\section{Auxiliary primes}\label{sec:aux}

Here we derive the condition (Q) of \cref{sec:sieve}.  For sixteen primes $q\equiv1\pmod5$, the splitting of $\Lo$ at $q$ shows that $q\nmid XYZ$ (\cref{cor:aux}).  At such a prime the residue of $\vt$ at each prime of $\Lo$ above $q$ is a root of $f_{\bar\eta}$, where $\bar\eta=\eta\bmod q$.  So the vector of fifth-power residue symbols of $E$ at the primes of $\Lt$ above $q$ lies in a finite set $\Iq$, which can be computed without knowing the solution (\cref{prop:A2}).

Throughout this section $q$ is a prime, $q\notin\{2,5,7\}$.  Then $q$ is unramified in $\Lo$ and in $\Lt$ (\cref{lem:fields}).  For a prime $\qq$ of $\Lo$ or $\qQ$ of $\Lt$ above $q$ we write $k(\qq)$, $k(\qQ)$ for the residue fields and $f(\qq)=[k(\qq):\Fq q]$.

\subsection{Definitions}

The \emph{type} of a nonzero polynomial over $\Fq q$ is the multiset of the degrees of its irreducible factors, counted with multiplicity.  The \emph{type of $\Lo$ at $q$} is the multiset $\{f(\qq):\qq\mid q\}$.  The \emph{type} of an unramified \'etale $\Qp q$-algebra is the multiset of residue degrees of its factor fields.  It is the multiset of orbit sizes of Frobenius on the roots of any defining polynomial.  If a defining polynomial has coefficients in $\Zp q$, a unit leading coefficient and a separable reduction modulo $q$, this multiset is the type of that reduction.

For $q\equiv1\pmod5$ let $\zeta_q\in\Fq q$ be the $((q-1)/5)$-th power of the least positive primitive root modulo $q$.  It is a primitive fifth root of unity in $\Fq q$.  For a prime $\qQ$ of $\Lt$ above $q$ and a $\qQ$-unit $\alpha$, the \emph{fifth-power residue symbol} $\sym_\qQ(\alpha)\in\F_5$ is defined by
\begin{equation}\label{eq:symbol}
\alpha^{(\#k(\qQ)-1)/5}\equiv\zeta_q^{\,\sym_\qQ(\alpha)}\pmod\qQ .
\end{equation}
It is a homomorphism from the $\qQ$-units to $\F_5$ that vanishes on fifth powers.  We use the same notation for an element $\bar\alpha\in k(\qQ)^\times$.  For a prime $\qq$ of $\Lo$ above $q$ the symbol $\sym_\qq$ is defined in the same way, with $k(\qq)$ in place of $k(\qQ)$.

\begin{definition}\label{def:Iq}
Let $q\equiv1\pmod5$.  Call $\bar\eta\in\Fq q\smallsetminus\{0,1\}$ \emph{admissible} if the type of $f_{\bar\eta}(t)=4t^5\psi(t)-\bar\eta(4t-1)\in\Fq q[t]$ is the type of $\Lo$ at $q$.  For admissible $\bar\eta$, an \emph{admissible family} is a choice, for each prime $\qq$ of $\Lo$ above $q$, of a root $t_\qq\in k(\qq)$ of $f_{\bar\eta}$ whose minimal polynomial over $\Fq q$ has degree $f(\qq)$, such that these minimal polynomials are pairwise distinct.  Let $\Iq\subseteq\F_5^m$, where $m$ is the number of primes $\qQ_1,\dots,\qQ_m$ of $\Lt$ above $q$, be the set of the vectors
\[
\Big(\sym_{\qQ_k}\big(80000\,(t_{\qq_k}-\bar b)\,\psi(t_{\qq_k})^3\big)\Big)_{k=1}^{m},\qquad \qq_k=\qQ_k\cap\Lo,
\]
over all admissible $\bar\eta$ and all admissible families.  Here $\bar b$ is the residue of $b$ modulo $\qQ_k$, and $t_{\qq_k}$ is viewed in $k(\qQ_k)$ through the inclusion $k(\qq_k)\subseteq k(\qQ_k)$ induced by $\Lo\subseteq\Lt$.
\end{definition}

The arguments of the symbols are nonzero: if $t$ is a root of $f_{\bar\eta}$ with $\bar\eta\ne0$ and $\psi(t)=0$ then $\bar\eta(4t-1)=0$, so $t=\frac14$.  But $\psi(\frac14)=5^2\cdot7^2/2^6\ne0$ in $\Fq q$.  So $\psi(t)\ne0$, and in particular $t\ne\bar b$.

