\section{The sieve, and the proof of \texorpdfstring{\cref{thm:L8}}{Theorem 1.1}}\label{sec:sieve}

We now write the conditions of \cref{sec:descent,sec:local,sec:aux} in the coordinates of \cref{sec:selmer}, show by a finite computation that no class satisfies all of them (\cref{thm:sieve}), and deduce \cref{thm:L8}.

For each $q\in\cQ$, number the primes of $\Lt$ above $q$ as $\qQ_1,\dots,\qQ_m$ in the order of \cref{app:primesq}, and let $\Mq\in M_{m\times24}(\F_5)$ be the matrix $\big(\sym_{\qQ_k}(B_j)\big)_{k,j}$, printed in \cref{app:Mq}.  The sets $\Iq\subseteq\F_5^m$ of \cref{def:Iq}, with coordinates in the same order, are printed in \cref{app:Iq}.

\begin{theorem}\label{thm:sieve}
There is no $c\in\F_5^{24}$ satisfying all of the following conditions:
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2.2em]
\item[\textup{(N)}] $\NM c\equiv\TT\pmod5$;
\item[\textup{(V)}] $c_3=2$, $c_5=3$ and $c_2=c_4=c_6=c_7=c_8=0$;
\item[\textup{(Q)}] $\Mq c\in\Iq$ for $q=181,311,131,251,101$.
\end{enumerate}
More precisely, the matrix $\NM$ has rank $10$ over $\F_5$, so \textup{(N)} defines an affine subspace of $\F_5^{24}$ of dimension $14$; adding \textup{(V)} gives an affine subspace $\mathcal A$ of dimension $10$, that is, $5^{10}=9\,765\,625$ vectors.  Imposing \textup{(Q)} for $q=181,311,131,251,101$ in turn leaves $36\,875$, $567$, $19$, $1$ and $0$ vectors.
\end{theorem}

\begin{proof}
Of the seven conditions in (V), only four are independent of (N).  Indeed, the rows of (N) for the three primes of $\Lo$ above $5$, generated by $u_2$, $u_3$ and $u_4$, say that the norm has valuation $0$ modulo $5$ there; since every prime of $\Lt$ above $5$ has residue degree $1$ over $\Lo$, these rows read $c_6+c_7\equiv0$, $c_3+c_5\equiv0$ and $c_2+c_4+c_8\equiv0$, and (V) satisfies them.  The affine space $\mathcal A$ is $\{c_0+\sum_{k=1}^{10}\lambda_kd_k:\lambda\in\F_5^{10}\}$ with $c_0$ and $d_1,\dots,d_{10}$ printed in~\eqref{eq:affine}.  All $5^{10}$ of its elements were listed and tested against (Q) by arithmetic modulo $5$ (\cref{app:computations}).  The $19$ vectors that satisfy (Q) for $q=181,311,131$ are printed in \cref{tab:survivors} together with their images under $M_{251}$.  Exactly one of these images lies in $I(251)$, and for that vector $c^\ast$,
\[
M_{101}\,c^\ast=(1,4,2,3,0,3,0)\notin I(101),
\]
by the list of $I(101)$ in \cref{app:Iq}.
\end{proof}

Each count is the number of classes of $\Sel$ satisfying the conditions imposed so far, so it does not depend on the basis $B_1,\dots,B_{24}$ or on the elements $u_i$.  The intermediate counts depend on the order in which the five primes are imposed, but the final count does not.  The five primes are those with the smallest proportion $\#\Iq/5^m$ (\cref{tab:aux}), taken in increasing order of that proportion.  A separately written program gives the same counts (\cref{app:independent}).

