Z^N$ and $\omega\in\R^N$ let $\max_\omega(A)$ be the set of $a\in A$ at which $\langle a,\omega\rangle$ is largest, and for $F=\sum_{a\in A}c_a\xi^a$ let $\mathrm{in}_\omega F=\sum_{a\in\max_\omega(A)}c_a\xi^a$. The \emph{prevariety} of the universal supports is the set of $\omega$ with $|\max_\omega(A_\nu)|\ge2$ for every $\nu$. Its \emph{cells} are the classes of its nonzero points with the same tuple of max-sets. On a cell $\sigma$ the initial forms $\mathrm{in}_\sigma F_\nu$ are well defined. Since the prevariety depends only on the supports, we compute it once for each of the five systems ($\mathrm{K14}$, and $\mathrm{K15}$ with $k=1,\dots,4$). We enumerate the cells by a search in which each linear-programming decision is proved in integer arithmetic, by an integer point or by nonnegative integer Gordan multipliers (the numbers of decisions are in \S\ref{ss:checks}). We compare the result with the prevariety computed by gfan~0.7: the two agree, and the lineality space is $0$. The prevariety of $\mathrm{K14}$ has $28$ cells, spanned by $16$ rays. That of each system $\mathrm{K15}$ has $43$ cells, spanned by $18$ rays.

\subsection{Algebraic mass points}\label{ss:KA}
Let $f\in\Z[t]$ be irreducible over $\Q$, and let $g=(g_i)_{i\in Q}$ with $g_i\in\Q[t]$. For a root $\theta$ of $f$ put $\mu(\theta)=(g_i(\theta))_{i\in Q}$. For a prime $\ell$ and $\bar\theta\in\mathbb F_\ell$ put $\bar\mu_i=g_i(\bar\theta)$ and $\bar F_\nu=\sum_ac_{\nu,a}(\bar\mu)\xi^a\in\mathbb F_\ell[\xi]$.

\begin{theorem}\label{thm:KA}
Let $\ell$ be a prime and $\bar\theta\in\mathbb F_\ell$ such that:
\begin{enumerate}[label=(P\arabic*)]
\item $\ell$ divides neither the leading coefficient of $f$ nor a denominator of a coefficient of a $g_i$; $\bar f=f\bmod\ell$ is squarefree of degree $\deg f$, and $\bar f(\bar\theta)=0$;
\item $\bar\mu_i\ne0$ for $i\in Q$, $\sum_i\bar\mu_i\ne0$, and $c_{\nu,a}(\bar\mu)\ne0$ for every generator and every $a\in A_\nu$;
\end{enumerate}
\begin{enumerate}[label=(\Alph*)]
\item the ideal of $\mathbb F_\ell[\xi_0,\xi]$ generated by $\bar F_1,\dots,\bar F_N$ and $\xi_0\xi_1\cdots\xi_N-1$ is the unit ideal;
\item for every cell $\sigma$, the ideal generated by $\mathrm{in}_\sigma\bar F_1,\dots,\mathrm{in}_\sigma\bar F_N$ and $\xi_0\xi_1\cdots\xi_N-1$ is the unit ideal.
\end{enumerate}
Then for every root $\theta\in\C$ of $f$ the masses $g_i(\theta)$ and their sum are nonzero, and $\mathrm{K14}(\mu(\theta))$, respectively $\mathrm{K15}(\mu(\theta),k)$, has no solution.
\end{theorem}

\begin{proof}
\emph{Step 1: one embedding suffices.} Let $K=\Q[t]/(f)$, a field, with $\vartheta$ the class of $t$, and let $I\subset K[\xi_0,\xi]$ be the ideal generated by the polynomials $\sum_ac_{\nu,a}(g(\vartheta))\xi^a$ and $\xi_0\xi_1\cdots\xi_N-1$. For an embedding $\iota$ of $K$ into an algebraically closed field $\Omega$, the system $\iota F$ has a zero in $(\Omega^*)^N$ if and only if $1\notin I$: by the Nullstellensatz it has one if and only if $1\notin\iota(I)\,\Omega[\xi_0,\xi]$, and a representation $1=\sum h_\nu\,\iota(P_\nu)$ with prescribed degrees is a linear system over $\iota(K)$, which is solvable over $\iota(K)$ if it is solvable over $\Omega$. The complex embeddings of $K$ send $\vartheta$ to the roots of $f$. It therefore suffices to find one embedding of $K$ under which the system has no zero in the torus.

