\section{The descent element and its norm}\label{sec:descent}

A solution with $\Aeta\isom\Lo$ gives an element $E$ of $\Lt$ whose class modulo fifth powers lies in a finite group $\Sel$ (\cref{prop:D1}) and whose norm to $\Lo$ is $2$ times a fifth power (\cref{prop:D2}).  The first fact reduces the problem to finitely many classes, and the second gives the norm condition (N) of \cref{sec:sieve}.

Let $S$ be the set of the twelve primes of $\Lt$ above $2$, $5$ and $7$, and let
\begin{equation}\label{eq:Sel}
\Sel=\bigl\{\alpha\in\Lt^\times/\Lt^{\times5}\ :\ v_{\PT}(\alpha)\equiv0\pmod5\ \text{for every prime }\PT\notin S\bigr\},
\end{equation}
where $v_{\PT}$ is the normalized valuation of $\PT$ ($v_\PT(\Lt^\times)=\Z$).  The group $\Sel$ is a finite-dimensional vector space over $\F_5$ (\cref{sec:selmer}).

Suppose $(X,Y,Z)$ is a solution with $\Aeta\isom\Lo$.  Fix such an isomorphism and let $\vt\in\Lo$ be the image of the class of $t$.  Then $\vt$ generates $\Lo$, and~\eqref{eq:theta} holds.  The next proposition uses much less: only coprime nonzero integers $X$, $Z$ and a root $\vt$ of $\feta$ in a number field containing $b$.  \Cref{sec:control} uses this generality.

\needspace{5\baselineskip}
\begin{proposition}\label{prop:D1}
Let $X,Z$ be coprime nonzero integers and $\eta=X^5/Z^7$.  Let $K$ be a number field containing $b$ and an element $\vt$ with $\feta(\vt)=0$, and put $E=80000(\vt-b)\psi(\vt)^3\in K$.  Then:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.2em]
\item $\psi(\vt)\ne0$ and $\vt\ne b$, so $E\ne0$;
\item $v_\PT(E)\equiv0\pmod5$ for every prime $\PT$ of $K$ not above $2$, $5$ or $7$;
\item $v_\PT(E)=0$ for every prime $\PT$ of $K$ above a prime $p\notin\{2,5,7\}$ with $p\nmid XZ$.
\end{enumerate}
In particular, if $(X,Y,Z)$ is a solution with $\Aeta\isom\Lo$ and $\vt\in\Lo$ is as above, then the class of $E$ lies in $\Sel$.
\end{proposition}

\begin{proof}
Since $\feta(\vt)=0$, the element $\vt$ satisfies~\eqref{eq:theta}.  (i) If $\psi(\vt)=0$ then~\eqref{eq:theta} gives $\eta(4\vt-1)=0$, so $\vt=1/4$.  But $\psi(1/4)=5^2\cdot7^2/2^6\ne0$.  Hence $\psi(\vt)\ne0$, and so $\vt\ne b$.

For (ii) and (iii), let $\PT$ be a prime of $K$ above $p\notin\{2,5,7\}$, write $v=v_\PT$ and $e=v(p)$.  Since $5b$ is integral and $p\ne5$, $b$ is $\PT$-integral, and so are $1-4b$ and $\psi'(b)$.  Their norms to $\Q$ in~\eqref{eq:bnorms} are prime to $p$, so all three are units at every prime above $p$: an element that is integral at every prime above $p$ and whose norm is prime to $p$ is a unit at each of these primes.  Write $\psi(t)=25(t-b)\psi_1(t)$, where
\[
\psi_1(t)=t^2+\big(b+\tfrac45\big)t+\big(b^2+\tfrac45b+\tfrac{14}{25}\big)\in\Q(b)[t]
\]
has $\PT$-integral coefficients and $\psi_1(b)=\psi'(b)/25$, a $\PT$-unit.  Finally $v(\eta)=e\,(5v_p(X)-7v_p(Z))$, and at most one of $v_p(X)$, $v_p(Z)$ is positive.