\begin{proof}[Proof of \cref{thm:L8}]
Suppose that $(X,Y,Z)$ is a solution with $\Aeta\isom\Lo$.  Let $\vt\in\Lo$ and $E$ be as in \cref{sec:descent}.
\begin{enumerate}[label=(\arabic*),leftmargin=2em]
\item By \cref{prop:D1}, the class of $E$ lies in $\Sel$.  By \cref{prop:basis}(ii) it has a coordinate vector $c\in\F_5^{24}$: $E=\beta^5\prod_jB_j^{c_j}$ with $\beta\in\Lt^\times$ and integers $c_j$ (read modulo $5$).\looseness=-1
\item Condition (N).  By \cref{prop:D2}, $\Norm{\Lt/\Lo}{E}\in2\,\Lo^{\times5}$.  Taking norms in (1) and using \cref{lem:normdata}(ii), $\prod_iu_i^{(\NM c-\TT)_i}\in\pm\Lo^{\times5}=\Lo^{\times5}$, so $\NM c\equiv\TT\pmod5$ by \cref{lem:normdata}(i).
\item Condition (V).  For $\PT\in S$ we have $v_\PT(E)=5v_\PT(\beta)+\sum_jc_jv_\PT(B_j)$, so $c_i\equiv v_{\PT_i}(E)\pmod5$ for $i\le12$.  By \cref{prop:five} and \cref{tab:Sprimes}, $v_{\PT_3}(E)\equiv2$, $v_{\PT_5}(E)\equiv3$ and $v_{\PT_i}(E)\equiv0$ for $i\in\{2,4,6,7,8\}$.
\item Condition (Q).  Let $q\in\cQ$.  By \cref{cor:aux}, $q\nmid XYZ$ and $\big(\sym_{\qQ_k}(E)\big)_k\in\Iq$.  By \cref{prop:D1}(iii) and since the $B_j$ are $S$-units, $E$ and the $B_j$ are units at every $\qQ_k$, hence so is $\beta$, and $\sym_{\qQ_k}(E)=\sum_jc_j\sym_{\qQ_k}(B_j)$.  So $\Mq c\in\Iq$.
\end{enumerate}
This contradicts \cref{thm:sieve}.
\end{proof}

\needspace{4\baselineskip}
\begin{remark}\label{rem:robust}
The following computations lie outside the proof.  They show how much each condition contributes (\cref{app:computations}).
\begin{enumerate}[label=(\alph*),leftmargin=2em]
\item On the affine space defined by (N), the first coordinate, which is the valuation at $\PT_1$ modulo $5$, is identically $3$, in agreement with \cref{lem:two}.
\item With all sixteen primes of $\cQ$ imposed, no vector survives.  If any single prime of $\cQ$ is omitted, still none survives, except that $2$, $2$ and $1$ vectors survive when $131$, $181$ or $311$, respectively, is omitted.
\item For each of the five primes $q=181,311,131,251,101$, every vector of $\Iq$ is attained as $\Mq c$ with $c\in\mathcal A$, so no element of $\Iq$ is excluded by (N) and (V) alone.
\item If the values $2$ and $3$ that (V) prescribes at $\PT_3$ and $\PT_5$ are exchanged, again no vector survives, so the conclusion does not depend on telling these two primes apart.  Without condition (V), exactly $6$ vectors satisfy (N) and (Q) for all sixteen primes, and none of them satisfies (V).
\end{enumerate}
\end{remark}

\needspace{5\baselineskip}
\begin{remark}\label{rem:whyfive}
Five primes can be expected to suffice for the following reason.  Let $q\in\cQ$, let $\qq$ be a prime of $\Lo$ above $q$, and let $\alpha\in\Lt$ be a unit at every prime above $\qq$.  Since $\sym_\qQ(\alpha)=\sym_\qq\big(\Nm_{k(\qQ)/k(\qq)}\bar\alpha\big)$ for $\qQ\mid\qq$, and $q$ is unramified in $\Lt$,
\[
\sum_{\qQ\mid\qq}\sym_\qQ(\alpha)=\sym_\qq\big(\Norm{\Lt/\Lo}{\alpha}\big).
\]
On $\mathcal A$ the norm class is that of $2$, by (N).  So for $c\in\mathcal A$ the symbols of $\prod_jB_j^{c_j}$ above each prime $\qq$ add up to $\sym_\qq(2)$, and $c\mapsto\Mq c$ takes at most $5^{m-m'}$ values on $\mathcal A$, where $m'$ is the number of primes of $\Lo$ above $q$.  When $m'=m$, as for $q=11$, $401$, $431$ and $541$, it takes a single value, which lies in $\Iq$, so the condition at $q$ gives no information, although $\#\Iq=1$ there.  For the five primes of \cref{thm:sieve} it takes exactly $5^{m-m'}$ values, and by \cref{rem:robust}(c) every element of $\Iq$ is one of them.  The fractions of these values that lie in $\Iq$ are
\[
\frac{59}{5^6},\qquad \frac{240}{5^6},\qquad \frac{112}{5^6},\qquad \frac{172}{5^6},\qquad \frac{16}{5^4}
\]
for $q=181$, $311$, $131$, $251$, $101$.  In particular the condition at $181$ keeps $59\cdot5^4=36\,875$ of the $5^{10}$ vectors of $\mathcal A$, as in \cref{thm:sieve}.  If the five conditions were independent, the expected number of surviving vectors would be $5^{10}\cdot59\cdot240\cdot112\cdot172\cdot16/5^{28}\approx0.001$.
\end{remark}