# Two results on $x^{r}+y^{r}=dz^{p}$

Nuno Freitas  Instituto de Ciencias Matemáticas, CSIC, Calle Nicolás Cabrera
13–15, 28049 Madrid, Spain<EMAIL_ADDRESS>and  Filip Najman
University of Zagreb, Faculty of Science, Department of Mathematics, Bijenička
cesta 30, 10000 Zagreb, Croatia<EMAIL_ADDRESS>
###### Abstract.

This note proves two theorems regarding Fermat-type equation
$x^{r}+y^{r}=dz^{p}$ where $r\geq 5$ is a prime. Our main result shows that,
for infinitely many integers $d$, the previous equation has no non-trivial
primitive solutions such that $2\mid x+y$ or $r\mid x+y$, for a set of
exponents $p$ of positive density. We use the modular method with a symplectic
argument to prove this result.

The second author was supported by the QuantiXLie Centre of Excellence, a
project co-financed by the Croatian Government and European Union through the
European Regional Development Fund - the Competitiveness and Cohesion
Operational Programme (Grant KK.01.1.1.01.0004) and by the Croatian Science
Foundation under the project no. IP-2018-01-1313. The first author is
partially supported by the PID2019-107297GB-I00 grant of the MICINN (Spain)

##  1\. Introduction

We consider Fermat-type equations of the form

(1.1) $x^{r}+y^{r}=dz^{p},$

where $r,p>3$ are primes and $d$ is an odd positive integer, not an $r$-th
power and $r\nmid d$. A solution $(a,b,c)$ to (1.1) is called primitive if
$\gcd(a,b,c)=1$ and non-trivial if $abc\neq 0$.

We will study the above equation via the modular method. Since $d$ is not a
$r$-th power, we have $d\neq 1$; this avoids the trivial solution $(1,0,1)$
which is a well known obstruction. Indeed, the modular method aims to obtain a
contradiction with the existence of a solution, which is usually not possible
once a solution does exist. More precisely, for $d=1$, in the elimination
stage of the method, we would have to distinguish the mod $p$ representation
of the Frey curve attached to a non-trivial putative solution $(a,b,c)$ from
the mod $p$ representation of the Frey curve attached to the solution
$(1,0,1)$. Currently, with the exception of very few special cases, there are
no techniques available to do this. The hypothesis that $d$ is odd avoids the
solution $(1,1,1)$ for $d=2$ but it also plays (together with $r\nmid d$) an
essential rôle in the proof of Theorem 1.2 below (see Remark 3.2).

Equation (1.1) has been the focus of various recent works. It has been
completely resolved for $(r,d)=(5,3)$, $(13,3)$ and all $p\geq 2$ in [3, 2],
and for $(r,d)=(7,3)$ in [4]; moreover, in [12] it is shown for $d=1$ and many
values of $r<150$ that it admits no non-trivial primitive solutions $(a,b,c)$
with $c$ even. Under the extra assumptions that $d$ is not divisible by $p$-th
powers and is divisible only by primes $q\not\equiv 1\pmod{r}$, an older
result due to Kraus implies that, fixed both $r$ and $p$, the set of
coefficients $d$ for which (1.1) has a non-trivial primitive solution is
finite (see [11, Théorème 1]).

Given that we will focus on asymptotic results, i.e. for large enough $p$, we
can also assume that $p$ is large compared to $r$ and $p\nmid d$. We will
prove the following theorem.

######  Theorem 1.2.

Let $r$ and $d$ be as above. For a set of prime exponents $p$ of positive
density, the equation (1.1) has no non-trivial primitive solutions $(a,b,c)$
such that $2\mid a+b$ or $r\mid a+b$.

This result has no constraints on the value of $r$ and $d$ beyond those
introduced in the first paragraph. Instead we have $2$-adic or $r$-adic
restrictions on the solutions, that naturally occur in the study of Fermat-
type equations. In particular, the condition $r\mid a+b$ is equivalent to
$r\mid c$ and this is analogous to what Darmon calls a first case solution to
the Fermat-type equations $x^{p}+y^{p}=z^{r}$ (see [5, Definition 3.6] and
results in §3 of loc. cit.). A result of similar flavor to ours regarding
$x^{p}+y^{p}=z^{r}$ is given in [1], more precisely, it is shown this equation
has no non-trivial primitive solutions with $r\mid ab$ and $2\nmid ab$. We
remark that although in loc. cit. the proof uses Darmon’s higher dimensional
abelian varieties instead of Frey elliptic curves, the condition $r\mid ab$
plays a similar role to ours $r\mid a+b$ in that it forces the Frey varieties
to have multiplicative reduction at $r$.