\emph{Case $p\nmid Z$.}  Then $v(\eta)=5e\,v_p(X)\ge0$.
\begin{enumerate}[label=(\alph*),leftmargin=2em]
\item $v(\vt)\ge0$.  Otherwise $v(\psi(\vt))=3v(\vt)$, as $25\vt^3$ has strictly smaller valuation than the other three terms of $\psi(\vt)$, and $v(4\vt-1)=v(\vt)$.  Then~\eqref{eq:theta} gives $8v(\vt)=v(\eta)+v(\vt)$, so $7v(\vt)=v(\eta)\ge0$, a contradiction.
\item $v(4\vt-1)=0$.  Otherwise $\vt\equiv1/4\pmod\PT$, so $\vt$ is a unit and $\psi(\vt)\equiv\psi(1/4)$ is a unit.  The left side of~\eqref{eq:theta} is then a unit, while the right side has positive valuation.
\item Hence $5v(\vt)+v(\psi(\vt))=v(\eta)$, that is,
\[
5v(\vt)+v(\vt-b)+v(\psi_1(\vt))=v(\eta),
\]
and the three terms on the left are $\ge0$, since $\vt$, $b$ and the coefficients of $\psi_1$ are $\PT$-integral.
\item At most one of these three terms is positive.  If $v(\vt)>0$ then $\vt-b\equiv-b$ and $\psi(\vt)\equiv\psi(0)=14$ are units, so $v(\vt-b)=v(\psi_1(\vt))=0$.  If $v(\vt-b)>0$ then $\vt\equiv b$ is a unit and $\psi_1(\vt)\equiv \psi_1(b)$ is a unit.
\end{enumerate}
So $v(\vt-b)$ and $v(\psi(\vt))=v(\vt-b)+v(\psi_1(\vt))$ both lie in $\{0,v(\eta)\}\subset5\Z$.  As $v(80000)=0$, we get $v(E)=v(\vt-b)+3v(\psi(\vt))\equiv0\pmod5$, and $v(E)=0$ if moreover $p\nmid X$, because then $v(\eta)=0$.

\emph{Case $p\mid Z$.}  Then $p\nmid X$, and $v(\eta)=-7em$ with $m=v_p(Z)\ge1$.
\begin{enumerate}[label=(\alph*),leftmargin=2em]
\item If $v(\vt)\ge0$, the left side of~\eqref{eq:theta} is integral, so $v(4\vt-1)\ge7em>0$.  Then $\vt\equiv1/4$, so $\vt-b\equiv(1-4b)/4$ and $\psi(\vt)\equiv\psi(1/4)$ are units, and $v(E)=0$.
\item If $v(\vt)<0$, then as before $v(\psi(\vt))=3v(\vt)$ and $v(4\vt-1)=v(\vt)$, and~\eqref{eq:theta} gives $7v(\vt)=v(\eta)=-7em$, so $v(\vt)=-em$.  Then $v(\vt-b)=-em$ and $v(\psi(\vt))=-3em$, so $v(E)=-10em\equiv0\pmod5$.
\end{enumerate}
This proves (ii) and (iii).  The last assertion is (ii) applied to $K=\Lt$.
\end{proof}

\needspace{5\baselineskip}
\begin{proposition}\label{prop:D2}
For every $\vt\in\Lo$ with $\psi(\vt)\ne0$,
\[
\Norm{\Lt/\Lo}{80000\,(\vt-b)\,\psi(\vt)^3}=2^{21}\cdot5^{10}\cdot\psi(\vt)^{10}=2\cdot\big(2^4\cdot5^2\cdot\psi(\vt)^2\big)^5 .
\]
In particular $\Norm{\Lt/\Lo}{E}\equiv2$ modulo $\Lo^{\times5}$.
\end{proposition}

\begin{proof}
By~\eqref{eq:normpsi}, $\Norm{\Lt/\Lo}{\vt-b}=\psi(\vt)/25$, while $80000=2^7\cdot5^4$ and $\psi(\vt)$ lie in $\Lo$, so their norms are their cubes.  Hence the norm is $80000^3\cdot(\psi(\vt)/25)\cdot\psi(\vt)^9=2^{21}\cdot5^{10}\cdot\psi(\vt)^{10}$.
\end{proof}

\begin{remark}\label{rem:putzE}
In the notation of~\cite[\S4.5.2]{Putz} recalled in \cref{sec:related}, put $w=10\vt Z$.  Then $g_1(w,Z)=w-10bZ=10Z(\vt-b)$ and $g(w,Z)=w^3+8w^2Z+56wZ^2+560Z^3=40Z^3\psi(\vt)$, so
\[
E=\frac{g_1(w,Z)\,g(w,Z)^3}{8Z^{10}} .
\]
Here $80000=10\cdot40^3/8$, and the factor $1/8$ turns the power $2^{10}$ in $10\cdot40^3=2^{10}\cdot5^4$ into the power $2^7$ chosen in \cref{sec:idea}.  The product $g_1g^3$ is homogeneous of degree $10$, so if $w=dw_0$ and $Z=dz_0$ with $d\in\Lo^\times$, then modulo fifth powers $E$ is $4\,g_1(w_0,z_0)\,g(w_0,z_0)^3$, that is, Putz's linear factor $g_1(w_0,z_0)$ multiplied by the element $4\,g(w_0,z_0)^3$ of $\Lo$.  Since $\Norm{\Lt/\Lo}{g_1(w_0,z_0)}=g(w_0,z_0)$, the norm of $g_1(w_0,z_0)g(w_0,z_0)^3$ is $g(w_0,z_0)^{10}$, which is why the class of the norm of $E$ does not depend on the solution.
\end{remark}