\section{Remarks}\label{sec:remarks}
We explain why the Catalan solution $1^3+2^3=3^2$ plays no part in the proofs, how the arguments of Parts~I--III carry over to the exponents $p\equiv2\pmod3$, and which questions remain open.

\subsection{The obstructing point}
The Catalan solution $1^3+2^3=3^2$, of exponent $2$, gives, with $(u,w)=(1,2)$, the parameter $s_0=9$, and with $(u,w)=(2,1)$ the parameter $s_0=9/8$; in both cases $v_3(s_0)=2$, outside the disc of \cref{thm:inertia3}.  So nothing in the proofs concerns these points, as must be the case, since $x^3+y^3=z^2$ has this solution.  The exponent enters through $v_3(s_0)=p\,v_3(c)$, and also through the conditions $p\ge17$ in Kraus's results, $p\notin\{2,3,7,13\}$ in \cref{thm:irreducible}, $p\notin\{2,3,13\}$ in \cref{lem:lifts-at-3}, and $p>8$ in the theorem of Calegari, Emerton and Gee.

\subsection{Exponents \texorpdfstring{$p\equiv2\pmod 3$}{p = 2 mod 3}}
For $p\equiv2\pmod 3$ the prime $p$ is inert in $K$.  For $x^3+y^3=z^p$ these exponents are settled by Freitas's theorem \cite[Theorem~3]{Freitas16}.  Part~IV carries out the arguments of Parts~I--III at an inert prime, for $x^3+y^3=3z^p$ (\crefrange{sec:3zp-frey}{sec:3zp-three}).  None of the arguments at $\p$, $\lp$ and $\lpb$, nor those at $5$ in Part~II, depends on whether $p$ splits in $K$.  This transfer also proves \cref{thm:main-uncond} for the prime exponents $p\ge17$ with $p\equiv2\pmod 3$, without Freitas's theorem: \cref{prop:kraus,lem:s0} hold for every $p\ge17$, and for such $p$ the proof of \cref{prop:local}, with \cref{lem:3zp-crys} in place of \cref{prop:crys}, together with \cref{lem:cubic-local,prop:3zp-irreducible}, shows that the Frey representation, defined as in \cref{def:frey}, has all the properties that \cref{prop:3zp-noF} excludes.  That proof rests on \cref{lem:3zp-lift}; \cref{cor:three-all} uses Freitas's theorem instead, so that \cref{thm:main-uncond} does not depend on Part~IV.

\subsection{Open questions}
Three questions remain open.  The first is to find all solutions of $x^3+y^3=9z^4$ (\cref{rem:3zp-open}).  The second concerns the equations $x^3+y^3=3^kz^p$ with $k\ge2$ that \cref{rem:3zp-k} leaves open, among them those with $3\le k\le7$, $3\nmid z$ and $p>10^6$, on whose parameters \cref{thm:inertia3} gives no information.  The third is whether the radius $15/2$ of \cref{thm:inertia3} is optimal (\cref{rem:radius}).  Beyond these questions, the method relies on features of this case: the divisibility $3\mid c$ (\cref{prop:kraus}, \cref{lem:3zp-elementary}), which puts the parameter of a solution in the disc of \cref{thm:inertia3}; a reasonable residual representation, so that a compatible system exists (\cref{thm:CEG,lem:3zp-lift}); the Hodge--Tate weights $\{0,1,2\}$, which lie in the Fontaine--Laffaille range at $5$ and make the member at $3$ ordinary once its monodromy operator is regular (\cref{prop:three-ordinary}); and two features of $K$: its single prime above $3$ and its class number one imply that an inertia group at $\p$ is the whole Galois group of the maximal pro-$3$ extension of $K$ unramified outside $\p$ (\cref{lem:local-generation}), and the ramification at $\p$ bounds the root discriminants of the fields cut out by the reductions of the member by $3^{449/162}<21.01$ (\cref{prop:three-rd}), below Odlyzko's bounds for fields of large degree.  \Cref{thm:inertia3} and \cref{prop:three-scope}, by contrast, are statements about $\VV$ and about representations of $G_K$, not about the equation.  For another equation the same route would need a family that separates the solutions from the obstructing ones at some prime, as \cref{thm:inertia3} does here at $\p$, and then analogues of the other features above.