\subsection{The local algebras when \texorpdfstring{$q$}{q} divides \texorpdfstring{$XYZ$}{XYZ}}

The next proposition is a version of~\cite[Proposition~4.11 and Table~4.12]{Putz}, proved here in the form used below.

\begin{proposition}\label{prop:A1}
Let $(X,Y,Z)$ be a solution and $q\notin\{2,5,7\}$.
\begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em]
\item If $q\mid X$, then $\Aeta\otimes\Qp q$ is unramified and its type is $\type(\psi\bmod q)\cup\type(s^5-c)$ for some $c\in\Fq q^\times$.
\item If $q\mid Z$, then $\Aeta\otimes\Qp q$ is unramified and its type is $\{1\}\cup\type(s^7-c)$ for some $c\in\Fq q^\times$.
\item Let $\Psi_i$ run over the irreducible factors of $\Psi\bmod q$, let $d_i=\deg\Psi_i$, and let $r_i\in\Fq{q^{d_i}}$ be a root of $\Psi_i$.  If $q\mid Y$, then $\Aeta\otimes\Qp q$ is unramified, and there is $\alpha\in\Fq q^\times$ such that its type is the union of the $T_i$, where $T_i=\{d_i,d_i\}$ if $\alpha(4r_i-1)$ is a square in $\Fq{q^{d_i}}$ and $T_i=\{2d_i\}$ otherwise.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) Here $q\nmid Z$ and $v_q(\eta)=5k$ with $k=v_q(X)\ge1$.  Write $\eta=q^{5k}\eta_0$ with $\eta_0\in\Zp q^\times$.  The coefficients of $\feta$ have valuations $0$ at $t^8,t^7,t^6,t^5$ and $5k$ at $t^1,t^0$, so the Newton polygon has vertices $(0,5k)$, $(5,0)$, $(8,0)$, and $\feta=GH$ over $\Qp q$, where $G$ has the five roots of valuation $k$ and $H$ the three unit roots.  Since $\feta\equiv4t^5\psi(t)\pmod q$ and $\psi(0)\ne0$, the unit roots reduce to the roots of $\psi\bmod q$, which is separable of degree $3$ ($\disc\psi=-2^2\cdot5^7\cdot7$, leading coefficient $25$).  So $\Qp q[t]/(H)$ is unramified of type $\type(\psi\bmod q)$.  For $G$, the polynomial $F(s)=q^{-5k}\feta(q^ks)=4s^5\psi(q^ks)-\eta_0(4q^ks-1)$ has integral coefficients and $F\equiv56s^5+\eta_0\pmod q$, which is separable of degree $5$.  So $F$ has five integral roots, distinct modulo $q$; they are the roots of $G$ divided by $q^k$.  Hence $\Qp q[t]/(G)$ is unramified of type $\type(s^5-c)$ with $c=-\eta_0/56\bmod q$.

(b) Here $q\nmid X$ and $v_q(\eta)=-7k$ with $k=v_q(Z)\ge1$.  Write $\eta=q^{-7k}\eta_1$ with $\eta_1\in\Zp q^\times$.  Then
\begin{multline*}
\Phi(t)=q^{8k}\feta(q^{-k}t)=100t^8+80q^kt^7+56q^{2k}t^6+56q^{3k}t^5-4\eta_1t+\eta_1q^k\\
\equiv4t\,(25t^7-\eta_1)\pmod q,
\end{multline*}
which is separable.  As $\Qp q[t]/(\feta)\isom\Qp q[t]/(\Phi)$, the algebra is unramified of type $\{1\}\cup\type(s^7-c)$ with $c=\eta_1/25\bmod q$.