\emph{Step 2: the Hensel embedding.} Since $\bar f$ is squarefree, $\bar\theta$ is a simple root. Hensel's lemma gives a unique $\theta_\ell\in\Z_\ell$ with $f(\theta_\ell)=0$ and $\theta_\ell\equiv\bar\theta\pmod\ell$. As $f$ is irreducible, $\vartheta\mapsto\theta_\ell$ defines an embedding $\tau\colon K\to\Q_\ell$. By (P1) each $g_i(\theta_\ell)$ lies in $\Z_\ell$ and reduces to $\bar\mu_i$. Hence each coefficient $c_{\nu,a}(g(\theta_\ell))$ lies in $\Z_\ell$ and reduces to $c_{\nu,a}(\bar\mu)$, which is nonzero by (P2). All coefficients are therefore $\ell$-adic units. In particular the masses and their sum are nonzero in $K$, hence at every root of $f$.

\emph{Step 3: valuations.} Suppose that $\tau F$ has a zero $\xi\in(\bar\Q_\ell^{\,*})^N$. Its coordinates lie in a finite extension $E$ of $\Q_\ell$, with valuation $v$, $v(\ell)=1$, a uniformizer $\pi$ of valuation $1/e$, and residue field $k_E$. Put $\omega=-v(\xi)$ and $\xi_\alpha=\pi^{ev(\xi_\alpha)}y_\alpha$ with units $y_\alpha$. As the coefficients are units, the terms of $\tau F_\nu(\xi)$ of least valuation are those with exponent in $\max_\omega(A_\nu)$. Dividing by the least power of $\pi$ and reducing, we obtain $\mathrm{in}_\omega\bar F_\nu(\bar y)=0$ for every $\nu$, with $\bar y\in(k_E^*)^N$. If $\omega=0$, then $\bar y$, with $\xi_0=(\bar y_1\cdots\bar y_N)^{-1}$, is a zero of the ideal of (A), which is impossible. If $\omega\ne0$, then each $\max_\omega(A_\nu)$ has at least two elements, since a single term with a nonzero coefficient does not vanish at a torus point. Hence $\omega$ is a nonzero point of the prevariety. It lies in a cell $\sigma$, where $\mathrm{in}_\omega\bar F_\nu=\mathrm{in}_\sigma\bar F_\nu$. So $\bar y$ gives a zero of the ideal of (B), which is impossible.

Thus $\tau F$ has no zero in the torus. By Step 1, no complex root $\theta$ of $f$ gives one. By Lemma~\ref{lem:torus} neither $\mathrm{K14}(\mu(\theta))$ nor $\mathrm{K15}(\mu(\theta),k)$ holds.
\end{proof}

\begin{remark}\label{rem:KAforms}
(a) One prime suffices, although the certificates of \S\ref{ss:pointsets} that use Theorem~\ref{thm:KA} use two. (b) The proof uses the masses only through Step~2. If they are given in homogeneous coordinates, $\mu_i=a_i(\theta)/b(\theta)$ with $a_i,b\in\Z[t]$ and $b(\theta)\ne0$, the theorem holds with (P1) replaced by: $\ell$ does not divide the leading coefficient of $f$, $\bar f$ is squarefree of degree $\deg f$ with $\bar f(\bar\theta)=0$, $b(\bar\theta)\ne0$, and $\bar\mu_i=a_i(\bar\theta)\,b(\bar\theta)^{-1}$. Indeed $b(\theta_\ell)$ is then a unit, and $a_i(\theta_\ell)/b(\theta_\ell)\in\Z_\ell$ reduces to $\bar\mu_i$. (c) Since $\mathrm{K14}$ and $\mathrm{K15}$ are unchanged under scaling of the masses, the theorem may be applied to any nonzero multiple of the masses, for instance to the numerators over a common denominator.
\end{remark}

