#  Asymptotic Fermat for signatures $(p,p,2)$ and $(p,p,3)$ over totally real
fields

Diana Mocanu  Diana Mocanu

###### Abstract

Let $K$ be a totally real number field and consider a Fermat-type equation
$Aa^{p}+Bb^{q}=Cc^{r}$ over $K$. We call the triple of exponents $(p,q,r)$ the
signature of the equation. We prove various results concerning the solutions
to the Fermat equation with signature $(p,p,2)$ and $(p,p,3)$ using a method
involving modularity, level lowering and image of inertia comparison. These
generalize and extend the recent work of Işik, Kara and Özman. For example,
consider $K$ a totally real field of degree $n$ with $2\nmid h_{K}^{+}$ and
$2$ inert. Moreover, suppose there is a prime $q\geq 5$ which totally ramifies
in $K$ and satisfies $\gcd(n,q-1)=1$, then we know that the equation
$a^{p}+b^{p}=c^{2}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $2|b$ for $p$ sufficiently large.

##  1 Introduction

###  1.1 Historical background

The study of Diophantine equations is of great interest in Mathematics and
goes back to antiquity. The most famous example of a Diophantine equation
appears in Fermat’s Last Theorem. This is the statement, asserted by Fermat in
1637 without proof, that the Diophantine equation $a^{n}+b^{n}=c^{n}$ has no
solutions in whole numbers when $n$ is at least $3$, other than the trivial
solutions which arise when $abc=0$. Andrew Wiles famously proved the Fermat’s
Last Theorem in 1995 in his paper "Modular elliptic curves and Fermat’s Last
Theorem" [35]. The proof is by contradiction employing techniques from
algebraic geometry and number theory to prove a special case of the modularity
theorem for elliptic curves, which together with Ribet’s level lowering
theorem gives the long-waited result. Since then, number theorists extensively
studied Diophantine equations using Wiles’ modularity approach. Siksek gives a
comprehensive survey about this method over the field of rationals in [26].

Even before Wiles announced his proof, various generalizations of Fermat’s
Last Theorem had already been considered, which are of the shape

$Aa^{p}+Bb^{q}=Cc^{r}$ (1)

for fixed integers $A,B$ and $C$. We call $(p,q,r)$ the signature of the
equation (1). A primitive solution $(a,b,c)$ is a solution where $a,b$ and $c$
are pairwise coprime and a non-trivial solution $(a,b,c)$ is a solution where
$abc\neq 0$.

In [16], Işik, Kara and Özman list all known cases where equation (1) has been
solved over the rational integers in two tables (p.4). Table 1 contains all
unconditional results for infinitely many primes. In Table 2, they give all
conditional results. We highlight here one relevant family of solutions,
namely $(n,n,k)$ where $k\in\\{2,3\\}$. Darmon and Merel [6] and Poonen [20]
proved the following theorem:

######  Theorem 1 (Darmon and Merel).

  1. (i)

The equation $a^{n}+b^{n}=c^{2}$ has no non-trivial primitive integer
solutions for $n\geq 4$.

  2. (ii)

The equation $a^{n}+b^{n}=c^{3}$ has no non-trivial primitive integer
solutions for $n\geq 3$.

Note that the above equations, typically have infinitely many non-primitive
solutions. For example, if $n$ is odd, and $a$ and $b$ are any two integers
with $a^{n}+b^{n}=c$, then

$(ac)^{n}+(bc)^{n}=(c^{\frac{n+1}{2}})^{2}$

giving a rather uninteresting supply of solutions. Thus, we would only study
the primitive solutions of the above equations.

A naive sketch of the proof of Theorem 1 is as follows. First note that it is
enough to prove the assumption for $n=p$ an odd prime. Suppose
$a,b,c\in\mathbb{Z}$ is a non-trivial, primitive solution to (i) or (ii). In
each of the cases, we can associate a so-called Frey elliptic curve
$E_{a,b,c}/\mathbb{Q}$ and let $\overline{\rho}_{E,p}$ be its$\mod p$ Galois
representation, where $E=E_{a,b,c}$. Then $\overline{\rho}_{E,p}$ is
irreducible by Mazur [19] and modular by Wiles and Taylor [35] and [30].
Applying Ribet’s level lowering theorem [22] one gets that that
$\overline{\rho}_{E,p}$ arises from a weight $2$ newform of level $32$ for (i)
and level $27$ for (ii). These are closely related to the modular curves
$X_{0}(32)$ and $X_{0}(27)$ which turn out to be elliptic curves with complex
multiplication. Darmon and Merel prove in [6], by using the theory of complex
multiplication that this implies $j_{E}\in\mathbb{Z}[\frac{1}{p}]$ for $p>7$,
which gives a contradiction. The cases when $p\leq 7$ are treated in a more
elementary way by Poonen [20].

Recently, important progress has been done towards generalisation of the
modularity approach over larger number fields. In [12] Freitas and Siksek
proved the asymptotic Fermat’s Last Theorem (AFLT) for certain totally real
fields $K.$ That is, they showed that there is a constant $B_{K}$ such that
for any prime $p>B_{K}$, the only solutions to the Fermat equation
$a^{p}+b^{p}+c^{p}=0$ where $a,b,c\in\mathcal{O}_{K}$ are the trivial ones
i.e. the ones satisfying $abc=0$. Then, Deconinck [7] extended the results of
Freitas and Siksek [12] to the generalized Fermat equation of the form
$Aa^{p}+Bb^{p}+Cc^{p}=0$ where $A,B,C$ are odd integers belonging to a totally
real field. Later in [25] Şengün and Siksek proved the asymptotic FLT for any
number field $K$ by assuming modularity. This result has been generalized by
Kara and Özman in [15] to the case of the generalized Fermat equation. Also,
recently in [31] and [32] Turcas studied Fermat equation over imaginary
quadratic field $\mathbb{Q}(\sqrt{-d})$ with class number one.

We now present a result by Işik, Kara and Özman, proved in [16] which serves
as the starting point of this paper. It gives a computable criteria of testing
if the asymptotic Fermat Last Theorem holds for certain type of solutions of
the equations with signatures $(p,p,2)$. To state it, we need the following
notation:

$S_{K}:=\\{\mathfrak{P}\>:\mathfrak{P}\text{ is a prime of }K\text{ above
}2\\},\>\>\>T_{K}:=\\{\mathfrak{P}\in S_{K}:f(\mathfrak{P}/2)=1\\},$
$W_{K}:=\\{(a,b,c)\in\mathcal{O}_{K}^{3}:a^{p}+b^{p}=c^{2}\text{ with
}\mathfrak{P}|b\text{ for every }\mathfrak{P}\in T_{K}\\}$

where $f(\mathfrak{P}/2)$ denotes the residual degree of $\mathfrak{P}$.

######  Theorem 2 (Işik, Kara and Özman).

Let $K$ be a totally real number field with narrow class number $h_{K}^{+}=1$.
For each $a\in K(S_{K},2)$, let $L=K(\sqrt{a})$.
(A):Suppose that for every solution $(\lambda,\mu)$ to the $S_{K}$-unit
equation

$\lambda+\mu=1,\>\lambda,\mu\in\mathcal{O}_{S_{K}}^{*}$

there is some $\mathfrak{P}\in T_{K}$ that satisfies
max{$|v_{\mathfrak{P}}(\lambda)|,|v_{\mathfrak{P}}(\mu)|$}$\leq
4v_{\mathfrak{P}}(2)$.
(B):Suppose also that for each $L$, for every solution $(\lambda,\mu)$ of the
$S_{L}$-unit equation $\lambda+\mu=1,\>\lambda,\mu\in\mathcal{O}_{S_{L}}^{*}$,
there is some $\mathfrak{P}^{\prime}\in T_{L}$ that satisfies
max{$|v_{\mathfrak{P^{\prime}}}(\lambda)|,|v_{\mathfrak{P^{\prime}}}(\mu)|$}$\leq
4v_{\mathfrak{P^{\prime}}}(2)$.
Then, there is a constant $B_{K}$ (depending only on $K$) such that for each
$p>B_{K}$, the equation $a^{p}+b^{p}=c^{2}$ has no primitive, non-trivial
solutions with $(a,b,c)\in W_{K}$ (i.e. the asymptotic Fermat holds for
$W_{K}$).

###  1.2 Our results

We start by using the methods pioneered by Freitas and Siksek in [12]
involving modularity, level lowering and image of inertia comparison to
generalize Işik, Kara and Özman’s Theorem 2. More precisely, we relax the
assumption on the class group from $h_{K}^{+}=1$ to
$Cl_{S_{K}}(K)[2]=\\{1\\}.$ We use $Cl_{S}(K)$ to mean
$Cl(K)/\langle{[\mathfrak{P}]}\rangle_{\mathfrak{P}\in S}$ for $S$ a finite
set of primes of $K$ and consequently, $Cl_{S}(K)[n]$ denotes its $n$-torsion
points. Note that when all $\mathfrak{P}\in S$ are principal, $Cl_{S}(K)$ is
the usual $Cl(K)$, and hence we will drop the $S$ in the notation. Moreover,
in this case, $Cl(K)[p]=\\{1\\}$ is equivalent to $p\nmid h_{K}$, for $p$
prime.

Our main theorem regarding the Asymptotic Fermat Last Theorem for signature
$(p,p,2)$ reads as follows:

######  Theorem 3 (Main Theorem for $(p,p,2)$).

Let $K$ be a totally real number field with $Cl_{S_{K}}(K)[2]=\\{1\\}$ where
$S_{K}:=\\{\mathfrak{P}\>:\mathfrak{P}\text{ is a prime of }K\text{ above
}2\\}$. Suppose that there exists some distinguished prime
$\tilde{\mathfrak{P}}\in S_{K}$, such that every solution
$(\alpha,\beta,\gamma)\in\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}$
to the equation

$\alpha+\beta=\gamma^{2}$

satisfies $|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
6v_{\tilde{\mathfrak{P}}}(2)$. Then, there is a constant $B_{K}$ (depending
only on $K$) such that for each rational prime $p>B_{K}$, the equation
$a^{p}+b^{p}=c^{2}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $\tilde{\mathfrak{P}}|b$.

######  Remark 4.

By Theorem 39 the equation

$\alpha+\beta=\gamma^{2},\qquad(\alpha,\beta,\gamma)\in\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}$

has finitely many solutions up to scaling by a square in
$\mathcal{O}_{S_{K}}^{*}$, and these are effectively computable. Hence the
criteria in Theorem 3 is testable in finite time.

Imposing local constraints, we get that for a totally real number field, in
which $2$ is inert, the following holds:

######  Theorem 5.

Let K be a totally real number field with $2\nmid h_{K}^{+}$ in which $2$ is
inert. Let $\mathfrak{P}$ be the only prime above $2$, and hence
$S_{K}=\\{\mathfrak{P}\\}$. Suppose that every solution
$(\alpha,\gamma)\in\mathcal{O}^{*}_{S_{K}}\times\mathcal{O}_{S_{K}}$ with
$v_{\mathfrak{P}}(\alpha)\geq 0$ to the equation

$\alpha+1=\gamma^{2}$ (2)

satisfies $v_{\mathfrak{P}}(\alpha)\leq 6$. Then, there is a constant $B_{K}$
(depending only on $K$) such that for each rational prime $p>B_{K}$, the
equation $a^{p}+b^{p}=c^{2}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $2|b$.

More concretely, for quadratic totally real number fields $K$, Theorem 5
becomes:

######  Theorem 6.

Let $d>5$ be a rational prime satisfying $d\equiv 5\mod{8}$. Write
$K=\mathbb{Q}(\sqrt{d})$. Then, there is a constant $B_{K}$ (depending only on
$K$) such that for each rational prime $p>B_{K}$, the equation
$a^{p}+b^{p}=c^{2}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $2|b$.

More generally, by employing additional local information, the following
holds.

######  Theorem 7.

Let $K$ be a totally real field of degree $n$, and let $q\geq 5$ be a rational
prime. Suppose

  1. (i)

$2\nmid h_{K}^{+}$,

  2. (ii)

$\gcd(n,q-1)=1$,

  3. (iii)

$2$ is inert in $K$,

  4. (iv)

$q$ totally ramifies in $K$.

Then, there is a constant $B_{K}$ (depending only on $K$) such that for each
rational prime $p>B_{K}$, the equation $a^{p}+b^{p}=c^{2}$ has no primitive,
non-trivial solutions $(a,b,c)\in\mathcal{O}_{K}^{3}$ with $2|b$.