(c) Here $v_q(\eta-1)=v_q(-Y^2/Z^7)=2k$ with $k=v_q(Y)\ge1$.  Write $\eta-1=q^{2k}\alpha$ with $\alpha\in\Zp q^\times$.  By~\eqref{eq:belyi}, $\feta=\Psi^2-q^{2k}\alpha(4t-1)$.  The polynomial $\Psi\bmod q$ is separable of degree $4$ ($\disc\Psi=-2^6\cdot5^2\cdot7^3$, leading coefficient $10$), and $\Psi(\frac14)=-35/128\ne0$.  Let $\tilde r$ be a root of $\Psi$ in the maximal unramified extension $W$ of $\Qp q$, with reduction $r$.  In $F_r(s)=q^{-2k}\feta(\tilde r+q^ks)$ the coefficient of $s^j$ is $q^{(j-2)k}\feta^{(j)}(\tilde r)/j!$: it is $-\alpha(4\tilde r-1)$ for $j=0$, $-4\alpha q^k$ for $j=1$, $\Psi'(\tilde r)^2$ for $j=2$, and divisible by $q$ for $j\ge3$.  So $F_r\equiv\Psi'(r)^2s^2-\alpha(4r-1)\pmod q$, a separable quadratic.  Its two roots lift to roots $s\in W$, and the eight numbers $\tilde r+q^ks$ (four $\tilde r$, two $s$ each) are distinct roots of $\feta$, hence all of them.  So the algebra is unramified, and Frobenius permutes the pairs $(r,\bar s)$, where $\bar s$ is the reduction of $s$.  If $r$ is a root of the irreducible factor $\Psi_i$ of $\Psi\bmod q$, the orbit of $(r,\bar s)$ has size $d_i$ if $\bar s\in\Fq{q^{d_i}}$, that is, if $\alpha(4r-1)/\Psi'(r)^2$, equivalently $\alpha(4r-1)$, is a square in $\Fq{q^{d_i}}$, and size $2d_i$ otherwise.
\end{proof}

\subsection{Primes not dividing \texorpdfstring{$XYZ$}{XYZ}}

\begin{proposition}\label{prop:A2}
Let $(X,Y,Z)$ be a solution with $\Aeta\isom\Lo$, and let $\vt$ and $E$ be as in \cref{sec:descent}.  Let $q\notin\{2,5,7\}$ with $q\nmid XYZ$, and $\bar\eta=\eta\bmod q$.  Then $\bar\eta\notin\{0,1\}$, the polynomial $f_{\bar\eta}$ is separable of degree $8$ and its type is the type of $\Lo$ at $q$.  For each prime $\qq$ of $\Lo$ above $q$, $\vt$ is $\qq$-integral, its residue $\bar\vt_\qq\in k(\qq)$ is a root of $f_{\bar\eta}$ with minimal polynomial of degree $f(\qq)$, and these minimal polynomials are pairwise distinct.  The element $E$ is a unit at every prime $\qQ$ of $\Lt$ above $q$, with
\[
E\equiv80000\,(\bar\vt_\qq-\bar b)\,\psi(\bar\vt_\qq)^3\pmod\qQ,\qquad\qq=\qQ\cap\Lo .
\]
Consequently, if also $q\equiv1\pmod5$, then $\big(\sym_\qQ(E)\big)_{\qQ\mid q}\in\Iq$.
\end{proposition}

\begin{proof}
Since $q\nmid XYZ$ and $1-\eta=Y^2/Z^7$, $\eta$ is $q$-integral with $\bar\eta\ne0,1$.  By \cref{lem:disc}, $f_{\bar\eta}$ is separable (its leading coefficient $100$ is a unit).  Then $\feta/100\in\Zp q[t]$ is monic with unit discriminant, so $\Zp q[t]/(\feta)$ is \'etale over $\Zp q$ and is the maximal order of $\Qp q[t]/(\feta)\isom\Lo\otimes\Qp q$ ($t\mapsto\vt$).  Hence $\vt$ is integral at every $\qq\mid q$, and reduction gives an isomorphism $\Fq q[t]/(f_{\bar\eta})\isom\prod_{\qq\mid q}k(\qq)$, $t\mapsto(\bar\vt_\qq)$.  The left side is a product of fields, one for each irreducible factor of $f_{\bar\eta}$.  An isomorphism between finite products of fields matches the factors bijectively.  So each $\bar\vt_\qq$ is a root of exactly one irreducible factor of $f_{\bar\eta}$, which is its minimal polynomial and has degree $f(\qq)$, and different primes $\qq$ give different factors.  In particular the type of $f_{\bar\eta}$ is the type of $\Lo$ at $q$.  By \cref{prop:D1}(iii), $E$ is a unit at every $\qQ\mid q$, and since the reductions of $\Lo$ and $\Lt$ are compatible with $k(\qq)\subseteq k(\qQ)$, $E$ reduces to $80000(\bar\vt_\qq-\bar b)\psi(\bar\vt_\qq)^3$.  So $\bar\eta$ is admissible and $(\bar\vt_\qq)_\qq$ is an admissible family, which gives the last assertion.
\end{proof}

