\section{The valuations of \texorpdfstring{$E$}{E} at the primes above 2 and 5}\label{sec:local}

At the prime of $\Lt$ above $2$ the valuation of $E$ modulo $5$ follows from the norm identity (\cref{lem:two}).  At the seven primes above $5$ the values are the same for every solution (\cref{prop:five}) and give the condition (V) of \cref{sec:sieve}.  Without (V) the sieve does not eliminate every class (\cref{rem:robust}(d)).

\subsection{The prime above 2}

\begin{lemma}\label{lem:two}
Let $\PT_1$ be the prime of $\Lt$ above $2$.  For every $\vt\in\Lo$ with $\psi(\vt)\ne0$,
\[
v_{\PT_1}\big(80000\,(\vt-b)\,\psi(\vt)^3\big)\equiv3\pmod5 .
\]
\end{lemma}

\begin{proof}
As $\PT_1$ is the only prime of $\Lt$ above $2$ and $f(\PT_1/2)=1$, we have $v_{\PT_1}(\alpha)=v_2\big(\Norm{\Lt/\Q}{\alpha}\big)$ for $\alpha\in\Lt^\times$.  By \cref{prop:D2},
\[
\Norm{\Lt/\Q}{E}=\Norm{\Lo/\Q}{2^{21}\cdot5^{10}\cdot\psi(\vt)^{10}}=2^{168}\cdot5^{80}\cdot\Norm{\Lo/\Q}{\psi(\vt)}^{10},
\]
so $v_{\PT_1}(E)=168+10\,v_2\big(\Norm{\Lo/\Q}{\psi(\vt)}\big)\equiv168\equiv3\pmod5$.
\end{proof}

So the norm condition (N) of \cref{sec:sieve} already fixes the valuation at $\PT_1$ (\cref{rem:robust}(a)).

\subsection{The primes above 5}

Recall from \cref{lem:fields,lem:labels,lem:psi5} the primes $\pe_A,\pe_B,\pe_C$ of $\Lo$ above $5$, with ramification indices $5$, $1$, $2$, and the roots $b_1\in\Zp5$ and $b_2,b_3$ of $\psi$ over $\Qp5$.  For a prime $\PT$ of $\Lt$ above $5$ we have $v_\PT(\alpha)=e(\PT/5)\,v_5(\alpha)$, where $\alpha$ is viewed in the completion $\Ltp{\PT}$ and $v_5$ is extended to it.

\begin{proposition}\label{prop:five}
Let $(X,Y,Z)$ be a solution with $\Aeta\isom\Lo$, and let $\vt$ and $E$ be as in \cref{sec:descent}.  Then $5\nmid YZ$.  Put $n=v_5(\eta)=5v_5(X)\ge0$.  The valuations of $E$ at the seven primes of $\Lt$ above $5$ are
\begin{center}
\begin{tabular}{@{}lccccc@{}}
\toprule
label of $\PT$ & $(1,1,1)$ & $(2,1,1)$ & $(5,1,5)$ & $(10,1,5)$ & $(2,1,2)$, three primes\\
\midrule
$v_\PT(E)$ & $4n+12$ & $6n+18$ & $20$ & $30$ & $6n+30$, $8n+35$, $6n+35$\\
$v_\PT(E)\bmod5$ & $2$ & $3$ & $0$ & $0$ & $0$, $0$, $0$\\
\bottomrule
\end{tabular}
\end{center}
With the numbering of \cref{tab:Sprimes}: $v_{\PT_3}(E)\equiv2$, $v_{\PT_5}(E)\equiv3$, and $v_{\PT_i}(E)\equiv0\pmod5$ for $i\in\{2,4,6,7,8\}$.
\end{proposition}

The proof uses the hypothesis $\Aeta\isom\Lo$ only through the fact that $\vt$ generates $\Lo$, so that
\begin{equation}\label{eq:L8Q5}
\Lo\otimes\Qp5\isom\Qp5[t]/(\feta),\qquad t\mapsto\vt .
\end{equation}
The left side is the product of the three fields $\Lop{\pe_A}$, $\Lop{\pe_B}=\Qp5$, $\Lop{\pe_C}$, of degrees $5$, $1$, $2$ and ramification indices $5$, $1$, $2$.  Under~\eqref{eq:L8Q5} the image $\vt_\pe$ of $\vt$ in $\Lop{\pe}$ generates $\Lop{\pe}$ and is a root of $\feta$.  Its minimal polynomial over $\Qp5$ is an irreducible factor of $\feta$ of degree $[\Lop{\pe}:\Qp5]$, and different primes give different factors.  We write $\vt_A,\vt_B,\vt_C$ for the images of $\vt$ in $\Lop{\pe_A},\Lop{\pe_B},\Lop{\pe_C}$.  From~\eqref{eq:theta}, since $v_5(4)=0$, for $\bullet\in\{A,B,C\}$
\begin{equation}\label{eq:v5theta}
v_5(\psi(\vt_\bullet))=v_5(\eta)+v_5(4\vt_\bullet-1)-5v_5(\vt_\bullet),\qquad \psi(\vt_\bullet)=25(\vt_\bullet-b_1)(\vt_\bullet-b_2)(\vt_\bullet-b_3).
\end{equation}