######  Remark 8.

A few examples of totally real fields $K$ satisfying the conditions above are
the degree $3$ extensions of narrow class number $1$, which have the following
defining polynomials and totally ramified prime $q$:

  * •

$p_{1}(x)=x^{3}-51x-85\>\>(q=17)$,

  * •

$p_{2}(x)=x^{3}-x^{2}-40x+13\>\>(q=11)$,

  * •

$p_{3}(x)=x^{3}-x^{2}-38x-75\>\>(q=23)$,

  * •

$p_{4}(x)=x^{3}-17x-17\>\>(q=17)$.

We use the same methods to study the asymptotic behaviour of the analogue
$(p,p,3)$ equation and we get the following:

######  Theorem 9 (Main Theorem for $(p,p,3)$).

Let $K$ be a totally real number field with $Cl_{S_{K}}(K)[3]=\\{1\\}$ where
$S_{K}:=\\{\mathfrak{P}\>:\mathfrak{P}\text{ is a prime of }K\text{ above
}3\\}$. Suppose that there exists some distinguished prime
$\tilde{\mathfrak{P}}\in S_{K}$ such that every solution
$(\alpha,\beta,\gamma)\in\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}$
to the $S_{K}$ equation

$\alpha+\beta=\gamma^{3}$

satisfies $|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
3v_{\tilde{\mathfrak{P}}}(3)$. Then, there is a constant $B_{K}$ (depending
only on $K$) such that for each rational prime $p>B_{K}$, the equation
$a^{p}+b^{p}=c^{3}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $\tilde{\mathfrak{P}}|b$.

######  Remark 10.

By Theorem 39 the equation

$\alpha+\beta=\gamma^{3},\qquad(\alpha,\beta,\gamma)\in\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}^{*}\times\mathcal{O}_{S_{K}}$

has finitely many solutions up to scaling by a cube in
$\mathcal{O}_{S_{K}}^{*}$, and these are effectively computable. Hence the
criteria in Theorem 9 is testable in finite time.

Similarly to the $(p,p,2)$ case, the following hold when employing local
information. We will consider various field extensions involving the primitive
cube root of unity $\omega:=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$.

######  Theorem 11.

Let $K$ be a totally real number field such that $3\nmid h_{K(\omega)}$,
$3\nmid h_{K}$ and in which $3$ is inert. Let $\mathfrak{P}$ be the only prime
above $3$, and hence $S_{K}=\\{\mathfrak{P}\\}$. Suppose that every solution
$(\alpha,\gamma)\in\mathcal{O}^{*}_{S_{K}}\times\mathcal{O}_{S_{K}}$ with
$v_{\mathfrak{P}}(\alpha)\geq 0$ to the equation

$\alpha+1=\gamma^{3}$ (3)

satisfies $v_{\mathfrak{P}}(\alpha)\leq 3$. Then, there is a constant $B_{K}$
(depending only on $K$) such that for each rational prime $p>B_{K}$, the
equation $a^{p}+b^{p}=c^{3}$ has no primitive, non-trivial solutions
$(a,b,c)\in\mathcal{O}_{K}^{3}$ with $3|b$.

######  Theorem 12.

Let $d$ a positive, square-free satisfying $d\equiv 2\mod{3}$. Write
$K=\mathbb{Q}(\sqrt{d})$ and suppose $3\nmid h_{K(\omega)}$, $3\nmid h_{K}$.
Then, there is a constant $B_{K}$ (depending only on $K$) such that for each
rational prime $p>B_{K}$, the equation $a^{p}+b^{p}=c^{3}$ has no primitive,
non-trivial solutions $(a,b,c)\in\mathcal{O}_{K}^{3}$ with $3|b$.

######  Theorem 13.

Let $K$ be a totally real field of degree $n$, and let $q\geq 5$ be a rational
prime. Suppose

  1. (i)

$3\nmid h_{K(\omega)}$ and $3\nmid h_{K}$,

  2. (ii)

$\gcd(n,q^{2}-1)=1$,

  3. (iii)

$3$ is inert in $K$,

  4. (iv)

$q$ totally ramifies in $K$.

Then, there is a constant $B_{K}$ (depending only on $K$) such that for each
rational prime $p>B_{K}$, the equation $a^{p}+b^{p}=c^{3}$ has no primitive,
non-trivial solutions $(a,b,c)\in\mathcal{O}_{K}^{3}$ with $3|b$.

######  Remark 14.

A few examples of totally real fields $K$ satisfying the conditions above are
the degree $5$ extensions with totally ramified prime $q=5$, which have the
following defining polynomials and corresponding $h_{K},h_{K(\omega)}$:

  * •

$p_{1}(x)=x^{5}-25x^{3}-10x^{2}+50x-20\>\>(h_{K}=1,h_{K(\omega)}=29)$,

  * •

$p_{2}(x)=x^{5}-30x^{3}-20x^{2}+160x+128\>\>(h_{K}=1,h_{K(\omega)}=29)$,

  * •

$p_{3}(x)=x^{5}-15x^{3}-10x^{2}+10x+4\>\>(h_{K}=1,h_{K(\omega)}=31)$,

  * •

$p_{4}(x)=x^{5}-20x^{3}-15x^{2}+10x+4\>\>(h_{K}=1,h_{K(\omega)}=361)$.

###  1.3 Recent progress

More recently, Işik, Kara and Özman proved in [17] a similar asymptotic result
for signature $(p,p,3)$ over general number fields $K$ with narrow class
number one satisfying some technical conditions. In the appendix, they show
how this result can be adapted to signature $(p,p,2)$. These results use
standard modularity conjectures and the study of Bianchi newforms.

###  1.4 Notational conventions

We will follow the notational conventions in [12]. Throughout $p$ denotes a
rational prime, and $K$ a totally real number field, with ring of integers
$\mathcal{O}_{K}$. For a non-zero ideal $I$ of $\mathcal{O}_{K}$ , we denote
by $[I]$ the class of $I$ in the class group Cl($K$).
Let $G_{K}=\text{Gal}(\bar{K}/K)$. For an elliptic curve $E/K$, we write

$\overline{\rho}_{E,p}:G_{K}\to\text{Aut}(E[p])\simeq\text{GL}_{2}(\mathbb{F}_{p})$

for the representation of $G_{K}$ on the $p$-torsion of $E$. For a Hilbert
eigenform $\mathfrak{f}$ over $K$, we let $\mathbb{Q}_{\mathfrak{f}}$ denote
the field generated by its eigenvalues. In this situation $\varpi$ will denote
a prime of $\mathbb{Q}_{\mathfrak{f}}$ above $p$; of course if
$\mathbb{Q}_{\mathfrak{f}}=\mathbb{Q}$ we write $p$ instead of $\varpi$. All
other primes we consider are primes of $K$. We reserve the symbol
$\mathfrak{P}$ for primes belonging to $S$. An arbitrary prime of $K$ is
denoted by $\mathfrak{q}$, and $G_{\mathfrak{q}}$ and $I_{\mathfrak{q}}$ are
the decomposition and inertia subgroups of $G_{K}$ at $\mathfrak{q}$.

Acknowledgments. I am sincerely grateful to my supervisor Samir Siksek for his
continuous support, useful comments and reviewing this paper.

##  2 Preliminaries

###  2.1 Elliptic Curves

We begin by collecting some useful results about elliptic curves, as they play
a key role in the modular approach of solving Diophantine equations.

######  Lemma 15.

Let $K$ be a field of char$(K)\neq 2,3$ and $E/K$ an elliptic curve. The
following holds:

  1. (i)

If $E$ has a $K$-rational point of order $2$, then E has a model of the form

$E:Y^{2}=X^{3}+aX^{2}+bX.$ (4)

Moreover, there is a bijection between

$\\{E/K\text{ with a }K\text{-torsion of order }2\text{ up to
}\bar{K}-\text{isomorphism}\\}\to\mathbb{P}^{1}(K)-\\{4,\infty\\}$

via the map $E\to\lambda:=\frac{a^{2}}{b}.$

  2. (ii)

If $E$ has a $K$-rational point of order $3$, then E has a model of the form

$E:Y^{2}+cXY+dY=X^{3}.$ (5)

Moreover, there is a bijection between

$\\{E/K\text{ with a }K\text{-torsion of order }3\text{ up to
}\bar{K}-\text{isomorphism}\\}\to\mathbb{P}^{1}(K)-\\{27,\infty\\}$

via the map $E\to\lambda:=\frac{c^{3}}{d}$.

###### Proof.

  1. (i)

The first part is a well-known result. For the second part, we are given an
elliptic curve $E/K$ with a $K$-torsion point of order $2$. After writing it
as in (4), we make the assignment $E\mapsto\lambda:=\frac{a^{2}}{b}.$ As
$\Delta_{E}=2^{4}b^{2}(a^{2}-4b)$, non-singularity of $E$ gives
$\lambda\in\mathbb{P}^{1}(K)-\\{4,\infty\\}$, which proves our map is well-
defined. Moreover, any $\lambda\in\mathbb{P}^{1}(K)-\\{4,\infty\\}$ can be
written as a ratio of the form $\frac{a^{2}}{b}$ with $b\neq 0$ and $a^{2}\neq
4b$, and hence comes from an elliptic curve with a $K$-rational $2$-torsion.
Thus, our map is surjective.

Injectivity follows from writing

$j_{E}=2^{8}\frac{(a^{2}-3b)^{3}}{b^{2}(a^{2}-4b)}=2^{8}\frac{(\lambda-3)^{3}}{\lambda-4}$

and noting that $\lambda=\lambda^{\prime}$ for given $E\to\lambda$,
$E^{\prime}\to\lambda^{\prime}$ implies $j_{E}=j_{E^{\prime}}$, which gives
$E\simeq E^{\prime}$.

  2. (ii)

If $E$ is in Weierstrass form we can translate the $K$-torsion point to
$(0,0)$. This will give a model of the form

$E:Y^{2}+a_{1}XY+a_{3}Y=X^{3}+a_{2}X^{2}+a_{4}X$

We now impose the condition that $(0,0)$ has order $3$. First, we compute
$-(0,0)=(0,-a_{3})$ and note that we require $(0,0)\neq-(0,0)=(0,-a_{3})$, so
$a_{3}\neq 0$. Now, by performing the change of variables

$\begin{cases}Y\to(Y+\frac{a_{4}}{a_{3}}X)\\\ X\to X\end{cases}$ (6)

we get a model of the form

$E:Y^{2}+cXY+dY=X^{3}+eX^{2}\text{ with }d=a_{3}\neq 0.$

Finally, we make use of the order $3$,

$\begin{cases}(0,0)+(0,0)=-(0,0)=(0,-d)\\\ (0,0)+(0,0)=(-e,-d)\end{cases}$ (7)

Hence, we need $e=0$, and we get the desired form: $E:Y^{2}+cXY+dY=X^{3}$.

For the second part, we are given an elliptic curve $E/K$ with a $K$-torsion
point of order $3$. After writing it as in (5), we make the assignment
$E\mapsto\lambda:=\frac{c^{3}}{d}.$ As $\Delta_{E}=d^{3}(c^{3}-27d)$, non-
singularity of $E$ gives $\lambda\in\mathbb{P}^{1}(K)-\\{27,\infty\\}$, which
proves our map is well-defined. Moreover, any
$\lambda\in\mathbb{P}^{1}(K)-\\{27,\infty\\}$ can be written as a ratio of the
form $\frac{c^{3}}{d}$ with $d\neq 0$ and $c^{3}\neq 27d$, and hence comes
from an elliptic curve with a $K$-rational $3$-torsion. Thus, our map is
surjective.

Injectivity follows from writing

$j_{E}=\frac{c^{3}(c^{3}-24d)^{3}}{d^{3}(c^{3}-27d)}=\frac{\lambda(\lambda-24)^{3}}{\lambda-27}$

and noting that $\lambda=\lambda^{\prime}$ for given $E\to\lambda$,
$E^{\prime}\to\lambda^{\prime}$ implies $j_{E}=j_{E^{\prime}}$, which gives
$E\simeq E^{\prime}$.

