\section{Putz's theorem and the fields \texorpdfstring{$\Lo$}{L8} and \texorpdfstring{$\Lt$}{L24}}\label{sec:putz} This section states Putz's theorem and collects the facts about $\Lo$ and $\Lt$ that the later sections use: their degrees, signatures and discriminants (\cref{lem:fields}), the primes of $\Lt$ above $2$, $5$ and $7$ (\cref{lem:labels}), and the roots of $\psi$ over $\Qp5$ (\cref{lem:psi5}), on which the $5$-adic computations of \cref{sec:local} rest. \subsection{The fibre algebra} We keep the notation $\psi$, $\Psi$, $\feta$, $\Aeta$ of~\eqref{eq:psi}--\eqref{eq:Aeta}. For a variable $s$ put $f_s(t)=4t^5\psi(t)-s(4t-1)$. \begin{lemma}\label{lem:disc} The discriminant of $f_s$ with respect to $t$ is $-2^{28}\cdot5^{12}\cdot7^7\cdot s^4(s-1)^4$. In particular, for $\eta\in\Q\smallsetminus\{0,1\}$ the polynomial $\feta$ is separable of degree $8$, and $\Aeta$ is an \'etale $\Q$-algebra of dimension $8$. \end{lemma} \begin{proof} A direct computation (\cref{app:computations}) gives this formula, which agrees with~\cite[Lemma~4.4]{Putz}. The leading coefficient of $f_s$ is $100$, so $\feta$ has degree $8$. For $\eta\notin\{0,1\}$ the discriminant is nonzero, so $\feta$ is separable, and a separable polynomial defines an \'etale algebra. \end{proof} A solution of~\eqref{eq:main} in our sense is what Putz~\cite[\S1.1 and \S4.1]{Putz} calls a nontrivial solution, written $(x,y,z)$ there. He attaches to it $\eta=X^5/Z^7$ and the algebra $\Q[t]/(\varphi_0(t)-\eta\varphi_\infty(t))$ with $\varphi_0(t)=4t^5\psi(t)$ and $\varphi_\infty(t)=4t-1$, which is our $\Aeta$, and proves: \begin{theorem}[Putz {\cite[Theorem~4.33, p.~141]{Putz}}]\label{thm:putz} Let $(X,Y,Z)$ be a solution of~\eqref{eq:main}. Then $\Aeta$ is isomorphic to the field $\Lo=\Q[x]/(h(x))$ of~\eqref{eq:L8}. \end{theorem} \begin{remark}\label{rem:putzinputs} This is the only result of~\cite{Putz} that we use. The congruence conditions of~\cite[Theorem~4.34 and Corollary~4.35]{Putz} play no role: \cref{prop:five} below recovers $5\nmid YZ$ from the isomorphism $\Aeta\isom\Lo$ alone, and the valuation of $E$ at the prime above $2$ follows from the norm identity of \cref{prop:D2} (\cref{lem:two}). The thesis states its $2$-adic conditions twice, and the two statements differ. In its cases (a) and (b), \cite[Theorem~4.2(i)]{Putz} has $Z\equiv3$ and $Z\equiv7\pmod8$, where~\cite[Theorem~4.34(i)]{Putz} has $Z\equiv5$ and $Z\equiv1\pmod8$. The proof of~\cite[Theorem~4.34(i)]{Putz} describes the two cases by $v_2(1-\eta)=2$ with $1-\eta\equiv5$ modulo squares, and by $v_2(1-\eta)\ge4$ with $1-\eta\equiv1$ modulo squares. In both cases $v_2(1-\eta)>0$, so $Z$ is odd, because $1-\eta=Y^2/Z^7$ and $Y$, $Z$ are coprime. Then $1-\eta$ is $Z$ times a square in $\Qp2$, and the two cases give $Z\equiv5$ and $Z\equiv1\pmod8$, as in~\cite[Theorem~4.34(i)]{Putz}. So the congruences in~\cite[Theorem~4.2(i)]{Putz} appear to be misprints. \end{remark} \subsection{The fields}\label{sec:fields} Let $a$ be a root of $h$, so that $\Lo=\Q(a)$, and let $b$ be a root of $\psi$ in an algebraic closure of $\Lo$. \begin{lemma}\label{lem:fields} \begin{enumerate}[label=\textup{(\alph*)},leftmargin=2.2em] \item The polynomial $h$ is irreducible. The field $\Lo$ has signature $(2,3)$ and discriminant $-2^{14}\cdot5^{6}\cdot7^{7}$, and the index of $\Z[a]$ in the ring of integers of $\Lo$ is $2^7\cdot5\cdot41$. \item The primes $2$, $5$, $7$ factor in $\Lo$ as \[ 