\begin{proof}
\emph{Step 1: $5\nmid Z$.}  Suppose $m=v_5(Z)\ge1$, so $5\nmid X$ and $v_5(\eta)=-7m$.  The $5$-adic valuations of the coefficients of $\feta=100t^8+80t^7+56t^6+56t^5-4\eta t+\eta$ at $t^8,t^7,t^6,t^5,t^1,t^0$ are $2,1,0,0,-7m,-7m$.  The points $(5,0)$, $(6,0)$, $(7,1)$ lie above the segment joining $(1,-7m)$ to $(8,2)$, so the Newton polygon has vertices $(0,-7m)$, $(1,-7m)$, $(8,2)$.  The second segment has length $7$ and slope $(7m+2)/7$, whose denominator in lowest terms is $7$.  So $\Qp5[t]/(\feta)$ has a factor field whose ramification index is divisible by $7$, whereas the factors of $\Lo\otimes\Qp5$ have ramification indices $5$, $1$, $2$.  This contradicts~\eqref{eq:L8Q5}.

So $5\nmid Z$ and $n=v_5(\eta)=5v_5(X)\ge0$.  We now determine $\vt_A,\vt_B,\vt_C$.

\emph{Step 2: the case $n>0$.}  The valuations of the coefficients are $2,1,0,0$ at $t^8,t^7,t^6,t^5$ and $n,n$ at $t^1,t^0$.  The Newton polygon has vertices $(0,n)$, $(5,0)$, $(6,0)$, $(8,2)$ (the point $(1,n)$ lies above the first segment and $(7,1)$ lies on the last).  So $\feta=g_Ag_Bg_C$ over $\Qp5$ with $\deg g_A=5$, $\deg g_B=1$, $\deg g_C=2$, and the roots of $g_A$, $g_B$, $g_C$ have valuations $n/5$, $0$, $-1$.  The algebra $\prod\Qp5[t]/(g_\bullet)$ has at least three factor fields, with equality only if every $g_\bullet$ is irreducible.  By~\eqref{eq:L8Q5} it has exactly three, of degrees $5,1,2$.  Hence $\vt_A$, $\vt_B$, $\vt_C$ are roots of $g_A$, $g_B$, $g_C$: $v_5(\vt_A)=n/5$, $\vt_B\in\Zp5^\times$, $v_5(\vt_C)=-1$.

\emph{Step 3: the case $n=0$.}  Now $\eta$ is a $5$-adic unit.  The Newton polygon has vertices $(0,0)$, $(6,0)$, $(8,2)$: six unit roots and two roots of valuation $-1$.  The two roots of valuation $-1$ form a factor of $\feta$ of degree $2$.  By~\eqref{eq:L8Q5} the corresponding factor of $\Qp5[t]/(\feta)$ is a product of some of the fields $\Lop{\pe}$, so it is $\Lop{\pe_C}$, the only one of degree $2$.  So $v_5(\vt_C)=-1$, while $\vt_B\in\Zp5^\times$ and $\vt_A$ is a unit generating the totally ramified quintic field $\Lop{\pe_A}$.  Modulo $5$,
\[
\feta\equiv t^6+t^5+\eta t+\eta=(t+1)(t^5+\eta)\equiv(t+1)(t+\eta)^5 ,
\]
so every unit root is congruent to $-1$ or to $-\eta$ modulo the maximal ideal.