∎

######  Lemma 16.

Let $K$ be a number field and $S$ a set of finite primes of $K$. Then:

  1. (i)

If $S$ contains the primes above $2$ we get the following bijection
$\left\\{\begin{tabular}[]{@{}l@{}}$E/K$\text{ with a }$K$\text{-torsion of
order }$2$ with potentially\\\ \text{good reduction outside }$S$\text{ up to
}$\bar{K}$-\text{isomorphism}\end{tabular}\right\\}\longmapsto\mathcal{O}^{*}_{S}$
via the map $E\to\mu:=\lambda-4\in\mathcal{O}^{*}_{S}$, where $\lambda$ is as
in Lemma 15 (i).

  2. (ii)

If $S$ contains the primes above $3$ we get the following bijection
$\left\\{\begin{tabular}[]{@{}l@{}}$E/K$\text{ with a }$K$\text{-torsion of
order }$3$ with potentially\\\ \text{good reduction outside }$S$\text{ up to
}$\bar{K}$-\text{isomorphism}\end{tabular}\right\\}\longmapsto\mathcal{O}^{*}_{S}$
via the map $E\to\mu:=\lambda-27\in\mathcal{O}^{*}_{S}$, where $\lambda$ is as
in Lemma 15 (ii).

###### Proof.

  1. (i)

Let $E$ be an elliptic curve with a $K$-torsion point of order $2$ with
potentially good reduction outside $S$. By Lemma 15 (i) $E$ has a model

$E:Y^{2}=X^{3}+aX^{2}+bX$

with $\lambda:=\frac{a^{2}}{b}$ and $\mu:=\lambda-4=\frac{a^{2}-4b}{b}$. Thus

$j_{E}=2^{8}\frac{(\lambda-3)^{3}}{\lambda-4}=2^{8}\frac{(\mu+1)^{3}}{\mu}.$
(8)

Potentially good reduction outside $S$ implies that
$v_{\mathfrak{q}}(j_{E})\geq 0$ for all $\mathfrak{q}\notin S$, in other words
$j_{E}\in\mathcal{O}_{S}$. Consequently both $\lambda$ and $\mu$ satisfy monic
equations with coefficients in $\mathcal{O}_{S}$. Thus, we can conclude that
$\lambda,\mu\in\mathcal{O}_{S}$. Moreover, by writing $j_{E}$ in terms of
$\mu^{-1}$ and using the same reasoning, we deduce that also
$\mu^{-1}\in\mathcal{O}_{S}$ and hence $\mu\in\mathcal{O}_{S}^{*}$ and so the
assignment $E\longmapsto\mu$ is well-defined.

Note that every $\mu\in\mathcal{O}^{*}_{S}$ can be written in the form
$\mu=\frac{a^{2}}{b}-4$ for some $a,b\in K$, thus coming from an elliptic
curve with $2$-torsion. Moreover, $\mu\in\mathcal{O}^{*}_{S}$ implies
$j_{E}\in\mathcal{O}_{S}$, thus this represents a curve with potentially good
reduction outside $S$, proving surjectivity.

Injectivity follows by noting that $\mu=\mu^{\prime}$ implies
$j_{E}=j_{E^{\prime}}$ which gives $E\simeq E^{\prime}$.

  2. (ii)

Let $E$ be an elliptic curve with a $K$-torsion point of order $3$ with
potentially good reduction outside $S$. By Lemma 15 (ii) $E$ has a model

$E:Y^{2}+cXY+dY=X^{3}$

with $\lambda:=\frac{c^{3}}{d}$ and $\mu=\lambda-27=\frac{c^{3}-27d}{d}$ Thus,

$j_{E}=\frac{\lambda(\lambda-24)^{3}}{\lambda-27}=\frac{(\mu+27)(\mu+3)^{3}}{\mu}.$
(9)

Same arguments as in the proof of (i) give $j_{E},\lambda\in\mathcal{O}_{S}$
and $\mu\in\mathcal{O}^{*}_{S}$, giving $E\longmapsto\mu$ is well-defined.

Surjectivity and injectivity follow exactly as in (i).

∎

We say that a fractional ideal is an $S$-ideal if its decomposition into
primes contains only primes in $S$.

######  Lemma 17.

Let $K$ be a number field and $S$ a set of finite primes of $K$. Let $E/K$ be
an elliptic curve with good reduction outside $S$.

  1. (i)

Suppose $S$ contains the primes above $2$ and $E$ has a $K$-torsion point of
order $2$. Let $(\lambda,\mu)\in\mathcal{O}_{S}\times\mathcal{O}_{S}^{*}$
correspond to $E$ as in Lemma 16 (i) and therefore satisfy $\lambda-\mu=4$.
Then $(\lambda)\mathcal{O}_{K}=I^{2}J$ where $I,J$ are fractional ideals with
$J$ being an $S$-ideal.

  2. (ii)

Suppose $S$ contains the primes above $3$ and $E$ has a $K$-torsion point of
order $3$. Let $(\lambda,\mu)\in\mathcal{O}_{S}\times\mathcal{O}_{S}^{*}$
correspond to $E$ as in Lemma 16 (ii) and therefore satisfy $\lambda-\mu=27$.
Then $(\lambda)\mathcal{O}_{K}=I^{3}J$ where $I,J$ are fractional ideals with
$J$ being an $S$-ideal.

###### Proof.

  1. (i)

By Lemma 15 (i) $E$ has a model

$E:Y^{2}=X^{3}+aX^{2}+bX$

with $\Delta_{E}=2^{4}b^{2}(a^{2}-4b)$ and $c_{4}=2^{4}(a^{2}-3b)$. Good
reduction outside $S$ implies that for a $\mathfrak{q}\notin S$ we have that
$v_{\mathfrak{q}}(\Delta_{\text{min}})=0$ (where $\Delta_{\text{min}}$ is the
minimal discriminant of $E$ viewed over the local field $K_{\mathfrak{q}}$).
Standard results about the minimal discriminant of an elliptic curve (e.g.
[27, Ch. VII.1.]) give $\mathfrak{q}^{12k}||\Delta_{E}$ and
$\mathfrak{q}^{4k}|c_{4}$. As $\mathfrak{q}$ is an odd prime, this yields to
the following two relations

$\mathfrak{q}^{12k}||b^{2}(a^{2}-4b),\quad\mathfrak{q}^{4k}|(a^{2}-3b).$

Now, we claim that $\mathfrak{q}^{4k}|b$. Suppose not, by the first relation
it follows that $\mathfrak{q}^{4k}|(a^{2}-4b)$ and combining this with the
second relation we get that $\mathfrak{q}^{4k}|(a^{2}-3b)-(a^{2}-4b)=b$, a
contradiction. Hence, $v_{\mathfrak{q}}(b):=t\geq 4k$. Observe that the second
relation implies $v_{\mathfrak{q}}(a^{2}-3b):=s\geq 4k$. By the first relation

$12k=v_{\mathfrak{q}}(b^{2}(a^{2}-4b))=2t+v_{\mathfrak{q}}(a^{2}-3b-b)\geq
2t+\min(s,t)\geq 2t+4k.$

This implies $t\leq 4k$, which gives $t=4k$ (as we have already shown $t\geq
4k$). Moreover the second relation implies $\mathfrak{q}^{4k}|a^{2}$.
Therefore, we can conclude $\mathfrak{q}^{2k}|a$ and $\mathfrak{q}^{4k}||b$.
Hence,

$(a)\mathcal{O}_{K}=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{2k_{\mathfrak{q}}+l_{\mathfrak{q}}}\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{a_{\mathfrak{P}}},\>(b)\mathcal{O}_{K}=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{4k_{\mathfrak{q}}}\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{b_{\mathfrak{P}}}$

for some positive integers
$a_{\mathfrak{P}},b_{\mathfrak{P}},k_{\mathfrak{q}},l_{\mathfrak{q}}$. Thus,
as $\lambda=\frac{a^{2}}{b}$, we get

$(\lambda)\mathcal{O}_{K}=I^{2}J\text{, where
}I:=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{l_{\mathfrak{q}}},\>J:=\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{2a_{\mathfrak{P}}-b_{\mathfrak{P}}}$

which makes $J$ an $S$-ideal.

  2. (ii)

By Lemma 15 (ii) $E$ has a model

$E:Y^{2}+cXY+dY=X^{3}$

with $\Delta_{E}=d^{3}(c^{3}-27d)$ and $c_{4}=c(c^{3}-24d)$. As before, good
reduction outside $S$ implies that for a $\mathfrak{q}\notin S$ we have that
$v_{\mathfrak{q}}(\Delta_{\text{min}})=0$. So $\mathfrak{q}^{12k}||\Delta_{E}$
and $\mathfrak{q}^{4k}|c_{4}$ for some positive integer $k$. This yields to
the following two relations

$\mathfrak{q}^{12k}||d^{3}(c^{3}-27d),\quad\mathfrak{q}^{4k}|c(c^{3}-24d).$

Now, we claim that $\mathfrak{q}^{k}|c$. Suppose not, by the second relation
it follows that $\mathfrak{q}^{3k}|(c^{3}-24d)$. If $\mathfrak{q}^{3k}|d$, we
get $\mathfrak{q}^{3k}|c^{3}$, which in turn gives $\mathfrak{q}^{k}|c$, a
contradiction. So, $\mathfrak{q}^{3k}\nmid d$. The first relation then gives
$\mathfrak{q}^{3k}|(c^{3}-27d)$. It follows that
$\mathfrak{q}^{3k}|(c^{3}-24d)-(c^{3}-27d)=3d$. Since $\mathfrak{q}\notin
S=\\{\text{primes above }3\\}$, we get $\mathfrak{q}^{3k}|d$, another
contradiction. As we exhausted all the possibilities, we can conclude that
$\mathfrak{q}^{k}|c$. In particular, this gives $\mathfrak{q}^{3k}|c^{3}$.
Secondly, we claim that $\mathfrak{q}^{3k}|d$. Suppose not, by the first
relation we get $\mathfrak{q}^{3k}|(c^{3}-27d)$ and using
$\mathfrak{q}^{3k}|c^{3}$ we get that $\mathfrak{q}^{3k}|d$, a contradiction.
Hence $v_{\mathfrak{q}}(d):=t\geq 3k.$ In particular, so far we can deduce
that $\mathfrak{q}^{3k}|(c^{3}-27d)$. By the first relation

$12k=v_{\mathfrak{q}}(d^{3}(c^{3}-27d))=3t+v_{\mathfrak{q}}(c^{3}-27d)\geq
3t+3k$

This implies $t\leq 3k$, which gives $t=3k$ (as we have already shown $t\geq
3k$). Therefore, we can conclude $\mathfrak{q}^{k}|c$ and
$\mathfrak{q}^{3k}||d$. Hence,

$(c)\mathcal{O}_{K}=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{k_{\mathfrak{q}}+l_{\mathfrak{q}}}\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{c_{\mathfrak{P}}},\>(d)\mathcal{O}_{K}=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{3k_{\mathfrak{q}}}\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{d_{\mathfrak{P}}}$

for some positive integers
$c_{\mathfrak{P}},d_{\mathfrak{P}},k_{\mathfrak{q}},l_{\mathfrak{q}}$. Thus,
as $\lambda=\frac{c^{3}}{d}$, we get

$(\lambda)\mathcal{O}_{K}=I^{3}J\text{, where
}I:=\displaystyle\prod_{\mathfrak{q}\notin
S_{K}}\mathfrak{q}^{l_{\mathfrak{q}}},\>J:=\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{3c_{\mathfrak{P}}-d_{\mathfrak{P}}}$

which makes $J$ an $S$-ideal.

∎

###  2.2 Modularity Results

We now carefully formulate modularity in the context of a totally real field.
Let us first recall that given $K$ a totally real number field, $G_{K}$ its
absolute Galois group and $E$ an elliptic curve over $K$, we say that $E$ is
modular if there exists a Hilbert cuspidal eigenform $\mathfrak{f}$ over $K$
of parallel weight $2$, with rational Hecke eigenvalues, such that the
Hasse–Weil L-function of $E$ is equal to the Hecke L-function of
$\mathfrak{f}$. A more conceptual way to phrase this is that there is an
isomorphism of compatible systems of Galois representations

$\rho_{E,p}\simeq\rho_{\mathfrak{f},p}$

where the left-hand side is the Galois representation arising from the action
of $G_{K}$ on the $p$-adic Tate module $T_{p}(E)$, while the right-hand side
is the Galois representation associated to $\mathfrak{f}$. A comprehensive
definition of Hilbert modular forms and their associated representation can be
found, for example in Wiles’ [34]. In this paper we are mainly interested in
the mod $p$ Galois representations and we denote their isomorphism by
$\overline{\rho}_{E,p}\sim\overline{\rho}_{\mathfrak{f},p}$.