The proof of Theorem 1.2 was inspired by [10] and relies on combining the
modular method with the symplectic argument. We refer the reader to [7] for a
quick introduction to Diophantine applications of the symplectic argument; for
a comprehensive study of symplectic criteria and further Diophantine
applications we refer to [8].

We will also prove the following theorem about (1.1) where the $2$-adic and
$r$-adic conditions are replaced by a $p$-adic condition.

######  Theorem 1.3.

Let $r$ and $d$ be as above. Then for prime exponents $p\equiv\pm 1\pmod{r}$
large enough the equation (1.1) has no non-trivial primitive solutions
$(a,b,c)$ such that $p\mid a+b$.

##  2\. The Frey curve

We first recall the construction of a Frey curve from [6]. Let $r\geq 5$ be a
prime and $\zeta=\zeta_{r}$ a fixed primitive $r$-th root of unity. Let
$K=\mathbb{Q}(\zeta_{r})^{+}$ be the maximal real subfield of
$\mathbb{Q}(\zeta_{r})$. Define the polynomials

$f_{1}:=x^{2}+(\zeta^{2}+\zeta^{-2})xy+y^{2}\quad\text{and}\quad
f_{2}:=x^{2}+(\zeta+\zeta^{-1})xy+y^{2}$

and the constants

$\alpha=\zeta(1-\zeta)(1-\zeta^{-3}),\quad\beta=(1-\zeta)(1-\zeta^{-1}),\quad\gamma=-(1-\zeta^{2})(1-\zeta^{-2}).$

We set also

$A=A_{a,b}=\alpha(a+b)^{2},\quad B=B_{a,b}=\beta f_{1}(a,b),\quad
C=C_{a,b}=\gamma f_{2}(a,b)$

and define the Frey curve

(2.1) $F_{a,b}:Y^{2}=X(X-A_{(a,b)})(X+B_{(a,b)}).$

By construction we have $A+B=C$ and the following standard invariants

$\displaystyle c_{4}(F_{a,b})$ $\displaystyle=$ $\displaystyle
2^{4}(A_{a,b}^{2}+A_{a,b}B_{a,b}+B_{a,b}^{2}),$ $\displaystyle c_{6}(F_{a,b})$
$\displaystyle=$ $\displaystyle
2^{5}(2A_{a,b}^{3}+3A_{a,b}^{2}B_{a,b}-3A_{a,b}B_{a,b}^{2}-2B_{a,b}^{3}),$
$\displaystyle\Delta(F_{a,b})$ $\displaystyle=$ $\displaystyle
2^{4}\left(A_{a,b}B_{a,b}C_{a,b}\right)^{2}.$

For $n\in\mathbb{Z}_{>0}$ and $x\in K$ denote by $\mbox{Rad}_{n}(x)$ the
product of the primes in $K$ not dividing $n$.

Also let $\mathfrak{p}_{r}$ denote the unique prime in $K$ above $r$.

######  Proposition 2.2.

Let $(a,b,c)$ a primitive solution to (1.1) such that $\gcd(a,b)=1$. Then the
conductor $N_{F}$ of the curve $F=F_{a,b}$ is of the form

$2^{s}\mathfrak{p}_{r}^{t}\mathfrak{c}\mbox{Rad}_{2r}(a+b)$

where $\mathfrak{c}$ is a squarefree product of primes in $K$ dividing $c$.

Moreover, $s=1$ if $2\mid a+b$ and $t=1$ if $r\mid a+b$.

###### Proof.

Since $\alpha,\beta,\gamma$ are of the form
$\pm\zeta^{s}(1-\zeta^{t})(1-\zeta^{u})$, where neither $t$ nor $u$ are
$\equiv 0\pmod{r}$, this means that the only prime dividing
$\alpha\beta\gamma$ is $\mathfrak{p}_{r}$ and
$\upsilon_{\mathfrak{p}_{r}}(\alpha\beta\gamma)=3$.