For a unit root $\vt_\bullet\in\{\vt_A,\vt_B\}$ we have $v_5(\vt_\bullet-b_2)=v_5(\vt_\bullet-b_3)=-1$, so $v_5(\psi(\vt_\bullet))=v_5(\vt_\bullet-b_1)$, and~\eqref{eq:v5theta} gives $v_5(\psi(\vt_\bullet))=v_5(4\vt_\bullet-1)$.  Hence
\begin{equation}\label{eq:unitroots}
v_5(\vt_\bullet-b_1)=v_5(\psi(\vt_\bullet))=v_5(\vt_\bullet-\tfrac14)\qquad\text{for }\vt_\bullet\in\{\vt_A,\vt_B\}\text{ (when }n=0\text{)}.
\end{equation}
Put $F(s)=\feta(\tfrac14+s)=\sum_jc_js^j$.  Its roots are the numbers $\vt-\tfrac14$ for the roots $\vt$ of $\feta$, and all $c_j$ are $5$-integral.  The constant term $c_0=\feta(\tfrac14)=\psi(\tfrac14)/256=1225/16384$ has $v_5(c_0)=2$, and
\[
c_1=\feta'(\tfrac14)=\tfrac{823}{512}-4\eta=-\tfrac{1225}{512}-4(\eta-1).
\]
The product of the roots of $F$ is $c_0/100$, of valuation $0$, and the two roots coming from $\vt_C$ and its conjugate have valuation $-1$ each.  So the valuations $v_5(\vt-\frac14)$ of the six unit roots add up to $2$.

\emph{Step 4: $5\nmid Y$.}  Suppose $5\mid Y$.  Then $n=0$ and $v_5(1-\eta)=v_5(Y^2/Z^7)=2v_5(Y)\ge2$.  So $\eta\equiv1\pmod5$, and all six unit roots are $\equiv-1\equiv\frac14$ modulo the maximal ideal.  Thus $w_B=v_5(\vt_B-\frac14)$ is a positive integer ($\vt_B\ne\frac14$ as $c_0\ne0$), and $w_A=v_5(\vt_A-\frac14)$, the same for the five conjugates of $\vt_A$ over $\Qp5$, is a positive element of $\frac15\Z$ because $\Lop{\pe_A}$ has ramification index $5$.  From $w_B+5w_A=2$ we get $w_B=1$ and $w_A=\frac15$.  So $F$ has exactly one root of valuation $1$ and no root of larger valuation, and the first segment of its Newton polygon joins $(0,2)$ to $(1,1)$, so $v_5(c_1)=1$.  But $v_5(1225)=2$ and $v_5(\eta-1)\ge2$ give $v_5(c_1)\ge2$.  This contradiction shows $5\nmid Y$.

\emph{Step 5: the case $n=0$, continued.}  Now $5\nmid XYZ$, so $\eta\not\equiv0,1\pmod5$.  The factors $t+1$ and $(t+\eta)^5$ are coprime modulo $5$, so by Hensel's lemma the factor of $\feta$ with the six unit roots is the product over $\Qp5$ of a linear factor with root $\equiv-1$ and a quintic factor with roots $\equiv-\eta$.  Comparing degrees with $\Lop{\pe_B}$ and $\Lop{\pe_A}$, we get $\vt_B\equiv-1$ and $\vt_A\equiv-\eta$.  Hence $\vt_A\not\equiv\frac14$, that is, $v_5(\vt_A-\frac14)=0$.  Also $v_5(c_1)=v_5(4(\eta-1))=0$.  The Newton polygon of $F$ therefore begins with the segment from $(0,2)$ to $(1,0)$: exactly one root of $F$ has positive valuation, and its valuation is $2$.  This root is $\vt_B-\frac14$, so by~\eqref{eq:unitroots}
\[
v_5(\vt_B-b_1)=v_5(\psi(\vt_B))=2,\qquad v_5(\vt_A-b_1)=v_5(\psi(\vt_A))=0 .
\]