We need the following theorem proved by Freitas, Hung and Siksek in [9]:

######  Theorem 18.

Let $K$ be a totally real field. There are at most finitely many $\bar{K}$\-
isomorphism classes of non-modular elliptic curves $E$ over $K$. Moreover, if
$K$ is real quadratic, then all elliptic curves over $K$ are modular.

Furthermore Derickx, Najman and Siksek have recently proved in [8]:

######  Theorem 19.

Let $K$ be a totally real cubic number field and $E$ be an elliptic curve over
$K$. Then $E$ is modular.

###  2.3 Irreductibility of$\mod{p}$ representations of elliptic curves

We need the following theorem in the level lowering step of our proof. This
was proved in [11, Theorem 2] and it is derived from the work of David and
Momose who in turn built on Merel’s Uniform Boundedness Theorem.

######  Theorem 20.

Let $K$ be a Galois totally real field. There is an effective constant
$C_{K}$, depending only on $K$, such that the following holds. If $p>C_{K}$ is
prime, and $E$ is an elliptic curve over $K$ which has multiplicative
reduction at all $\mathfrak{q}|p$, then $\overline{\rho}_{E,p}$ is
irreducible.

######  Remark 21.

The above theorem is also true for any totally real field by replacing $K$ by
its Galois closure.

###  2.4 Level lowering

We present a level lowering result proved by Freitas and Siksek in [12]
derived from the work of Fujira [13], Jarvis [14], and Rajaei [21]. Let $K$ be
a totally real field and $E/K$ be an elliptic curve of conductor
$\mathcal{N}_{E}$. Let $p$ be a rational prime. Define the following
quantities:

$\mathcal{M}_{p}=\prod_{\begin{subarray}{c}\mathfrak{q}||\mathcal{N}_{E}\\\
p|v_{\mathfrak{q}}(\Delta_{\mathfrak{q}})\end{subarray}}\mathfrak{q},\text{
and }\mathcal{N}_{p}=\frac{\mathcal{N}_{E}}{\mathcal{M}_{p}}$ (10)

where $\Delta_{\mathfrak{q}}$ is the minimal discriminant of a local minimal
model for $E$ at $\mathfrak{q}$. For a Hilbert eigenform $\mathfrak{f}$ over
$K$, we write $\mathbb{Q}_{\mathfrak{f}}$ for the field generated by its
eigenvalues.

######  Theorem 22.

With the notation above, suppose the following statements hold:

  1. (i)

$p\geq 5$, the ramification index $e(\mathfrak{q}/p)<p-1$ for all
$\mathfrak{q}|p$, and $\mathbb{Q}(\zeta_{p})^{+}\nsubseteq K$,

  2. (ii)

E is modular,

  3. (iii)

$\overline{\rho}_{E,p}$ is irreducible,

  4. (iv)

$E$ is semistable at all $\mathfrak{q}|p$,

  5. (v)

$p|v_{\mathfrak{q}}(\Delta_{\mathfrak{q}})$ for all $\mathfrak{q}|p$.

Then, there is a Hilbert eigenform $\mathfrak{f}$ of parallel weight $2$ that
is new at level $\mathcal{N}_{p}$ and some prime $\varpi$ of
$\mathbb{Q}_{\mathfrak{f}}$ such that $\varpi|p$ and
$\overline{\rho}_{E,p}\sim\overline{\rho}_{\mathfrak{f},\varpi}$.

###### Proof.

A proof is given in [12, p. 8]. ∎

###  2.5 Eichler-Shimura

For totally real fields, modularity reads as follows.

######  Conjecture 23 (Eichler-Shimura).

Let $K$ be a totally real field. Let $\mathfrak{f}$ be a Hilbert newform of
level $\mathcal{N}$ and parallel weight $2$, with rational eigenvalues. Then
there is an elliptic curve $E_{\mathfrak{f}}/K$ with conductor $\mathcal{N}$
having the same L-function as $\mathfrak{f}$.

Freitas and Siksek [12] obtained the following theorem from works of Blasius
[3], Darmon [5] and Zhang [36].

######  Theorem 24.

Let $E$ be an elliptic curve over a totally real field $K$ , and $p$ be an odd
prime. Suppose that $\overline{\rho}_{E,p}$ is irreducible, and
$\overline{\rho}_{E,p}\sim\overline{\rho}_{\mathfrak{f},\varpi}$ for some
Hilbert newform $\mathfrak{f}$ over $K$ of level $\mathcal{N}$ and parallel
weight $2$ which satisfies $\mathbb{Q}_{\mathfrak{f}}=\mathbb{Q}$. Let
$\mathfrak{q}\nmid p$ be a prime ideal of $\mathcal{O}_{K}$ such that:

  1. (i)

$E$ has potentially multiplicative reduction at $\mathfrak{q}$,

  2. (ii)

$p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{q}})$,

  3. (iii)

$p\nmid(\text{Norm}_{K/\mathbb{Q}}(\mathfrak{q})\pm 1)$.

Then there is an elliptic curve $E_{\mathfrak{f}}/K$ of conductor
$\mathcal{N}$ with the same L-function as $\mathfrak{f}$.

##  3 Signature $(p,p,2)$

Let $K$ be a totally real field. Recall the set
$S_{K}=\\{\mathfrak{P}\>:\mathfrak{P}\text{ is a prime of }K\text{ above
}2\\}.$ Throughout this section we denote by $(a,b,c)\in\mathcal{O}_{K}^{3}$ a
non-trivial, primitive solution of $a^{p}+b^{p}=c^{2}$.

###  3.1 Frey Curve

For $(a,b,c)\in\mathcal{O}_{K}^{3}$ as described above we associate the
following Frey elliptic curve defined over $K$:

$E:Y^{2}=X^{3}+4cX^{2}+4a^{p}X.$ (11)

We compute the arithmetic invariants:

$\Delta_{E}=2^{12}(a^{2}b)^{p},c_{4}=2^{6}(4b^{p}+a^{p})\text{ and
}j_{E}=2^{6}\frac{(4b^{p}+a^{p})^{3}}{(a^{2}b)^{p}}.$

######  Lemma 25.

Let $(a,b,c)$ be the non-trivial, primitive solution to the equation
$a^{p}+b^{p}=c^{2}$. Let $E$ be the associated Frey curve (11) with conductor
$\mathcal{N}_{E}$. Then, for all primes $\mathfrak{q}\notin S_{K}$, the model
$E$ is minimal, semistable and satisfies $p|v_{\mathfrak{q}}(\Delta_{E})$.
Moreover

$\mathcal{N}_{E}=\displaystyle\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{r_{\mathfrak{P}}}\prod_{\begin{subarray}{c}\mathfrak{q}|ab\\\
\mathfrak{q}\notin
S_{K}\end{subarray}}\mathfrak{q},\qquad\mathcal{N}_{p}=\displaystyle\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{r^{\prime}_{\mathfrak{P}}}$ (12)

where $0\leq r^{\prime}_{\mathfrak{P}}\leq r_{\mathfrak{P}}\leq
2+6v_{\mathfrak{P}}(2)$.

###### Proof.

Let $\mathfrak{q}$ be an odd prime of $K$. The invariants of the model $E$ are
$\Delta_{E}=2^{12}(a^{2}b)^{p}\text{ and }c_{4}=2^{6}(4b^{p}+a^{p}).$ Suppose
that $\mathfrak{q}$ divides $\Delta_{E}$, so $\mathfrak{q}|ab$. Since $a$ and
$b$ are relatively prime, $\mathfrak{q}$ divides exactly one of $a$ and $b$.
Therefore, $\mathfrak{q}$ does not divide $c_{4}$. In particular, the model is
minimal at $\mathfrak{q}$ and has multiplicative reduction. Hence
$p|v_{\mathfrak{q}}(\Delta_{E})=v_{\mathfrak{q}}(\Delta_{\mathfrak{q}})$. On
the other hand $\mathfrak{P}\in S_{K}$ is an even prime, so we have
$r_{\mathfrak{P}}=v_{\mathfrak{P}}(\mathcal{N}_{E})\leq
2+6v_{\mathfrak{P}}(2)$ by [28, Theorem IV.10.4]. The definition of
$\mathcal{N}_{E}$ gives the desired form in (12). Then, use (10) to get
$\mathcal{N}_{p}$ and observe that
$r^{\prime}_{\mathfrak{P}}=r_{\mathfrak{P}}$ unless $E$ has multiplicative
reduction at $\mathfrak{P}$ and $p|v_{\mathfrak{P}}(\Delta_{\mathfrak{P}})$ in
which case $r_{\mathfrak{P}}=1$ and $r^{\prime}_{\mathfrak{P}}=0$. ∎

######  Lemma 26.

Let $K$ be a totally real field. There is some constant $A_{K}$ depending only
on $K$, such that for any non-trivial, primitive solution $(a,b,c)$ of
$a^{p}+b^{p}=c^{2}$ and $p>A_{K}$, the Frey curve given by (11) is modular.

###### Proof.

By Theorem 18, there are at most finitely many possible $\bar{K}$-isomorphism
classes of elliptic curves over $E$ which are not modular. Let
$j_{1},j_{2},\dots,j_{n}\in K$ be the $j$-invariants of these classes. Define
$\lambda:=b^{p}/a^{p}$. The $j$-invariant of $E$ is

$j(\lambda)=2^{6}(4\lambda+1)^{3}\lambda^{-1}.$

We can assume $\lambda\notin\\{0,\pm 1\\}$ as these $\lambda$ lead to
$j(\lambda)\in\mathbb{Q}$ and we know that all rational elliptic curves are
modular. Each equation $j(\lambda)=j_{i}$ has at most three solutions
$\lambda\in K$. Thus there are values $\lambda_{1},\dots,\lambda_{m}\in
K\>(\text{where }m\leq 3n)$ such that if $\lambda\neq\lambda_{k}$ for all $k$,
then the elliptic curve $E$ with $j$-invariant $j(\lambda)$ is modular.

If $\lambda=\lambda_{k}$ then $(b/a)^{p}=\lambda_{k}$, but the polynomial
$x^{p}+\lambda_{k}$ has a root in $K$ if and only if
$\lambda_{k}\in(K^{*})^{p}$ because $K$ is totally real and
$\lambda_{k}\notin\\{0,\pm 1\\}.$ Hence we get a lower bound on $p$ for each
$k$, and by taking the maximum of these bounds we get $A_{K}$.

######  Remark 27.

The constant $A_{K}$ is ineffective as the finiteness of Theorem 18 relies on
Falting’s Theorem (which is ineffective). See [9] for more details. Note that
if $K$ is quadratic or cubic we get $A_{K}=0$ (by the last part of Theorem 18
and Theorem 19).

∎

###  3.2 Images of Inertia

We gather information about the images of inertia
$\overline{\rho}_{E,p}(I_{\mathfrak{q}})$. This is a crucial step in applying
Theorem 24 and for controlling the behaviour at the primes in $S_{K}$ of the
newform obtained by level lowering.

######  Lemma 28.

Let $E$ be an elliptic curve over $K$ with $j$-invariant $j_{E}$. Let $p\geq
5$ and let $\mathfrak{q}\nmid p$ be a prime of $K$. Then
$p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{q}})$ if and only if $E$ has
potentially multiplicative reduction at $\mathfrak{q}$ (i.e.
$v_{\mathfrak{q}}(j_{E})<0$) and $p\nmid v_{\mathfrak{q}}(j_{E})$.

###### Proof.

See [12, Lemma 3.4]. ∎

######  Lemma 29.

Let ${\mathfrak{P}}\in S_{K}$ and $(a,b,c)$ a non-trivial, primitive solution
to $a^{p}+b^{p}=c^{2}$ with $\mathfrak{P}|b$ and prime exponent
$p>6v_{\mathfrak{P}}(2)$. Let $E$ be the Frey curve in $(\ref{Frey2})$ with
$j$-invariant $j_{E}$. Then $E$ has potentially multiplicative reduction at
${\mathfrak{P}}$ and $p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{P}})$.