The certificate also bounds the masses at which a cluster system can hold along an algebraic curve.

\begin{corollary}\label{cor:GF}
Let $X$ be an irreducible curve over $\Q$ with a rational map $x\mapsto\mu(x)$ to the masses of a cluster, and suppose that the hypotheses of Theorem~\ref{thm:KA} hold, for $\mathrm{K14}$ or for $\mathrm{K15}(\cdot,k)$ and with some prime $\ell$, at $\mu(x_0)$ for a point $x_0\in X$ defined over a number field. Then the set of $x\in X(\C)$ at which the system has a zero in the torus is finite.
\end{corollary}

\begin{proof}
Since the set $Z$ in question is constructible and defined over $\Q$, it is finite or contains a nonempty Zariski open subset of $X$. Step~3 of the proof of Theorem~\ref{thm:KA} uses only that the masses lie in $\mathcal O_E^4$ for a finite extension $E$ of $\Q_\ell$ and reduce to $\bar\mu=\mu(x_0)\bmod\ell$, where (P2), (A) and (B) are conditions on $\bar\mu$. The points of $X$ over finite extensions of $\Q_\ell$ that are $\ell$-adically close to $x_0$ have such masses. They are infinitely many and hence Zariski dense in $X$. None of them lies in $Z$. Since membership in $Z$ is a first-order statement over $\Q$, it transfers between algebraically closed fields of characteristic $0$. Therefore $Z$ is not cofinite and hence finite.
\end{proof}

For masses along a curve we can make the parameter a further unknown. Let $G_1,\dots,G_n$ be nonzero polynomials in $\xi=(\xi_1,\dots,\xi_N)$ and a parameter $s$, with coefficients $e_{\nu,a}(\theta)$, where $e_{\nu,a}\in\Z[t]$ is not divisible by $f$. We now take the prevariety and its cells in the variables $(\xi,s)$.

\begin{theorem}\label{thm:KApar}
Let $\ell$ be a prime and $\bar\theta\in\mathbb F_\ell$ such that $\ell$ does not divide the leading coefficient of $f$, $\bar f$ is squarefree of degree $\deg f$, $\bar f(\bar\theta)=0$, and $e_{\nu,a}(\bar\theta)\ne0$ for all $\nu$ and $a$.
\begin{enumerate}[label=(\alph*)]
\item If $1$ lies in the ideal of $\mathbb F_\ell[\xi_0,\xi,s]$ generated by $\bar G_1,\dots,\bar G_n$ and $\xi_0\xi_1\cdots\xi_Ns-1$, and in the corresponding ideal of the initial systems of every cell, then for every root $\theta$ of $f$ and every $s\in\C^*$ the system $G(\cdot,s)$ has no zero in $(\C^*)^N$.
\item Let $h\in\Z[s]$ with $\bar h=h\bmod\ell$ of degree $\deg h$ and $\bar h(0)\ne0$. If over $\bar{\mathbb F}_\ell$ the full system, together with $\bar h(s)=0$, has no zero with all coordinates nonzero, and the same holds for the initial system of every cell whose relative interior contains a point with $s$-coordinate $0$, then for every root $\theta$ of $f$ and every root $s_0$ of $h$ the system $G(\cdot,s_0)$ has no zero in $(\C^*)^N$.
\end{enumerate}
\end{theorem}

\begin{proof}
(a) is the proof of Theorem~\ref{thm:KA} with $s$ as one more coordinate of the torus. (b) We extend the Hensel embedding of Step~2 to $\Q(\theta,s_0)$. Since $\bar h$ has full degree and $\bar h(0)\ne0$, all roots of $h$ in $\bar\Q_\ell$ are units. In Step~3 the vector $\omega=-v(\xi,s_0)$ therefore has $s$-coordinate $0$. The residue of $s_0$ is a root of $\bar h$. The reduction is a zero of the full system with $\bar h(s)=0$ (if $\omega=0$) or of the initial system of the cell containing $\omega$.
\end{proof}