2=\pe_2^{8},\qquad 5=\pe_A^{5}\,\pe_B\,\pe_C^{2},\qquad 7=\pe_7\,\pe_7'^{\,7}, \] where all the primes shown have residue degree $1$. \item The polynomial $\psi$ is irreducible over $\Lo$. Hence $\Lt=\Lo(b)$ has degree $24$, and for $t\in\Lo$ \begin{equation}\label{eq:normpsi} \Norm{\Lt/\Lo}{t-b}=(t-b_1)(t-b_2)(t-b_3)=\psi(t)/25, \end{equation} where $b_1,b_2,b_3$ are the roots of $\psi$. \item The field $\Lt$ has signature $(2,11)$ and discriminant $-2^{44}\cdot5^{20}\cdot7^{23}$. The element $5b$ is an algebraic integer, a root of $x^3+4x^2+14x+70$. \end{enumerate} \end{lemma} \begin{proof} Parts (a), (b) and the discriminant in (d) are computations with PARI/GP, listed in \cref{app:computations}: \texttt{nfinit} and \texttt{idealprimedec} compute the rings of integers and the prime decompositions, and \texttt{nfcertify} proves that the computed integral bases are bases of the rings of integers (we then call the maximal orders certified). For (c): the coefficients $20$, $14$, $14$ of $\psi$ are even, $14$ is not divisible by $4$ and $25$ is odd, so $\psi$ is Eisenstein at $2$ and $\Q(b)$ is a cubic field. If $\psi$ were reducible over $\Lo$, it would have a root there, and the cubic field $\Q(b)$ would be a subfield of the octic field $\Lo$, which is impossible as $3\nmid8$. Since the roots of $\psi$ are the conjugates of $b$ over $\Lo$, \eqref{eq:normpsi} follows. For (d): $\psi$ has negative discriminant $-2^2\cdot5^7\cdot7$, so it has one real root. Each of the two real embeddings of $\Lo$ therefore extends to exactly one real embedding of $\Lt$, and $\Lt$ has $2$ real and $(24-2)/2=11$ pairs of complex embeddings. Finally $5\,\psi(x/5)=x^3+4x^2+14x+70$. \end{proof} We will use the following norms from $\Q(b)$ to $\Q$, all read off from $\psi$ (or its resultants): \begin{equation}\label{eq:bnorms} \Norm{}{b}=-\tfrac{14}{25},\qquad \Norm{}{1-4b}=\tfrac{64\,\psi(1/4)}{25}=49,\qquad \Norm{}{\psi'(b)}=2^2\cdot5^5\cdot7, \end{equation} together with $\psi(0)=14$ and $\psi(1/4)=5^2\cdot7^2/2^6$. \subsection{The primes of \texorpdfstring{$\Lt$}{L24} above 2, 5 and 7}\label{sec:Sprimes} For a prime $\PT$ of $\Lt$ above a rational prime $p$, with $\pe=\PT\cap\Lo$, we call the triple \[ \big(e(\PT/p),\ f(\PT/p),\ e(\pe/p)\big) \] the \emph{label} of $\PT$. \begin{lemma}\label{lem:labels} There are twelve primes of $\Lt$ above $2$, $5$ and $7$, with the following labels: \begin{center} \begin{tabular}{@{}cl@{}} \toprule $p$ & labels of the primes of $\Lt$ above $p$\\ \midrule $2$ & $(24,1,8)$\\ $5$ & $(1,1,1)$, $(2,1,1)$, $(5,1,5)$, $(10,1,5)$, and $(2,1,2)$ three times\\ $7$ & $(1,1,1)$, $(2,1,1)$, $(7,1,7)$, $(14,1,7)$\\ \bottomrule \end{tabular} \end{center} In particular the prime $\pe_C$ of $\Lo$ above $5$ splits into three primes of $\Lt$, each of ramification index and residue degree $1$ over $\pe_C$. So $\psi$ splits completely over the completion $\Lop{\pe_C}$. \end{lemma} \begin{proof} The labels are computed by \texttt{idealprimedec} (\cref{app:computations}). They can also be seen by hand from the local structure of $\psi$ and $\Lo$. Since $\Lt=\Lo\otimes_\Q\Q(b)$, for each prime $p$ we have $\Lt\otimes\Qp{p}=(\Lo\otimes\Qp{p})\otimes_{\Qp{p}}(\Q(b)\otimes\Qp{p})$. At $2$, $\psi$ is Eisenstein and $\Lo\otimes\Qp2$ is a totally ramified field of degree $8$. The ramification indices $3$ and $8$ are coprime, so the tensor product is a field with $e=24$. At $5$, $\Q(b)\otimes\Qp5=\Qp5\times\Qp5(b_2)$ with $\Qp5(b_2)$ ramified quadratic (\cref{lem:psi5} below), and $\Lo\otimes\Qp5=\Lop{\pe_A}\times\Lop{\pe_B}\times\Lop{\pe_C}$ with $\Lop{\pe_B}=\Qp5$ and $\Lop{\pe_A}$, $\Lop{\pe_C}$ totally ramified of degrees $5$ and $2$. The factors are $\Lop{\pe_B}\otimes\Qp5$ (label $(1,1,1)$), $\Lop{\pe_B}\otimes\Qp5(b_2)$ (label $(2,1,1)$), $\Lop{\pe_A}\otimes\Qp5$ (label $(5,1,5)$), $\Lop{\pe_A}\otimes\Qp5(b_2)$ (a field with $e=10$, label $(10,1,5)$), $\Lop{\pe_C}\otimes\Qp5$ (label $(2,1,2)$), and $\Lop{\pe_C}\otimes\Qp5(b_2)$. For the last one, $\Lop{\pe_C}\isom\Qp5(b_2)$, so it is a product of two fields, each isomorphic to $\Lop{\pe_C}$ (as the labels computed by \texttt{idealprimedec} also show). Indeed, $h(y-3)=y^8-20y^7+140y^6-420y^5+490y^4-28y^3-420y^2+420y-175$ has $5$-adic Newton polygon with vertices $(0,2)$, $(1,1)$, $(3,0)$, $(8,0)$, so its two roots of valuation $1/2$, which generate $\Lop{\pe_C}$, satisfy $y^2=15u$ with $u\equiv1$ modulo the maximal ideal (the terms $420y$ and $-28y^3$ dominate), and $\Lop{\pe_C}=\Qp5(\sqrt{15})$; and since $b_1\in\Qp5$ and $\disc\psi=25^2\psi'(b_1)^2(b_2-b_3)^2$, we have $\Qp5(b_2)=\Qp5(\sqrt{\disc\psi})=\Qp5(\sqrt{-35})=\Qp5(\sqrt{15})$, because $-35/15=-7/3\equiv1\pmod5$ is a square in $\Qp5$. At $7$, $\psi$ has one root in $\Zp7$ and two roots generating a ramified quadratic extension, and $\Lo\otimes\Qp7$ is the product of $\Qp7$ and a totally ramified field of degree $7$. The four products give the four labels. \end{proof} We number the twelve primes $\PT_1,\dots,\PT_{12}$ as in \cref{tab:Sprimes}. Each $\PT_i$ is generated by the element $B_i$ of \cref{tab:B} (\cref{prop:basis}(i)). The two primes above $5$ with labels $(1,1,1)$ and $(2,1,1)$ are $\PT_3$ and $\PT_5$. \begin{table}[ht] \centering \small \begin{tabular}{@{}c c c@{\qquad\qquad}c c c@{}} \toprule prime & $p$ & label & prime & $p$ & label\\ \midrule $\PT_{1}$ & $2$ & $(24,1,8)$ & $\PT_{7}$ & $5$ & $(5,1,5)$\\ $\PT_{2}$ & $5$ & $(2,1,2)$ & $\PT_{8}$ & $5$ & $(2,1,2)$\\ $\PT_{3}$ & $5$ & $(1,1,1)$ & $\PT_{9}$ & $7$ & $(1,1,1)$\\ $\PT_{4}$ & $5$ & $(2,1,2)$ & $\PT_{10}$ & $7$ & $(7,1,7)$\\ $\PT_{5}$ & $5$ & $(2,1,1)$ & $\PT_{11}$ & $7$ & $(14,1,7)$\\ $\PT_{6}$ & $5$ & $(10,1,5)$ & $\PT_{12}$ & $7$ & $(2,1,1)$\\ \bottomrule \end{tabular} \vspace{3pt} \caption{The primes $\PT_i=B_i\mathcal O_{\Lt}$ of $\Lt$ above $2$, $5$ and $7$ and their labels $(e(\PT_i/p),f(\PT_i/p),e(\PT_i\cap\Lo/p))$.}\label{tab:Sprimes} \end{table} From now on $v_5$ denotes the valuation on an algebraic closure of $\Qp5$ with $v_5(5)=1$. \begin{lemma}\label{lem:psi5} Over $\Qp5$, the polynomial $\psi$ has exactly one root $b_1\in\Zp5$; it satisfies $b_1\equiv-1\pmod5$, $v_5(4b_1-1)=2$ and $v_5(\psi'(b_1))=0$. The other two roots $b_2$, $b_3$ have $v_5(b_2)=v_5(b_3)=-1$ and $v_5(b_2-b_3)=3/2$, and $\Qp5(b_2)=\Qp5(b_3)$ is a ramified quadratic extension of $\Qp5$. \end{lemma} \begin{proof} The $5$-adic valuations of the coefficients of $\psi$ at $t^0,t^1,t^2,t^3$ are $0,0,1,2$, so the Newton polygon has vertices $(0,0)$, $(1,0)$, $(3,2)$: one root of valuation $0$ and two of valuation $-1$. Modulo $5$, $\psi\equiv14(t+1)$, so the unit root is a simple root modulo $5$, lies in $\Zp5$ by Hensel's lemma, and is $\equiv-1$. Since $\prod_i(1-4b_i)=49$ is a $5$-adic unit and $v_5(1-4b_2)=v_5(1-4b_3)=-1$, we get $v_5(1-4b_1)=2$. Next, $\psi'(b_1)=25(b_1-b_2)(b_1-b_3)$ has valuation $2-1-1=0$. From $\disc\psi=-2^2\cdot5^7\cdot7=25^4\prod_{i