Recall the factorisation

(2.3) $a^{r}+b^{r}=(a+b)\Phi_{r}(a,b)=dc^{p}$

where $\Phi_{r}(a,b)$ is the $r$-th cyclotomic polynomial; we have
$\gcd(a+b,\Phi_{r}(a,b))=1$ or $r$ since $a,b$ are coprime. Moreover, the
polynomials $f_{1}(a,b)$ and $f_{2}(a,b)$ in the definitions of $B$ and $C$
are factors of $\Phi_{r}(a,b)$ over $K$; see [6, §2] for proofs of these
elementary properties (in particular, Corollary 2.2 and the discussion after
Corollary 2.6 in loc. cit.).

Let $\mathfrak{q}\nmid 2r$ be a prime in $K$. Then
$\upsilon_{\mathfrak{q}}(\Delta(F))=2(\upsilon_{\mathfrak{q}}(A)+\upsilon_{\mathfrak{q}}(B)+\upsilon_{\mathfrak{q}}(C)).$

If $\mathfrak{q}\nmid a^{r}+b^{r}$ we have
$\upsilon_{\mathfrak{q}}(\Delta(F))=0$, so $\mathfrak{q}$ is a prime of good
reduction.

Assume $\mathfrak{q}\mid a^{r}+b^{r}$. Then $\mathfrak{q}\mid A$ or
$\mathfrak{q}\mid\Phi_{r}(a,b)$ and not both simultaneously.

If $\mathfrak{q}\mid\Phi_{r}(a,b)$ and $\mathfrak{q}\nmid BC$, then
$\mathfrak{q}$ is a prime of good reduction.

Otherwise, either $\mathfrak{q}\mid A$ and $\mathfrak{q}\nmid BC$ or
$\mathfrak{q}\nmid A$ and $\mathfrak{q}\mid BC$. In both cases, we get
$v_{\mathfrak{q}}(\Delta(F))>0$ and $v_{\mathfrak{q}}(c_{4}(F))=0$, so
$\mathfrak{q}$ is a prime of multiplicative reduction.

We now consider the prime $\mathfrak{p}_{r}$ above $r$. Assume $r\mid a+b$. We
have

$\upsilon_{\mathfrak{p}_{r}}(c_{4})=4,\quad\upsilon_{\mathfrak{p}_{r}}(c_{6})=6,\quad\upsilon_{\mathfrak{p}_{r}}(\Delta)=10+4\upsilon_{\mathfrak{p}_{r}}(a+b).$

Since $\upsilon_{\mathfrak{p}_{r}}(a+b)\geq(r-1)/2$ the equation is non-
minimal and after a coordinate change we have
$\upsilon_{\mathfrak{p}_{r}}(c_{4})=0$ and
$\upsilon_{\mathfrak{p}_{r}}(\Delta)>0$. Thus $F$ has bad multiplicative
reduction, i.e. $\upsilon_{\mathfrak{p}_{r}}(N_{F})=1$.

Finally we consider a prime $\mathfrak{q}_{2}\mid 2$. As 2 is unramified in
$K$ we will use [13, Table IV] to read the conductor at $\mathfrak{q}_{2}$ in
terms of the $\mathfrak{q}_{2}$-adic valuations of the standard invariants
$(c_{4}(F),c_{6}(F),\Delta(F))$.

Assume $2\mid a+b$ so $2\nmid\Phi(a,b)$. Since $p>3$, the shape of equation
(1.1) implies $8\mid a+b$. Moreover,

$\upsilon_{\mathfrak{q}_{2}}(c_{4})=4,\quad\upsilon_{\mathfrak{q}_{2}}(c_{6})=6,\quad\upsilon_{\mathfrak{q}_{2}}(\Delta)=4+4\upsilon_{\mathfrak{q}_{2}}(a+b).$

The corresponding entries in [13, Table IV] give us Tate case 7 with
$\upsilon_{\mathfrak{q}_{2}}(N_{F})=4$ or the equation for $F$ is non-minimal
as $\upsilon_{\mathfrak{q}_{2}}(a+b)\geq 3$. We will apply [13, Prop. 4] to
show that we are in the non-minimal case; we state it here for convenience.