\emph{Step 6: the valuations of $\vt-b_i$ and $\psi(\vt)$.}  We show, for all $n\ge0$:
\begin{enumerate}[label=(\alph*),leftmargin=2em]
\item $v_5(\vt_A-b_1)=0$, $v_5(\vt_A-b_2)=v_5(\vt_A-b_3)=-1$ and $v_5(\psi(\vt_A))=0$;
\item $v_5(\vt_B-b_1)=v_5(\psi(\vt_B))=n+2$, and $v_5(\vt_B-b_2)=v_5(\vt_B-b_3)=-1$;
\item $v_5(\vt_C-b_1)=-1$, $v_5(\psi(\vt_C))=n+4$, and $\{v_5(\vt_C-b_2),v_5(\vt_C-b_3)\}=\{n+\tfrac32,\tfrac32\}$.
\end{enumerate}
For $n=0$, (a) and the first part of (b) are Step 5.  For $n>0$: in (a), $v_5(\vt_A)=n/5>0$, so $\vt_A-b_1\equiv-b_1$ is a unit and $\psi(\vt_A)\equiv14$ is a unit.  In (b), put $s=\vt_B-b_1$.  By~\eqref{eq:v5theta}, $v_5(s)=v_5(\psi(\vt_B))=n+v_5(4\vt_B-1)$, where $4\vt_B-1=(4b_1-1)+4s$ and $v_5(4b_1-1)=2$.  If $v_5(s)<2$ then $v_5(4\vt_B-1)=v_5(s)$ and $n=0$.  If $v_5(s)=2$ then $v_5(4\vt_B-1)\ge2$ and $v_5(s)\ge n+2>2$.  Both are impossible, so $v_5(s)>2$, $v_5(4\vt_B-1)=2$ and $v_5(s)=n+2$.  The remaining statements in (a) and (b) hold because $\vt_A,\vt_B$ are integral and $v_5(b_2)=v_5(b_3)=-1$.  In (c), for all $n\ge0$: $v_5(\vt_C)=-1$ gives $v_5(4\vt_C-1)=-1$ and $v_5(\vt_C-b_1)=-1$, so by~\eqref{eq:v5theta} $v_5(\psi(\vt_C))=n-1+5=n+4$ and
\[
v_5(\vt_C-b_2)+v_5(\vt_C-b_3)=n+4-2+1=n+3 .
\]
Since $v_5(b_2-b_3)=\frac32$ (\cref{lem:psi5}), if one of the two valuations exceeds $\frac32$ the other equals $\frac32$.  If neither does, their sum is at most $3$.  For $n>0$ this gives $\{n+\frac32,\frac32\}$.  For $n=0$ the sum is $3$ and both equal $\frac32$.

\emph{Step 7: the primes of $\Lt$ above $5$.}  As in the proof of \cref{lem:labels}, each prime $\PT$ of $\Lt$ above $5$ corresponds to a factor of $\Lop{\pe}\otimes(\Qp5\times\Qp5(b_2))$ for $\pe\in\{\pe_A,\pe_B,\pe_C\}$, and in $\Ltp{\PT}$ the elements $\vt$ and $b$ map to $\vt_\pe$ and to $b_1$ (in the factor $\Lop{\pe}\otimes\Qp5$) or to $b_2$ or $b_3$ (in the other).  By \cref{lem:labels}, $\Lop{\pe_C}\isom\Qp5(b_2)$, so $\Lop{\pe_C}\otimes\Qp5(b_2)$ is a product of two copies of $\Lop{\pe_C}$, in which $b$ maps to $b_2$ and to $b_3$ respectively.  All three primes above $\pe_C$ have $e(\PT/5)=2$.  Using $v_\PT(E)=e(\PT/5)\big(v_5(80000)+v_5(\vt_\pe-b_i)+3v_5(\psi(\vt_\pe))\big)$, where $b_i$ is the image of $b$ in $\Ltp{\PT}$, with $v_5(80000)=4$, Step 6 gives:
\begin{itemize}[leftmargin=2em]
\item above $\pe_B$: at $(1,1,1)$, $b\mapsto b_1$ and $v_\PT(E)=4+(n+2)+3(n+2)=4n+12$; at $(2,1,1)$, $b\mapsto b_2$ and $v_\PT(E)=2\,(4-1+3(n+2))=6n+18$;
\item above $\pe_A$: at $(5,1,5)$, $b\mapsto b_1$ and $v_\PT(E)=5\cdot4=20$; at $(10,1,5)$, $b\mapsto b_2$ and $v_\PT(E)=10\,(4-1)=30$;
\item above $\pe_C$: with $b\mapsto b_1$, $v_\PT(E)=2\,(4-1+3(n+4))=6n+30$; with $b$ mapped to a root $b_i\in\{b_2,b_3\}$ with $v_5(\vt_C-b_i)=n+\frac32$, $v_\PT(E)=2\,(4+n+\frac32+3(n+4))=8n+35$; with $b$ mapped to the other one, $v_\PT(E)=2\,(4+\frac32+3(n+4))=6n+35$.
\end{itemize}
Since $n=5v_5(X)\equiv0\pmod5$, the residues modulo $5$ are as stated.
\end{proof}

\begin{remark}
The labels do not distinguish the three primes with label $(2,1,2)$, and the proof does not determine which of them carries which of the values $6n+30$, $8n+35$, $6n+35$.  All three values are $\equiv0\pmod5$, so condition (V) is the same for every assignment.  Condition (V) concerns only the primes above $5$: the valuations of $E$ at the primes above $7$ are left free.
\end{remark}