\section{A test on fibres where the descent element exists}\label{sec:control}

A sieve that ends with no survivors could in principle do so because a condition was implemented wrongly.  To test the implementation, we applied the same programs to fibres of $\varphi$ above rational points of a twisted equation, where the element $E$ is defined and its class must survive every condition.  \Cref{prop:D1,prop:D2} use only coprime nonzero integers $X$, $Z$ and a root $\vt$ of $\feta$, not a solution of~\eqref{eq:main}, so they apply to such fibres (for \cref{prop:D2} because $\psi$ is irreducible over any field of degree $8$, as in \cref{lem:fields}(c)).  We use the twisted equation
\[
X^5+2Y^2=Z^7,
\]
because the fibres above its points with $X$, $Z$ coprime and $Y\ne0$ are, like $\Lo$, unramified outside $2$, $5$ and $7$ (the arguments of \cref{prop:A1,prop:A2} apply).  It has the points $(X,Y,Z)=(-1,1,1)$ and $(-3,11,-1)$, which give $\eta=X^5/Z^7=-1$ and $\eta=243$.  For each of them, let $K_8=\Aeta$ and $K_{24}=K_8(b)$.

\begin{itemize}[leftmargin=2em]
\item In both cases $\feta$ is irreducible with Galois group $S_8$, and $K_8$ is an octic field of signature $(2,3)$ and discriminant $-2^{18}\cdot5^6\cdot7^7$, not isomorphic to $\Lo$.  The field $K_{24}$ has degree $24$, signature $(2,11)$ and discriminant $-2^{56}\cdot5^{20}\cdot7^{23}$, and eleven primes above $2$, $5$ and $7$: one above $2$, six above $5$ and four above $7$.  Its class number, computed assuming the generalized Riemann hypothesis, is $2$ for $\eta=-1$ and $1$ for $\eta=243$.  Let $S$ now be the set of these eleven primes.  In both cases the class number is prime to $5$, so, under the same hypothesis, $K_{24}(S,5)$ has dimension $23$, the rank of the $S$-units of $K_{24}$.  These class numbers are used only in this test.
\item Let $\vt\in K_8$ be the class of $t$ and $E=80000(\vt-b)\psi(\vt)^3\in K_{24}$.  Its valuations at the primes outside $S$ are divisible by $5$: for $\eta=-1$ it is an $S$-unit, and for $\eta=243$ (where $3\mid X$) it has three prime factors above $3$, with exponents $20$, $15$ and $15$.  Its coordinate vector $c_E$ on a basis of $K_{24}(S,5)$ satisfies the norm condition, formed as (N) is in \cref{sec:sieve} but with this basis and with $S$-units of $K_8$.
\item At the six primes of $K_{24}$ above $5$, the valuations of $E$ modulo $5$ are $2$ at the prime with label $(1,1,1)$, $3$ at the prime with label $(2,1,1)$ and $0$ at the other four, and its valuation at the prime above $2$ is $\equiv3$: the values that \cref{lem:two,prop:five} give in the case $5\nmid XYZ$.  We impose these valuations at the primes above $5$ in place of (V).  Together with the norm condition they define an affine subspace of dimension $10$.
\item For the first twelve primes $q\equiv1\pmod5$ not dividing $XYZ$, the set $I_{K_8}(q)$, defined as in \cref{def:Iq} with $K_8$ in place of $\Lo$, contains the vector of residue symbols of $E$, and this vector equals $M_qc_E$, where $M_q$ now denotes the matrix of residue symbols of the basis of $K_{24}(S,5)$.
\item The sieve with the norm condition, these valuations and the twelve primes leaves exactly one of the $5^{10}$ classes of this affine subspace, and it is $c_E$.  With the primes taken in increasing order of $\#I_{K_8}(q)/5^m$, as in \cref{sec:sieve}, one class remains after five primes for $\eta=-1$ (the counts are $350$, $31$, $3$, $2$, $1$) and after six primes for $\eta=243$ (the counts are $30\,000$, $30\,000$, $514$, $40$, $9$, $1$).
\end{itemize}