######  Proposition 2.4.

Let $K/\mathbb{Q}_{2}$ be a finite extension with ring of integers
$\mathcal{O}$ and uniformizer $\pi$. Let $W/K$ be an elliptic curve given by a
Weierstrass model $(W)$ with standard invariants $a_{i}$ and $b_{i}$. Assume
that $(W)$ is in a Tate case $\geq 7$.

(a) There is $r\in\mathcal{O}$ such that

$b_{8}+3rb_{6}+3r^{2}b_{4}+r^{3}b_{2}+3r^{4}\equiv 0\pmod{\pi^{5}}$

(b) Choose any $r$ satisfying (a). Then we are in a Tate case $\geq 8$ if and
only if there is $s\in\mathcal{O}$ satisfying

$a_{2}+3r-sa_{1}-s^{2}\equiv 0\pmod{\pi^{2}}$

For our curve $F$, we have $a_{1}=a_{3}=a_{6}=0$, $a_{2}=B-A$ and $a_{4}=-AB$
and $b_{8}=-(AB)^{2}$. Since $\upsilon_{\mathfrak{q}_{2}}(A)\geq 3$, taking
$r=0$ the condition in part (a) of Proposition 2.4 is satisfied and the
condition of part (b) becomes $B\equiv s^{2}\pmod{\mathfrak{q}_{2}^{2}}$, that
is $B$ is a square mod $\mathfrak{q}_{2}^{2}$.

Let $z=-\zeta_{r}-\zeta_{r}^{-1}$. Using
$a\equiv-b\pmod{\mathfrak{q}_{2}^{2}}$ we get

$\displaystyle B$
$\displaystyle=(2+z)(a^{2}+(z^{2}-2)ab+b^{2})=(2+z)((a+b)^{2}-(z^{2}-4)ab)$
$\displaystyle\equiv(z+2)(z^{2}-4)ab\equiv-b^{2}(z+2)^{2}(z-2)\pmod{\mathfrak{q}_{2}^{2}}.$

Moreover,

$2-z=\zeta_{r}+\zeta_{r}^{-1}+2=(\zeta_{r}^{\frac{r+1}{2}}+\zeta_{r}^{-\frac{r+1}{2}})^{2}$

with $\zeta_{r}^{\frac{r+1}{2}}+\zeta_{r}^{-\frac{r+1}{2}}\in\mathcal{O}_{K}$
because it is fixed by complex conjugation; thus $B$ is a square mod
$\mathfrak{q}_{2}^{2}$ so part (b) is satisfied and we are in the non-minimal
case. After a change of variables we get

(2.5)
$\upsilon_{\mathfrak{q}_{2}}(c_{4})=\upsilon_{\mathfrak{q}_{2}}(c_{6})=0,\quad\upsilon_{\mathfrak{q}_{2}}(\Delta(F))=-8+4\upsilon_{\mathfrak{q}_{2}}(a+b)>0,$

from which we see that $F$ has multiplicative reduction, i.e
$\upsilon_{\mathfrak{q}_{2}}(N_{F})=1$. ∎

Recall from [6, §2] that $\gcd(a+b,\Phi_{r}(a,b))=1$ or $r$ and
$\Phi_{r}(a,b)$ is divisible only by $r$ and primes $q\equiv 1\pmod{r}$. Write
$d=d_{0}d_{1}$ where a prime $q\mid d_{1}$ if and only if $q\equiv 1\pmod{r}$.

######  Proposition 2.6.

For large enough $p$ the following holds. For all primitive solutions
$(a,b,c)$ of (1.1) with $\gcd(a,b)=1$ the representation $\bar{\rho}_{F,p}$ is
irreducible and modular of weight $2$. Moreover, it’s Serre level is
$2^{s}\mathfrak{p}_{r}^{t}\mbox{Rad}_{2r}(d_{0}d_{1}^{\prime})$ where
$\mathfrak{q}\mid d_{1}^{\prime}$ if and only if $\mathfrak{q}\mid d_{1}$ and
$\mathfrak{q}\mid\mathfrak{c}$ where $s$, $t$ and $\mathfrak{c}$ are as in
Proposition 2.2.