The hypotheses of (a) fail for $\mathrm{K14}$ and $\mathrm{K15}$ along a family on which these systems have solutions; on such a family, however, Corollary~\ref{cor:GF} shows that the solutions occur at finitely many points if the hypotheses of Theorem~\ref{thm:KA} hold at one of its points. We will use part (b) for $\mathrm{K14}$ at the points of the family $T_9$ at which the label set changes (\S\ref{sec:T9}).

\subsection{The certified points}\label{ss:pointsets}

\begin{definition}\label{def:Kcert}
Let $m\in(\R_{>0})^5$ and $T\subseteq\{1,\dots,5\}$. The point $m$ is \emph{K-certified with transversal $T$} if
\begin{enumerate}[label=(\roman*)]
\item for every $j\in T$, $\mathrm{K14}(m_{Q_j})$ has no solution; and
\item there are a set of labels $S$ with $\mathrm{pieces}(S(m))\subseteq\mathrm{pieces}(S)$ and a D5 hull $V$ valid at $m$ (Definition~\ref{def:validhull}) such that the final check of $(S,T)$ holds: some $l_1,\dots,l_r\in\Z^{10}$ satisfy (a) and (b) of Definition~\ref{def:cert} for the attained faces of $\mathrm{pieces}(S)$, built with $V$, without those of all pieces $\pc{d2}$ and $\pc{d9}$, and of $\pc{d14a}(j)$, $\pc{d14b}(j,k)$ and $\pc{d15}(j,k)$, $j\in T$, $k\in Q_j$.
\end{enumerate}
\end{definition}

Note that this is the notion of \S\ref{sec:lex}, with the set of removed clusters made explicit.

\begin{proposition}\label{prop:Kcert}
A K-certified point is settled.
\end{proposition}

\begin{proof}
Let $C\subseteq Y_m$ be a nontrivial curve. By Corollary~\ref{cor:Kdrop} and (i), no singular place of $C$ has its type in the attained set of a piece $\pc{d14a}(j)$ with $j\in T$. By \S\ref{ss:certs} none has its type in that of a piece $\pc{d2}$, $\pc{d9}$, $\pc{d14b}$ or $\pc{d15}$. The proof of Proposition~\ref{prop:cert}, with these faces removed and Lemma~\ref{lem:lex} applied to the functionals of (ii), shows that there is no such curve.
\end{proof}

\emph{Exact points.} Every point below is given exactly, by an irreducible $f\in\Z[t]$, a rational isolating interval of a root $\theta$ (one sign change and either $f'\ne0$ on the interval in ball arithmetic or one sign variation in Descartes' rule), and masses $g_i(\theta)$ with $g_i\in\Q[t]$. We compute its label set $S(m)$. A label polynomial vanishes if it reduces to $0$ modulo $f$. It does not vanish if a ball value excludes $0$ or its reduction modulo $f$ is nonzero. This gives a set $S$ with $\mathrm{pieces}(S(m))\subseteq\mathrm{pieces}(S)$, as Definition~\ref{def:Kcert} requires.