\subsection{The auxiliary primes}

We searched the primes $q\equiv1\pmod5$ up to $700$ for primes at which the splitting of $\Lo$ excludes $q\mid XYZ$.  Sixteen of them qualify, more than the sieve of \cref{sec:sieve} needs.

\begin{proposition}\label{prop:A3}
Among the primes $q\equiv1\pmod5$ with $11\le q\le700$, the type of $\Lo$ at $q$ is not of any of the forms \textup{(a)}, \textup{(b)}, \textup{(c)} of \cref{prop:A1} \textup{(}for any $c$, $\alpha$\textup{)} exactly for the sixteen primes
\[
q\in\cQ=\{11,101,131,181,211,251,271,311,331,401,431,461,541,601,631,661\}.
\]
For these primes the types, the numbers of admissible residues $\bar\eta$, the numbers $m$ of primes of $\Lt$ above $q$, and the sizes of the sets $\Iq$ are those of \cref{tab:aux}; the sets $\Iq$ are listed in \cref{app:Iq}.
\end{proposition}

\begin{table}[ht]
\centering
\small
\begin{tabular}{@{}rlrrr@{\qquad}rlrrr@{}}
\toprule
$q$ & type of $\Lo$ & $\#\bar\eta$ & $m$ & $\#\Iq$ & $q$ & type of $\Lo$ & $\#\bar\eta$ & $m$ & $\#\Iq$\\
\midrule
11 & 1,1,2,4 & 1 & 4 & 1 & 331 & 4,4 & 23 & 6 & 21\\
101 & 2,3,3 & 5 & 7 & 16 & 401 & 1,7 & 102 & 2 & 1\\
131 & 2,2,4 & 9 & 9 & 112 & 431 & 1,7 & 144 & 2 & 1\\
181 & 1,1,1,2,3 & 9 & 11 & 59 & 461 & 1,3,4 & 70 & 7 & 84\\
211 & 2,6 & 37 & 4 & 24 & 541 & 1,7 & 162 & 2 & 1\\
251 & 2,2,4 & 16 & 9 & 172 & 601 & 1,1,6 & 105 & 7 & 120\\
271 & 8 & 71 & 3 & 5 & 631 & 1,2,2,3 & 56 & 6 & 25\\
311 & 1,1,1,1,4 & 6 & 11 & 240 & 661 & 2,3,3 & 28 & 7 & 68\\
\bottomrule
\end{tabular}
\vspace{3pt}
\caption{The sixteen auxiliary primes: the type of $\Lo$ at $q$, the number of admissible residues $\bar\eta\in\Fq q$, the number $m$ of primes of $\Lt$ above $q$, and the size of $\Iq$.}\label{tab:aux}
\end{table}

\begin{proof}
A finite computation (\cref{app:computations}): for each prime $q\equiv1\pmod5$ up to $700$, list all the types of the forms (a), (b), (c) over all $c,\alpha\in\Fq q^\times$ and compare with the type of $\Lo$ at $q$.  For $q\in\cQ$, enumerate the admissible $\bar\eta$ and families and compute the symbols.  The type of $\Lo$ at $q$ is read off from the prime decomposition of $q$ in $\Lo$.  For $q\in\cQ$ the prime $q$ does not divide the index $2^7\cdot5\cdot41$ of $\Z[a]$, so by the Dedekind--Kummer theorem it is also the type of $h\bmod q$.  For $q=41$, which divides the index, the type of $\Lo$ at $q$ is $\{1,2,5\}$, which is of the form (a).
\end{proof}

\needspace{4\baselineskip}
\begin{corollary}\label{cor:aux}
Let $(X,Y,Z)$ be a solution with $\Aeta\isom\Lo$.  For every $q\in\cQ$ we have $q\nmid XYZ$ and $\big(\sym_\qQ(E)\big)_{\qQ\mid q}\in\Iq$.
\end{corollary}

\begin{proof}
If $q\mid XYZ$, then by \cref{prop:A1} the type of $\Aeta\isom\Lo$ at $q$ would be of one of the forms (a), (b), (c), which \cref{prop:A3} excludes.  So $q\nmid XYZ$, and \cref{prop:A2} applies.
\end{proof}

\needspace{4\baselineskip}
\begin{remark}
The single admissible residue at $q=11$ is $\bar\eta=7$, in agreement with~\cite[Theorem~4.34(v)]{Putz}.
\end{remark}