###### Proof.

Assume that $\mathfrak{P}\in S_{K}$ with $v_{\mathfrak{P}}(b)=k$. Then
$v_{\mathfrak{P}}(j_{E})=6v_{\mathfrak{P}}(2)-pk$. Since
$p>6v_{\mathfrak{P}}(2)$, it follows that $v_{\mathfrak{P}}(j_{E})<0$ and
clearly $p\nmid v_{\mathfrak{P}}(j_{E})$. This implies that $E$ has
potentially multiplicative reduction at $\mathfrak{P}$ and by Lemma 28 we get
$p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{P}})$. ∎

###  3.3 Level Lowering and Eichler Shimura

This is a key result in the proof of Theorem 3, for which we have prepared the
ingredients in the previous sections. We will follow the corresponding proofs
in [12] and [16].

######  Theorem 30.

Let $K$ be a totally real number field and assume it has a distinguished prime
$\tilde{\mathfrak{P}}\in S_{K}$. Then there is a constant $B_{K}$ depending
only on $K$ such that the following hold. Suppose
$(a,b,c)\in\mathcal{O}_{K}^{3}$ is a non-trivial, primitive solution to
$a^{p}+b^{p}=c^{2}$ with prime exponent $p>B_{K}$ such that
$\tilde{\mathfrak{P}}|b$. Write $E$ for the Frey curve (11). Then, there is an
elliptic curve $E^{\prime}$ over $K$ such that:

  1. (i)

the elliptic curve $E^{\prime}$ has good reduction outside $S_{K}$;

  2. (ii)

$\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$;

  3. (iii)

$E^{\prime}$ has a $K$-rational point of order $2$;

  4. (iv)

$E^{\prime}$ has potentially multiplicative reduction at
$\tilde{\mathfrak{P}}$ $(v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})<0$ where
$j_{E^{\prime}}$ is the $j$-invariant of $E^{\prime})$.

###### Proof.

We first observe by Lemma 25 that $E$ has multiplicative reduction outside
$S_{K}$. By taking $B_{K}$ sufficiently large, we see from Lemma 26 that $E$
is modular and by Theorem 20 that $\overline{\rho}_{E,p}$ is irreducible.
Applying Theorem 22 and Lemma 25 we see that
$\overline{\rho}_{E,p}\sim\overline{\rho}_{\mathfrak{f},\varpi}$ for a Hilbert
newform $\mathfrak{f}$ of level $\mathcal{N}_{p}$ and some prime $\varpi|p$ of
$\mathbb{Q}_{\mathfrak{f}}$. Here $\mathbb{Q}_{\mathfrak{f}}$ denotes the
field generated by the Hecke eigenvalues $\mathfrak{f}$. Next we reduce to the
case when $\mathbb{Q}_{\mathfrak{f}}=\mathbb{Q}$, after possibly enlarging
$B_{K}$. This step uses standard ideas originally due to Mazur that can be
found in [2, Section 4], [4, Proposition 15.4.2], and so we omit the details.

Next we want to show that there is some elliptic curve $E^{\prime}/K$ of
conductor $\mathcal{N}_{p}$ having the same L-function as $\mathfrak{f}$.We
apply Lemma 29 with $\mathfrak{P}=\tilde{\mathfrak{P}}$ and get that $E$ has
potentially multiplicative reduction at $\tilde{\mathfrak{P}}$ and
$p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{P}})$. The existence of $E^{\prime}$
follows from Theorem 24 after possibly enlarging $B_{K}$ to ensure that
$p\nmid(\text{Norm}_{K/\mathbb{Q}}(\tilde{\mathfrak{\mathfrak{P}}})\pm 1)$. By
putting all the pieces together we can conclude that there is an elliptic
curve $E^{\prime}/K$ of conductor $\mathcal{N}_{p}$ satisfying
$\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$. This proves (i) and
(ii).

To prove (iii) we use that
$\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$ for some
$E^{\prime}/K$ with conductor $\mathcal{N}_{p}$. After enlarging $B_{K}$ by an
effective amount, and possibly replacing $E^{\prime}$ by an isogenous curve,
we may assume that $E^{\prime}$ has a $K$-rational point of order $2$. This
uses standard ideas which can be found, for example, in [23, Section IV-6].

Now let $j_{E^{\prime}}$ be the $j$-invariant of $E^{\prime}$. As we have
already seen, Lemma 29 implies
$p|\\#\overline{\rho}_{E,p}(I_{\tilde{\mathfrak{P}}})$, hence
$p|\\#\overline{\rho}_{E^{\prime},p}(I_{\tilde{\mathfrak{P}}})$, thus by Lemma
28 we get that $E^{\prime}$ has potentially multiplicative reduction at
$\tilde{\mathfrak{P}}$ and so $v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})<0$. ∎

###  3.4 Proof of Theorem 3

###### Proof.

Given a primitive, non-trivial solution $(a,b,c)$ such that
$\tilde{\mathfrak{P}}|b$ with a prime exponent $p$ we associate the Frey
elliptic curve in (11). By Theorem 30 for there exists $B_{K}$ such that for
all $p>B_{K}$ we can find an elliptic curve $E^{\prime}$ which is related to
$E$ by $\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$ and has a
$K$-rational point of order $2$. Hence by Lemma 15 (i) we get a model

$E^{\prime}:Y^{2}=X^{3}+a^{\prime}X^{2}+b^{\prime}X$

with arithmetic invariants $\Delta_{E^{\prime}}=2^{4}b^{\prime 2}(a^{\prime
2}-4b^{\prime})$, $j_{E^{\prime}}=2^{8}\frac{(a^{\prime
2}-3b^{\prime})^{3}}{b^{\prime 2}(a^{\prime 2}-4b^{\prime})}$. Moreover, by
Theorem 30 (i), we know that $E^{\prime}$ has good reduction outside $S_{K}$
which implies that $v_{\mathfrak{q}}(j_{E^{\prime}})\geq 0$ for
$\mathfrak{q}\notin S_{K}$. Therefore, $j_{E^{\prime}}\in\mathcal{O}_{S_{K}}$.
Consider $\lambda:=\frac{a^{\prime 2}}{b^{\prime}}$ and
$\mu:=\lambda-4=\frac{a^{\prime 2}-4b^{\prime}}{b^{\prime}}$. Next, we need to
show that $\lambda$ can be written as $\lambda=u\gamma^{2}$, where $u$ is an
$S_{K}$-unit. By Lemma 17 (i) applied to $E^{\prime}$ we get that

$(\lambda)\mathcal{O}_{K}=I^{2}J\text{ where }J\text{ is an
}S_{K}\text{-ideal}.$

Thus $[I]^{2}=[J]$ as elements of the class group Cl($K$) and
$[J]\in\langle[{\mathfrak{P}}]\rangle_{\mathfrak{P}\in S_{K}}$. This implies
that $[I]\in Cl_{S_{K}}(K)[2]$ and by our assumption on $K$ that
$Cl_{S_{K}}(K)[2]$ is trivial, we get that
$[I]\in\langle[{\mathfrak{P}}]\rangle_{\mathfrak{P}\in S_{K}}$, i.e.
$I:=\gamma\tilde{I}$, where $\tilde{I}$ is an $S_{K}$-ideal and
$\gamma\in\mathcal{O}_{K}$. Consequently,

$(\lambda)\mathcal{O}_{K}=(\gamma)^{2}\tilde{I}^{2}J\text{ where both
}\tilde{I}\text{ and }J\text{ are }S_{K}\text{-ideals.}$

Finally, $(\frac{\lambda}{\gamma^{2}})\mathcal{O}_{K}$ is an $S_{K}$-ideal,
which implies that $u:=\frac{\lambda}{\gamma^{2}}$ is an $S_{K}$-unit. Now, by
dividing $\mu+4=\lambda\text{ by }u$, we get

$\alpha+\beta=\gamma^{2},\qquad\alpha:=\frac{\mu}{u}\in\mathcal{O}_{S_{K}}^{*},\qquad\beta:=\frac{4}{u}\in\mathcal{O}_{S_{K}}^{*}.$
(13)

Now, suppose that there is some $\tilde{\mathfrak{P}}\in S_{K}$ that satisfies
$|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
6v_{\tilde{\mathfrak{P}}}(2)$. We will show that
$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$, contradicting Theorem 30
(iv) and hence we can conclude the proof. By using (13) we can rewrite the
assumption $|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
6v_{\tilde{\mathfrak{P}}}(2)$ in terms of the valuation of $\mu$, using that
$\frac{\alpha}{\beta}=\frac{\mu}{4}$:

$-4v_{\tilde{\mathfrak{P}}}(2)\leq v_{\tilde{\mathfrak{P}}}(\mu)\leq
8v_{\tilde{\mathfrak{P}}}(2).$

Note that $j_{E^{\prime}}=2^{8}(\mu+1)^{3}\mu^{-1}$, hence

$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})=8v_{\tilde{\mathfrak{P}}}(2)+3v_{\tilde{\mathfrak{P}}}(\mu+1)-v_{\tilde{\mathfrak{P}}}(\mu).$

There are three cases according to the valuation of $\tilde{\mathfrak{P}}$ at
$\mu$:
Case (1): Suppose $v_{\tilde{\mathfrak{P}}}(\mu)=0$. This implies that
$v_{\tilde{\mathfrak{P}}}(\mu+1)\geq 0$, thus
$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$, a contradiction.
Case (2): Suppose $v_{\tilde{\mathfrak{P}}}(\mu)>0$. This implies
$v_{\tilde{\mathfrak{P}}}(\mu+1)=0$, thus, by using
$v_{\tilde{\mathfrak{P}}}(\mu)\leq 8v_{\tilde{\mathfrak{P}}}(2)$ we get again
$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$.
Case (3): Finally, suppose $v_{\tilde{\mathfrak{P}}}(\mu)<0$. This implies
$v_{\tilde{\mathfrak{P}}}(\mu+1)=v_{\tilde{\mathfrak{P}}}(\mu)$, thus, by
using $-4v_{\tilde{\mathfrak{P}}}(2)\leq v_{\tilde{\mathfrak{P}}}(\mu)$, we
get one last time $v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$.
All three cases lead to contradictions and hence we conclude the proof. ∎

###  3.5 Proof of Theorem 5

###### Proof.

We want to apply Theorem 3 with $\tilde{\mathfrak{P}}=\mathfrak{P}$ and
$S_{K}=\\{\mathfrak{P}\\}$. Note that $2\nmid h_{K}^{+}$ implies that
$Cl_{S_{K}}(K)[2]$ is trivial. As $2$ is inert, we get
$v_{\mathfrak{P}}(2)=1$.

Now, let us consider the equation $\alpha+\beta=\gamma^{2}$, with
$\alpha,\beta\in\mathcal{O}_{S_{K}}^{*}$. By scaling the equation by even
powers of $2$ and swapping $\alpha$ and $\beta$ if necessary, we may assume
$0\leq v_{\mathfrak{P}}(\beta)\leq v_{\mathfrak{P}}(\alpha)$ with
$v_{\mathfrak{P}}(\beta)\in\\{0,1\\}$.
Case (1): Suppose $v_{\mathfrak{P}}(\beta)=1$. If
$v_{\mathfrak{P}}(\alpha)\geq 2$, then
$v_{\mathfrak{P}}(\gamma^{2})=v_{\mathfrak{P}}(\alpha+\beta)=1$, which leads
to a contradiction as $v_{\mathfrak{P}}(\gamma^{2})$ must be even. Thus,
$v_{\mathfrak{P}}(\alpha)=v_{\mathfrak{P}}(\beta)=1$ and
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=0\leq 6$.
Case (2): Suppose $v_{\mathfrak{P}}(\beta)=0$ with $\beta$ not a square. If
$v_{\mathfrak{P}}(\alpha)>6$, then
$v_{\mathfrak{P}}(\gamma^{2})=v_{\mathfrak{P}}(\alpha+\beta)=0$ and
$\beta\equiv\gamma^{2}\mod 2^{6}$. Consider the field extension
$L=K(\sqrt{\beta})$. We will show that $L$ is unramified at $2$, hence
contradicting $2\nmid h_{K}^{+}$. Consider the element
$\delta:=\frac{\gamma+\sqrt{\beta}}{2}$. Its minimal polynomial is