Table~\ref{tab:points} lists the points, with their point records in the supplementary material. They are the jump points of the curve families of \S\ref{sec:curves} that are taken from this section (Table~\ref{tab:curvecounts}), in other labellings and scalings. Among them is the point $\hthree=\big(u_*^{3/2}(1+u_*)^{-2},u_*,u_*,1,1\big)$, with $u_*$ the root in $(0,1)$ of $u^4-4u^3-6u^2-4u+1$. After a relabelling and a scaling it is a jump point of $T_1$. The point records in one row of the table are conjugate: they belong to roots of one polynomial $f$ with the same $g_i$.

\begin{table}[ht]
\centering\small
\begin{tabular}{@{}llrrll@{}}
\toprule
point & used in & transversal & runs & final check & point records\\
\midrule
$\hthree$ & $T_1$ & $\{1\}$ & $5$ & $\Vtwo^{\mathrm w}$ & $48_0$, $48_1$\\
$(1,1,1,1,1)$ & $T_3$, \S\ref{sec:curves} & $\{1,\dots,5\}$ & $5$ & $\Vtwo^{\mathrm w}$ & $53_0$\\
two points & $T_5$ & $\{3\}$ & $5$ & $\VN$ & $20_0$, $20_1$\\
$(\frac19,\frac19,\frac14,1,1)$ & $T_7$ & $\{1\}$ & $5$ & $\Vtwo^{\mathrm w}$ & $22_0$\\
\midrule
& & & $20$ & $4$ on $\Vtwo^{\mathrm w}$, $2$ on $\VN$ & $6$ point records\\
\bottomrule
\end{tabular}
\caption{The K-certified points, with the point, sorted and scaled to $m_5=1$, where it is short. The last column gives their point records, the names of their files in the supplementary material. Where a row names one point and lists several point records, these are that point in other labellings and scalings. A run of Theorem~\ref{thm:KA} certifies, for one cluster, one of its five systems for one choice of $f$ and the cluster masses, and the point records of one row share their runs. Each row has one run of $\mathrm{K14}$, which is used in the proof, and four runs of $\mathrm{K15}$, which are cross-checks (\S\ref{ss:crosschecks}). At $(1,1,1,1,1)$ the five clusters have the same masses and share these runs. Each point record has one final check. It holds relative to the hull listed, with the functionals stored in the supplementary material.}\label{tab:points}
\end{table}

\begin{theorem}\label{thm:points}
\begin{enumerate}[label=(\alph*)]
\item At the point of every point record of Table~\ref{tab:points} and every $j$ in its transversal, $\mathrm{K14}(m_{Q_j})$ has no solution, and the pieces $\pc{d14a}(j)$ and $\pc{d14b}(j,k)$, $k\in Q_j$, are therefore attained at no such point.
\item The point of every point record of Table~\ref{tab:points} is K-certified with the transversal listed. Hence it is settled.
\end{enumerate}
\end{theorem}

\begin{proof}
(a) For each point and each $j$ in its transversal, the hypotheses of Theorem~\ref{thm:KA} hold for $\mathrm{K14}$. In each run of Table~\ref{tab:points} we obtain the conclusion ``no solution''. A run applies to the point of the table, because its input data are those of the exact point. Theorem~\ref{thm:Kgating} turns (i) of Definition~\ref{def:Kcert} into the statement on the pieces.

(b) Table~\ref{tab:points} gives, for each point record, the hull of Remark~\ref{rem:hulls} relative to which its final check holds. The hull $\VN$ is valid at the two point records at which it is used (\S\ref{ss:d5}), by the check of \S\ref{ss:checks}. Proposition~\ref{prop:Kcert} completes the proof.
\end{proof}

The point $(1,1,1,1,1)$ of Table~\ref{tab:points} has the transversal $\{1,\dots,5\}$. Each cluster has the masses $(1,1,1,1)$ there. Theorem~\ref{thm:points}(a) shows that $\mathrm{K14}(1,1,1,1)$ has no solution, a fact that will be used in \S\ref{sec:curves}. We will apply Theorems~\ref{thm:KA} and~\ref{thm:KApar} at further points of the curve families, among them the jump points of $T_9$ (\S\ref{sec:T9}), and count these points there.