$m_{\delta}(X)=X^{2}-\gamma X+\frac{\gamma^{2}-\beta}{4}.$

This belongs to $\mathcal{O}_{K}[X]$ and has odd discriminant $\Delta=\beta$,
proving that $L$ is unramified at $2$. Thus, we must have
$v_{\mathfrak{P}}(\alpha)\leq 6$, giving
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=v_{\mathfrak{P}}(\alpha)\leq 6$.
Case (3): Suppose $\beta$ is a square. By dividing everything through $\beta$,
we may assume $\beta=1$. Then, by the hypothesis of the theorem we get

$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=v_{\mathfrak{P}}(\alpha)\leq 6.$

All of the possible three cases lead to
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})\leq 6=6v_{\mathfrak{P}}(2)$, so we can
conclude the proof by Theorem 3. ∎

###  3.6 Proof of Theorem 6

###### Proof.

Note that the assumption $d\equiv 5\mod 8$ gives that $2$ is inert in the
quadratic field $K=\mathbb{Q}(\sqrt{d})$, take $\mathfrak{P}$ to be the unique
prime above $2$ and denote $S_{K}=\\{\mathfrak{P}\\}$. Moreover, $d$ prime is
equivalent to $2\nmid h_{K}^{+}$ [18, Section 1.3.1]. By Theorem 5 it is
enough to check that every solution
$(\alpha,\gamma)\in\mathcal{O}^{*}_{S_{K}}\times\mathcal{O}_{S_{K}}$ with
$v_{\mathfrak{P}}(\alpha)\geq 0$ to the equation $\alpha+1=\gamma^{2}$
satisfies $v_{\mathfrak{P}}(\alpha)\leq 6$. Rearranging the above we get that
$(\gamma+1)(\gamma-1)=\alpha$. Denote $x=\frac{(\gamma+1)}{2}$ and
$y=\frac{(1-\gamma)}{2}$. Note that since
$(\gamma+1),(\gamma-1)\in\mathcal{O}_{S_{K}}$ and they are factors of the
$S_{K}$-unit $\alpha$, they must be $S_{K}$-units, consequently
$x,y\in\mathcal{O}_{S_{K}}^{*}$.

In [12, p. 15], it is proved that the only solutions of $S_{K}$-unit equation
$x+y=1$, where $K=\mathbb{Q}(\sqrt{d})$ with $d\equiv 5\mod 8$, $d>5$ and
$S_{K}=\\{\mathfrak{P}\\}$ are the so-called irrelevant solutions
$(-1,2),(1/2,1/2),(2,-1)$. This leads to $\alpha\in\\{-1,8\\}$, and hence
$v_{\mathfrak{P}}(\alpha)\in\\{0,3\\}$, proving $v_{\mathfrak{P}}(\alpha)\leq
6$. Thus we can conclude the proof by Theorem 5. ∎

###  3.7 Proof of Theorem 7

###### Proof.

We will take $\mathfrak{P}$ to be the unique prime above $2$ and denote
$S_{K}=\\{\mathfrak{P}\\}$. By Theorem 5 it is enough to check that every
solution $(\alpha,\gamma)\in\mathcal{O}^{*}_{S_{K}}\times\mathcal{O}_{S_{K}}$
with $v_{\mathfrak{P}}(\alpha)\geq 0$ to the equation $\alpha+1=\gamma^{2}$
satisfies $v_{\mathfrak{P}}(\alpha)\leq 6$. Rearranging as in (3.6) we get an
$S_{K}$-unit equation $x+y=1$ such that $\alpha=-4xy$.

We will now use a result proved in [10, p.5]. saying that if $K$ satisfies the
hypothesis of Theorem 7, it follows that every solution $(x,y)$ of the
$S_{K}$-unit equation satisfies
$\max\\{v_{\mathfrak{P}}(x),v_{\mathfrak{P}}(y)\\}<2v_{\mathfrak{P}}(2)=2$.
Thus,

$v_{\mathfrak{P}}(\alpha)=2v_{\mathfrak{P}}(2)+v_{\mathfrak{P}}(x)+v_{\mathfrak{P}}(y)<2+2+2=6.$

Hence we can conclude the proof by Theorem 5. ∎

##  4 Signature $(p,p,3)$

Let $K$ be a totally real field. Recall the set
$S_{K}=\\{\mathfrak{P}\>:\mathfrak{P}\text{ is a prime of }K\text{ above
}3\\}.$ Throughout this section we denote by $(a,b,c)\in\mathcal{O}_{K}^{3}$ a
non-trivial, primitive solution of $a^{p}+b^{p}=c^{3}$.

###  4.1 Frey Curve

For $(a,b,c)\in\mathcal{O}_{K}^{3}$ as described above we associate the
following Frey elliptic curve defined over $K$:

$E:Y^{2}+3cXY+a^{p}Y=X^{3}.$ (14)

We compute the arithmetic invariants:

$\Delta_{E}=3^{3}(a^{3}b)^{p},\>c_{4}=3^{2}c(9b^{p}+a^{p})\text{ and
}j_{E}=3^{3}\frac{c^{3}(9b^{p}+a^{p})^{3}}{(a^{3}b)^{p}}.$

######  Lemma 31.

Let $(a,b,c)$ be the non-trivial, primitive solution to the equation
$a^{p}+b^{p}=c^{3}$. Let $E$ be the associated Frey curve (14) with conductor
$\mathcal{N}_{E}$. Then, for all primes $\mathfrak{q}\notin S_{K}$, the model
$E$ is minimal, semistable and satisfies $p|v_{\mathfrak{q}}(\Delta_{E})$.
Moreover

$\mathcal{N}_{E}=\displaystyle\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{r_{\mathfrak{P}}}\prod_{\begin{subarray}{c}\mathfrak{q}|ab\\\
\mathfrak{q}\notin
S_{K}\end{subarray}}\mathfrak{q},\qquad\mathcal{N}_{p}=\displaystyle\prod_{\mathfrak{P}\in
S_{K}}\mathfrak{P}^{r^{\prime}_{\mathfrak{P}}}$ (15)

where $0\leq r^{\prime}_{\mathfrak{P}}\leq r_{\mathfrak{P}}\leq
2+3v_{\mathfrak{P}}(3)$.

###### Proof.

The proof follows exactly like the proof of Lemma 25. ∎

######  Lemma 32.

Let $K$ be a totally real field. There is some constant $A_{K}$ depending only
on $K$, such that for any non-trivial, primitive solution $(a,b,c)$ of
$a^{p}+b^{p}=c^{3}$ the Frey curve given by (14) is modular.

###### Proof.

The proof follows exactly like the proof of Lemma 26. ∎

###  4.2 Images of Inertia

We need the following result about images of inertia whose prove follows
exactly like the proof of Lemma 29, hence it is omitted.

######  Lemma 33.

Let ${\mathfrak{P}}\in S_{K}$ and $(a,b,c)$ with $\mathfrak{P}|b$ and prime
exponent $p>3v_{\mathfrak{P}}(3)$. Let $E$ be the Frey curve in
$(\ref{Frey3})$ with $j$-invariant $j_{E}$. Then $E$ has potentially
multiplicative reduction at ${\mathfrak{P}}$ and
$p|\\#\overline{\rho}_{E,p}(I_{\mathfrak{P}})$.

###  4.3 Level Lowering and Eichler Shimura

As in the previous section, the crucial level lowering theorem reads as
follows:

######  Theorem 34.

Let $K$ be a totally real number field and assume it has a distinguished prime
$\tilde{\mathfrak{P}}\in S_{K}$. Then there is a constant $B_{K}$ depending
only on $K$ such that the following hold. Suppose
$(a,b,c)\in\mathcal{O}_{K}^{3}$ is a non-trivial, primitive solution to
$a^{p}+b^{p}=c^{3}$ with prime exponent $p>B_{K}$ such that
$\tilde{\mathfrak{P}}|b$. Write $E$ for the Frey curve (14). Then, there is an
elliptic curve $E^{\prime}$ over $K$ such that:

  1. (i)

the elliptic curve $E^{\prime}$ has good reduction outside $S_{K}$,

  2. (ii)

$\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$,

  3. (iii)

$E^{\prime}$ has a $K$-rational point of order $3$,

  4. (iv)

$E^{\prime}$ has potentially multiplicative reduction at
$\tilde{\mathfrak{P}}$ $(v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})<0$ where
$j_{E^{\prime}}$ is the $j$-invariant of $E^{\prime})$.

###### Proof.

The proof follows exactly like the proof of Theorem 30. ∎

###  4.4 Proof of Theorem 9

###### Proof.

Given a primitive, non-trivial solution $(a,b,c)$ such that
$\tilde{\mathfrak{P}}|b$ with a prime exponent $p$ we associate the Frey
elliptic curve in (14). By Theorem 34 for $p>B_{K}$ we can find an elliptic
curve $E^{\prime}$ which is related to $E$ by
$\overline{\rho}_{E,p}\sim\overline{\rho}_{E^{\prime},p}$ and has a
$K$-rational point of order $3$. Hence by Lemma 15 (ii) we get a model

$E^{\prime}:Y^{2}+c^{\prime}XY+d^{\prime}Y=X^{3}$

with arithmetic invariants $\Delta_{E^{\prime}}=d^{\prime 3}(c^{\prime
3}-27d^{\prime})$ and $j_{E^{\prime}}=\frac{c^{\prime 3}(c^{\prime
3}-24d^{\prime})^{3}}{d^{\prime 3}(c^{\prime 3}-27d^{\prime})}$.

Moreover, by Theorem 30 (i), we know that $E^{\prime}$ has good reduction
outside $S_{K}$ which implies that $v_{\mathfrak{q}}(j_{E^{\prime}})\geq 0$
for $\mathfrak{q}\notin S_{K}$. Therefore,
$j_{E^{\prime}}\in\mathcal{O}_{S_{K}}$. Consider $\lambda:=\frac{c^{\prime
3}}{d^{\prime}}$ and $\mu:=\lambda-27=\frac{c^{\prime
3}-27d^{\prime}}{d^{\prime}}$. Next, we need to show that $\lambda$ can be
written as $\lambda=u\gamma^{3}$, where $u$ is an $S_{K}$-unit. By Lemma 17
(ii) applied to $E^{\prime}$ we get that

$(\lambda)\mathcal{O}_{K}=I^{3}J\text{ where }J\text{ is an
}S_{K}\text{-ideal}.$

Thus $[I]^{3}=[J]$ as elements of the class group Cl($K$) and
$[J]\in\langle[{\mathfrak{P}}]\rangle_{\mathfrak{P}\in S_{K}}$. This implies
that $[I]\in Cl_{S_{K}}(K)[3]$ and by our assumption on $K$ that
$Cl_{S_{K}}(K)[3]$ is trivial, we get that
$[I]\in\langle[{\mathfrak{P}}]\rangle_{\mathfrak{P}\in S_{K}}$, i.e.
$I:=\gamma\tilde{I}$, where $\tilde{I}$ is an $S_{K}$-ideal and
$\gamma\in\mathcal{O}_{K}$. Consequently,

$(\lambda)\mathcal{O}_{K}=(\gamma)^{3}\tilde{I}^{3}J\text{ where both
}\tilde{I}\text{ and }J\text{ are }S_{K}\text{-ideals.}$

Finally, $(\frac{\lambda}{\gamma^{3}})\mathcal{O}_{K}$ is an $S_{K}$-ideal,
which implies that $u:=\frac{\lambda}{\gamma^{3}}$ is an $S_{K}$-unit. Now, by
dividing $\mu+27=\lambda\text{ by }u$, we get

$\alpha+\beta=\gamma^{3},\qquad\alpha:=\frac{\mu}{u}\in\mathcal{O}_{S_{K}}^{*},\qquad\beta:=\frac{27}{u}\in\mathcal{O}_{S_{K}}^{*}$
(16)

Now, suppose that there is some $\tilde{\mathfrak{P}}\in S_{K}$ that satisfies
$|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
3v_{\tilde{\mathfrak{P}}}(3)$. We will show that
$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$, contradicting Theorem 34
(iv) and hence we can conclude the proof. By using (16) we can rewrite the
assumption $|v_{\tilde{\mathfrak{P}}}(\frac{\alpha}{\beta})|\leq
3v_{\tilde{\mathfrak{P}}}(3)$ in terms of the valuation of $\mu$, using that
$\frac{\alpha}{\beta}=\frac{\mu}{27}$:

$0\leq v_{\tilde{\mathfrak{P}}}(\mu)\leq 6v_{\tilde{\mathfrak{P}}}(3).$

Note that $j_{E^{\prime}}=(\mu+27)(\mu+3)^{3}\mu^{-1}$, hence

$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})=v_{\mathfrak{P}}(\mu+27)+3v_{\mathfrak{P}}(\mu+3)-v_{\mathfrak{P}}(\mu).$

There are three cases according to the valuation of $\tilde{\mathfrak{P}}$ at
$\mu$:
Case (1): Suppose $0\leq v_{\tilde{\mathfrak{P}}}(\mu)\leq
v_{\tilde{\mathfrak{P}}}(3)$. This implies that
$v_{\tilde{\mathfrak{P}}}(\mu+27)=v_{\tilde{\mathfrak{P}}}(\mu)$ and
$v_{\tilde{\mathfrak{P}}}(\mu+3)\geq v_{\tilde{\mathfrak{P}}}(\mu)$, thus
$v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$.
Case (2): Suppose
$v_{\tilde{\mathfrak{P}}}(3)<v_{\tilde{\mathfrak{P}}}(\mu)\leq
3v_{\tilde{\mathfrak{P}}}(3)$. This implies that
$v_{\tilde{\mathfrak{P}}}(\mu+27)\geq v_{\tilde{\mathfrak{P}}}(\mu)$ and
$v_{\tilde{\mathfrak{P}}}(\mu+3)=v_{\tilde{\mathfrak{P}}}(3)$, thus we get
again $v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$.
Case (3): Suppose
$3v_{\tilde{\mathfrak{P}}}(3)<v_{\tilde{\mathfrak{P}}}(\mu)\leq
6v_{\tilde{\mathfrak{P}}}(3)$. This implies that
$v_{\tilde{\mathfrak{P}}}(\mu+27)=3v_{\tilde{\mathfrak{P}}}(3)$ and
$v_{\tilde{\mathfrak{P}}}(\mu+3)=v_{\tilde{\mathfrak{P}}}(3)$, thus we get one
last time $v_{\tilde{\mathfrak{P}}}(j_{E^{\prime}})\geq 0$.
All three cases lead to contradictions and hence we conclude the proof. ∎

###  4.5 Proof of Theorem 11

###### Proof.

We want to apply Theorem 9 with $\tilde{\mathfrak{P}}=\mathfrak{P}$ and
$S_{K}=\\{\mathfrak{P}\\}$. As $3$ is inert, we get $v_{\mathfrak{P}}(3)=1$.

Now, let us consider the equation $\alpha+\beta=\gamma^{3}$, with
$\alpha,\beta\in\mathcal{O}_{S_{K}}^{*}$. By scaling the equation by triple
powers of $3$ and swapping $\alpha$ and $\beta$ if necessary, we may assume
$0\leq v_{\mathfrak{P}}(\beta)\leq v_{\mathfrak{P}}(\alpha)$ with
$v_{\mathfrak{P}}(\beta)\in\\{0,1,2\\}$. Also, we can assume that $\beta$ is
positive, otherwise we multiply everything by $-1$.
Case (1): Suppose $v_{\mathfrak{P}}(\beta)=2$. If
$v_{\mathfrak{P}}(\alpha)\geq 3$, then
$v_{\mathfrak{P}}(\gamma^{3})=v_{\mathfrak{P}}(\alpha+\beta)=2$, which leads
to a contradiction as $v_{\mathfrak{P}}(\gamma^{3})$ must be a multiple of
$3$. Thus, $v_{\mathfrak{P}}(\alpha)=v_{\mathfrak{P}}(\beta)=2$ and
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=0<3$.
Case (2): Suppose $v_{\mathfrak{P}}(\beta)=1$. If
$v_{\mathfrak{P}}(\alpha)\geq 2$, then
$v_{\mathfrak{P}}(\gamma^{3})=v_{\mathfrak{P}}(\alpha+\beta)=1$, which leads
to a contradiction as $v_{\mathfrak{P}}(\gamma^{3})$ must be a multiple of
$3$. Thus, $v_{\mathfrak{P}}(\alpha)=v_{\mathfrak{P}}(\beta)=1$ and
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=0<3$.
Case (3): Suppose $v_{\mathfrak{P}}(\beta)=0$ with $\beta$ not a cube. If
$v_{\mathfrak{P}}(\alpha)>3$, then $v_{\mathfrak{P}}(\gamma^{3})=0$ and
$\beta\equiv\gamma^{3}\mod 3^{4}$. Consider the field extension
$L=K(\sqrt[3]{\beta},\omega)$ of $K(\omega)$. We will show that $L$ is
unramified at $3$, hence contradicting $3\nmid h_{K(\omega)}$.

Consider the element
$\delta:=\frac{\gamma^{2}+\gamma\omega\sqrt[3]{\beta}+\omega^{2}\sqrt[3]{\beta}}{3}$.
Its minimal polynomial is

$m_{\delta}(X)=X^{3}+\gamma\frac{\gamma^{3}-\beta}{3}X^{2}-\gamma^{2}X+\frac{(\gamma^{3}-\beta)^{2}}{27}.$

This belongs to $\mathcal{O}_{K}[X]$ and has discriminant

$\Delta=-2\gamma^{3}\frac{(\gamma^{3}-\beta)^{3}}{3^{5}}-4\gamma^{3}\frac{(\gamma^{3}-\beta)^{5}}{3^{9}}+\gamma^{6}\frac{(\gamma^{3}-\beta)^{2}}{3^{2}}-4\gamma^{6}-\frac{(\gamma^{3}-\beta)^{4}}{3^{3}}.$

We can deduce that $\Delta\equiv-4\gamma^{6}\mod 3$, proving that $L$ is
unramified at $3$. Thus, we must have $v_{\mathfrak{P}}(\alpha)\leq 3$, giving
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=v_{\mathfrak{P}}(\alpha)\leq 3$.
Case (4): Suppose $\beta$ is a cube. By dividing everything through $\beta$,
we can assume that $\beta=1$. Then by the hypothesis of the theorem, we get
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})=v_{\mathfrak{P}}(\alpha)\leq 3$.
All of the possible four cases lead to
$v_{\mathfrak{P}}(\frac{\alpha}{\beta})\leq 3=3v_{\mathfrak{P}}(3)$, so we can
conclude the proof by Theorem 9. ∎

###  4.6 Proof of Theorem 12

###### Proof.

Note that $d\equiv 2\mod 3$ gives that $3$ is inert in the quadratic field
$K=\mathbb{Q}(\sqrt{d})$, take $\mathfrak{P}$ to be the unique prime above $3$
and denote $S_{K}=\\{\mathfrak{P}\\}$. By Theorem 11 it is enough to check
that every solution
$(\alpha,\gamma)\in\mathcal{O}^{*}_{S_{K}}\times\mathcal{O}_{S_{K}}$ with
$v_{\mathfrak{P}}(\alpha)\geq 0$ to the equation $\alpha+1=\gamma^{3}$
satisfies $v_{\mathfrak{P}}(\alpha)\leq 3$.

Assume by a contradiction that we have a solution $\alpha$ to the above
equation such that $v_{\mathfrak{P}}(\alpha)>3$. This implies that
$v_{\mathfrak{P}}(\gamma)=0$, giving $\gamma\in\mathcal{O}_{K}$.
Rearranging we get that $(\gamma-1)(\gamma-\omega)(\gamma-\omega^{2})=\alpha$
when viewed over $L:=K(\omega)$. In the new field extension $L$ we have that
$(3)\mathcal{O}_{L}=(\omega-1)^{2}\mathcal{O}_{L}$. We take
$\mathfrak{p}=(\omega-1)\mathcal{O}_{L}$ and $S_{L}=\\{\mathfrak{p}\\}$.
Denote $x:=\gamma-1,\>y:=\gamma-\omega,\>z:=\gamma-\omega^{2}$ and observe
that

$\begin{cases}x-y=(\omega-1)\\\ y-z=\omega(\omega-1)\\\ \end{cases}$ (17)

Note that $x,y,z\in\mathcal{O}_{S_{L}}$ and they are factors of the
$S_{K}$-unit $\alpha$, hence they must be $S_{L}$-units.

Consider $\tau\in\text{Gal}(L/K)$ such that $\tau(\omega)=\omega^{2}$. It is
easy to see that

$\tau(x)=x,\>\>\>\tau(y)=z\text{ and }\tau(\mathfrak{p})=\mathfrak{p}.$

This implies that $v_{\mathfrak{p}}(y)=v_{\mathfrak{p}}(z)=:r$. We will show
that $r=1$. Firstly note that by (17) we get that
$1=v_{\mathfrak{p}}(\omega(\omega-1))=v_{\mathfrak{p}}(y-z)\geq r$. Suppose
$r\leq 0$. Then $v_{\mathfrak{p}}(x)\geq
v_{\mathfrak{p}}(xyz)=v_{\mathfrak{p}}(\alpha)\geq 8$ since $3^{4}|\alpha$.
Then, by using (17) again, we will get
$1=v_{\mathfrak{p}}(\omega-1)=v_{\mathfrak{p}}(x-y)=r\leq 0$, a contradiction.
So, $r$ must be exactly $1$. As
$v_{\mathfrak{p}}(\alpha)=v_{\mathfrak{p}}(xyz)=8$, we must have
$v_{\mathfrak{p}}(x)=6$. Consider now

$u:=\frac{x}{\omega-1}\text{ and }v=\frac{-y}{\omega-1}.$

By the above discussion, we will get that $\mathfrak{p}^{5}|u$ and
$v\in\mathcal{O}_{L}^{*}$. Denote $F:=\mathbb{Q}(\omega)$. As $v$ is a unit,
we must have

$\text{Norm}_{L/F}(v)\in\mathcal{O}_{F}^{*}=\langle\omega+1\rangle$ (18)

As $u+v=1$, we get that $v\equiv 1\mod 3$. Let $\sigma$ be the generator of
$\text{Gal}(L/F)$. By noting that $3|\sigma(u)$, we get that $\sigma(v)\equiv
1\mod 3$ and consequently $\text{Norm}_{L/F}(v)=v\sigma(v)\equiv 1\mod 3$.
This and (18) give $\text{Norm}_{L/F}(v)=1$. Suppose that
$v\in\mathcal{O}_{L}^{*}\setminus\mathcal{O}_{K}^{*}=\omega\mathcal{O}_{K}^{*}$,
then $\omega|\text{Norm}_{L/F}(v)$ contradicting $\text{Norm}_{L/F}(v)=1$.
Thus $v\in\mathcal{O}_{K}^{*}$ giving $u=1-v\in\mathcal{O}_{K}$ which is a
contradiction as $u$ is a ratio of a $K$-integer and $\omega-1\notin K$. ∎

###  4.7 Proof of Theorem 13

We first need to prove some preliminary lemmas. Throughout this section, $K$
denotes a totally real field of degree $n$, $L:=K(\omega)$ and
$F:=\mathbb{Q}(\omega)$. Moreover, $K$ satisfies the conditions (i), (ii),
(iii) and (iv) in the statement of Theorem 13. More precisely let $q$ be the
prime which totally ramifies in $K$. Note that $q\geq 5$ so it is inert in
$F$. Denote $\tilde{\mathfrak{q}}:=(q)\mathcal{O}_{F}$ and take $\mathfrak{q}$
to be the unique prime above $q$ in $L$, so that
$(q)\mathcal{O}_{L}=\mathfrak{q}^{n}\mathcal{O}_{L}$. Take $\mathfrak{P}$ to
be the unique prime above $3$ in $K$ and denote $S_{K}=\\{\mathfrak{P}\\}$. In
$L$ we have that $(3)\mathcal{O}_{L}=(\omega-1)^{2}\mathcal{O}_{L}$. We take
$\mathfrak{p}=(\omega-1)\mathcal{O}_{L}$ and $S_{L}=\\{\mathfrak{p}\\}$.

######  Lemma 35.

Let $\lambda\in\mathcal{O}_{L}$, then there exists
$\beta\in\mathbb{Z}[\omega]$ such that $\lambda\equiv b\mod\mathfrak{q}$ and

$\textup{Norm}_{L/F}(\lambda)\equiv b^{n}\mod\tilde{\mathfrak{q}}.$ (19)

###### Proof.

Note that
$\mathcal{O}_{L}/\mathfrak{q}\mathcal{O}_{L}\cong\mathbb{F}_{q}(\omega)\cong\mathbb{Z}[\omega]/q\mathbb{Z}[\omega]$.
Thus, there exists $b\in\mathbb{Z}[\omega]$ such that $\lambda\equiv
b\mod\mathfrak{q}$. Let $\bar{L}$ be the normal closure of $L$. Take
$\sigma\in\text{Gal}(\bar{L}/F)$. Note that

$(\sigma(\mathfrak{q}\mathcal{O}_{\bar{L}}))^{n}=\sigma(q\mathcal{O}_{\bar{L}})=q\mathcal{O}_{\bar{L}}=(\mathfrak{q}\mathcal{O}_{\bar{L}})^{n}.$

Thus, by the unique factorisation of ideals we get
$\sigma(\mathfrak{q}\mathcal{O}_{\bar{L}})=\mathfrak{q}\mathcal{O}_{\bar{L}}$.
Moreover, by applying $\sigma$ to $\lambda\equiv b\mod\mathfrak{q}$ we get
that $\sigma(\lambda)\equiv b\mod\mathfrak{q}\mathcal{O}_{\bar{L}}$. Finally
multiplying everything together

$\text{Norm}_{L/F}(\lambda)=\prod_{\sigma}\sigma(\lambda)\equiv
b^{n}\mod\mathfrak{q}\mathcal{O}_{\bar{L}}.$

As $\lambda\in\mathcal{O}_{L}$, it follows that
$\text{Norm}_{L/F}(\lambda)\in\mathcal{O}_{F}$. Also
$b^{n}\in\mathcal{O}_{F}$. Thus,
$\text{Norm}_{L/F}(\lambda)-b^{n}\in\mathcal{O}_{F}\cap\mathfrak{q}\mathcal{O}_{\bar{L}}=\tilde{q}\mathcal{O}_{F}.$
Hence (19) holds. ∎

######  Lemma 36.

Suppose $\lambda\in\mathcal{O}_{L}^{*}$ and (ii) holds, i.e.
$\gcd(n,q^{2}-1)=1$. Then
$(\lambda\mod\mathfrak{q})\in\langle\omega+1\rangle=\\{\pm
1,\pm(\omega+1),\pm\omega\\}$.

###### Proof.

Let $b\in\mathbb{Z}[\omega]$ with $\lambda\equiv b\mod\mathfrak{q}$ as in
Lemma 35. This gives us $\text{Norm}_{L/F}(\lambda)\equiv
b^{n}\mod\tilde{\mathfrak{q}}$. However, as $\lambda$ is a unit, we must have

$\text{Norm}_{L/F}(\lambda)\in\mathcal{O}_{F}^{*}=\langle\omega+1\rangle.$

Putting these together we get that
$b^{n}\equiv(\omega+1)^{i}\mod\tilde{\mathfrak{q}}$. On the other hand,
$b\in\mathcal{O}_{F}$ and maps to a non-zero element of
$\mathcal{O}_{F}/\tilde{\mathfrak{q}}\mathcal{O}_{F}\cong\mathbb{F}_{q^{2}}$
thus $b^{q^{2}-1}\equiv 1\mod\tilde{\mathfrak{q}}$. The assumption
$\gcd(n,q^{2}-1)=1$ is equivalent to the existence of integers $u,v$ so that
$un+v(q^{2}-1)=1$. It follows that

$b=(b^{n})^{u}(b^{q^{2}-1})^{v}\equiv(\omega+1)^{iu}\mod\tilde{\mathfrak{q}}.$

Thus, $(\lambda\mod\mathfrak{q})\in\langle\omega+1\rangle=\\{\pm
1,\pm(\omega+1),\pm\omega\\}$. ∎

###### Proof of Theorem 13.

We will reduce the problem to a simpler one as described in Section 4.6. More
precisely, by using Theorem 11 and then rewriting the equation into an
$S_{K}$-unit equation, we get that it is enough to show that there are no
solutions to

$u+v=1$ (20)

with $(u,v)\in\mathcal{O}_{S_{L}}^{*}\times\mathcal{O}_{L}^{*}$ such that
$\mathfrak{p}^{5}|u$. We will prove the slightly stronger statement that there
are no solutions to (20) such that $9|u$.
Note that by (20) it follows that $v\equiv 1\mod 9$. Thus $\sigma(v)\equiv
1\mod 9$ for all conjugates $\sigma(v)$ of $v$ in $\text{Gal}(\bar{L}/F)$,
where $\bar{L}$ is the normal closure of $L$. Hence,
$\text{Norm}_{L/F}(v)\equiv 1\mod 9$. As $v$ is a unit, we get
$\text{Norm}_{L/F}(v)\in\mathcal{O}_{F}^{*}=\langle\omega+1\rangle.$ Thus, the
only possibility is

$\text{Norm}_{L/F}(v)=1.$ (21)

By Lemma 36 applied with $\lambda=v$ we get that

$(v\mod\mathfrak{q})\in\langle\omega+1\rangle=\\{\pm
1,\pm(\omega+1),\pm\omega\\}.$ (22)

If $v\equiv 1\mod\mathfrak{q}$, then $u=1-v\equiv 0\mod\mathfrak{q}$, so
$\mathfrak{q}|u$, but this is false as $u$ is an $S_{L}$-unit and
$\mathfrak{p}$ and $\mathfrak{q}$ are different primes.
Thus $(v\mod\mathfrak{q})\in\\{-1,\pm(\omega+1),\pm\omega\\}$. Then

$(\text{Norm}_{L/F}(v)\mod\mathfrak{q})\in\\{(-1)^{n},(\pm(\omega+1))^{n},(\pm\omega)^{n}\\}.$
(23)

Since $\gcd(n,q^{2}-1)=1$ and $q\geq 5$ is a prime, it follows in particular
that $2\nmid n$ and $3\nmid n$. This observation along with (23) proves that
$\text{Norm}_{L/F}(v)\mod\mathfrak{q}\not\equiv 1$, contradicting (21).

∎

##  5 $S$-unit equations and computability

Finally, we will describe how to algorithmically check the hypothesises in our
two main Theorems 3 and 9 by studying how to compute solutions of certain
(linear) $S$-unit equations over the number field $K$, i.e. equations of the
form

$ax+by=1\text{ where }a,b\in K^{*}\text{ with solutions
}x,y\in\mathcal{O}_{S}^{*}.$

Throughout this section $S$ denotes a finite set of prime ideals of $K$.

######  Theorem 37 (Siegel).

Let $K$ be a number field and $S\subset\mathcal{O}_{K}$ a finite set of prime
ideals, and let $a,b\in K^{*}$. Then, the equation

$ax+by=1$

has only finitely many solutions in $\mathcal{O}_{S}^{*}.$

######  Remark 38.

Methods of effectively computing solutions to $S$-unit were pioneered by De
Weger’s famous thesis [33] for $K=\mathbb{Q}$. His method of lattice
approximation reduction algorithms was later generalized for all number fields
by others, see for example Smart’s [29]. Moreover, an $S$-unit solver for
$a=b=1$ has been implemented in the free open-source mathematics software,
Sage by A. Alvarado, A. Koutsianas, B. Malmskog, C. Rasmussen, D. Roe, C.
Vincent, M. West in [1].

We will now study two non-linear equations involving $S$-units which are going
to play a crucial role in checking the hypothesis of our Theorems 3 and 9. Let
$K$ be a number field and $S$ a finite set of prime ideals. Consider the
equation

$\alpha+\beta=\gamma^{i},\>\alpha,\beta\in\mathcal{O}_{S}^{*},\>\>\gamma\in\mathcal{O}_{S}.$

There is a natural scaling action of $\mathcal{O}_{S}^{*}$ on the solutions.
We regard two solutions
$(\alpha_{1},\beta_{1},\gamma_{1})\sim_{i}(\alpha_{2},\beta_{2},\gamma_{2})$
as equivalent if there is some $\epsilon\in\mathcal{O}_{S}^{*}$ such that
$\alpha_{2}=\epsilon^{i}\alpha_{1}$, $\beta_{2}=\epsilon^{i}\beta_{1}$ and
$\gamma_{2}=\epsilon\gamma_{1}$.

######  Theorem 39.

Let $K$ be a number field and $S$ a finite set of prime ideals. Consider the
equation

$\alpha+\beta=\gamma^{i},\>\alpha,\beta\in\mathcal{O}_{S}^{*},\>\>\gamma\in\mathcal{O}_{S}.$

For $i=2,3$, the equation has a finite number of solutions up to the
equivalence relation $\sim_{i}$. Moreover, these are effectively computable.

###### Proof.

Let $i=2$ and
$(\alpha,\beta,\gamma)\in\mathcal{O}^{*}_{S}\times\mathcal{O}^{*}_{S}\times\mathcal{O}_{S}$
a solution to $\alpha+\beta=\gamma^{2}$. By Dirichlet Unit Theorem
$\mathcal{O}^{*}_{S}$ is finitely generated, and hence
$\mathcal{O}^{*}_{S}/(\mathcal{O}^{*}_{S})^{2}$ is finite. Fix a set of
representatives $\beta_{1},\beta_{2},...,\beta_{l}$. We may scale our solution
so that $\beta\in\\{\beta_{1},\beta_{2},...,\beta_{l}\\}$. Thus, there are
finitely many choices of $\beta$ (up to $\sim_{2}$ equivalence) and we fix one
of them. We next show that for each such choice of $\beta$, there is a finite
number of distinct $\alpha$, and thus, a finite number of triples
$(\alpha,\beta,\gamma)$ up to $\sim_{2}$ equivalence.

We rewrite the equation as

$(\gamma+\sqrt{\beta})(\gamma-{\sqrt{\beta}})=\alpha\>\text{over }L.$ (24)

where $L:=K(\sqrt{\beta})$. Denote by
$x:=\gamma+\sqrt{\beta},y:=\gamma-\sqrt{\beta}$ and consider
$S^{\prime}:=\\{\mathfrak{P}_{L}\text{ prime of
}L:\mathfrak{P}_{L}|\mathfrak{P}_{K},\text{ for some }\mathfrak{P}_{K}\in
S\\}.$ We claim that $x,y$ are both $S^{\prime}$-units in $L$. This follows by
considering the valuation of the product in (24) at the primes of $L$ outside
the set $S^{\prime}$. Then, we use the definition of $S^{\prime}$ and the fact
that $\alpha$ is an $S$-unit in $K$. Notice that

$\frac{1}{2\sqrt{\beta}}x-\frac{1}{2\sqrt{\beta}}y=1.$

By Theorem 37, we get finitely many $S^{\prime}$-unit solutions $x,y$, and
thus finitely many possibilities for $\alpha=xy$. Moreover, these are
computable by Remark 38.

For $i=3$, the argument works in a similar manner. Fixing a representative
$\beta$ of the finite quotient
$\mathcal{O}^{*}_{S}/(\mathcal{O}^{*}_{S})^{3}$, we rewrite the equation as

$(\gamma-\sqrt[3]{\beta})(\gamma-\omega{\sqrt[3]{\beta}})(\gamma-\omega^{2}\sqrt[3]{\beta})=\alpha\>\text{over
}L$ (25)

where $L=K(\omega,\sqrt[3]{\beta})$ and $\beta\neq-1$. Denote by
$x:=\gamma-\sqrt[3]{\beta},y:=\gamma-\omega{\sqrt[3]{\beta}},$
$S^{\prime}:=\\{\mathfrak{P}_{L}\text{ prime of
}L:\mathfrak{P}_{L}|\mathfrak{P}_{K},\text{for some }\mathfrak{P}_{K}\in
S\\}.$

We make the quick note that for $\beta=-1$ we take $x:=\gamma+1$,
$y=\gamma+\omega$, $L:=K(\omega)$ and the rest of the argument follows the
same, so it is omitted.

As in the previous case, by examining the product in (25) we get that $x,y$
are both $S^{\prime}$-units in $L$ and

$\frac{1}{(\omega-1)\sqrt[3]{\beta}}x-\frac{1}{(\omega-1)\sqrt[3]{\beta}}y=1.$

Thus, by Theorem 37, Remark 38 and the observation that
$\alpha=xy(y-\omega(\omega-1)\sqrt[3]{\beta})$, giving finely many numbers
$\alpha$ for a fixed $\beta$ and so we conclude the proof.

∎

######  Remark 40.

In the hypotheses of Theorems 3 and 9 one needs to examine the local behaviour
of $\frac{\alpha}{\beta}$ which, by the above theorem, can only take a finite,
computable number of values.

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