###### Proof.

From [6, Theorems 4.3 and 4.4], it follows that there exists a constant
$C_{r}$, depending only on $r$, such that for all $p>C_{r}$ the representation
$\bar{\rho}_{F,p}$ is modular of weight 2 and absolutely irreducible.

Let $\mathfrak{q}\nmid 2r$ be a prime dividing $a+b$. From Proposition 2.2, we
know $\mathfrak{q}$ is a prime of bad multiplicative reduction.

From (2.3) we have $a+b=2^{sp}\cdot r^{k}\cdot d_{0}\cdot c_{0}^{p}$ with
$s,k\geq 0$ and $c_{0}\mid c$. Thus

(2.7)
$\upsilon_{\mathfrak{q}}(\Delta(F))=2\upsilon_{\mathfrak{q}}(A)=4\upsilon_{\mathfrak{q}}(a+b)=4\upsilon_{\mathfrak{q}}(d_{0})+4p\upsilon_{\mathfrak{q}}(c_{0})\equiv
4\upsilon_{\mathfrak{q}}(d_{0})\pmod{p}.$

When $\upsilon_{\mathfrak{q}}(d_{0})\neq 0$ then
$4\upsilon_{\mathfrak{q}}(d_{0})\not\equiv 0\pmod{p}$ for large enough $p$;
further enlarging $p$ we can assume $p\nmid d$. Hence the primes
$\mathfrak{q}\nmid 2r$ dividing $a+b$ that also divide the Serre level of
${\overline{\rho}}_{E,p}$ are precisely those dividing $d_{0}$.

Finally, the multiplicative primes $\mathfrak{q}\mid\mathfrak{c}$ divide the
Serre level if and only if $\mathfrak{q}\mid d_{1}$. Indeed, recall that
$\mathfrak{p}_{r}$ is the only prime that can divide both $B$ and $C$, and
that they are both coprime to $a+b$ as well (and hence to $d_{0}$) so both $B$
and $C$ are $p$-th powers times a divisor in $\mathcal{O}_{K}$ of $d_{1}r$. We
have

(2.8) $\upsilon_{\mathfrak{q}}(\Delta(F))\equiv
2\upsilon_{\mathfrak{q}}(\Delta(d_{1}))\pmod{p},$

hence, for large $p$, a prime $\mathfrak{q}\mid\mathfrak{c}$ satisfies
$\upsilon_{\mathfrak{q}}(\Delta(F))\not\equiv 0\pmod{p}$ if and only if
$\mathfrak{q}\mid d_{1}$. ∎

##  3\. Proof of Theorem 1.2

We will use the following lemma several times, so we state it here before
proceeding further.

######  Lemma 3.1 ([10, Lemme 1.6]).

Let $E$ and $E^{\prime}$ be two elliptic curves over $\mathbb{Q}$ and
$p,\ell_{1},\ell_{2}$ three distinct primes with $p>2$. Suppose $E[p]$ and
$E^{\prime}[p]$ are isomorphic. Suppose $E$ and $E^{\prime}$ have
multiplicative reduction at $\ell_{i}$ and $p$ does not divide
$v_{\ell_{i}}(\Delta(E))$, which implies that $p$ does not divide
$v_{\ell_{i}}(\Delta(E^{\prime}))$. Then the reduction mod $p$ of
$v_{\ell_{1}}(\Delta(E))v_{\ell_{2}}(\Delta(E))v_{\ell_{1}}(\Delta(E^{\prime}))v_{\ell_{2}}(\Delta(E^{\prime}))$
is a square in $\mathbb{F}_{p}$.

Note that from a non-trivial solution $(a,b,c)$ to equation (1.1) with
$m=\gcd(a,b)>1$ and $\gcd(a,b,c)=1$, we obtain a non-trivial primitive
solution to $x^{r}+y^{r}=d^{\prime}z^{p}$ with $\gcd(a,b)=1$ where
$d^{\prime}=\frac{d}{m^{r}}$. Since $d$ is not an $r$-power, the same is true
for $d^{\prime}$. Clearly $d^{\prime}$ is also odd and $r\nmid d^{\prime}$,
hence we are reduced to the case of solutions satisfying $\gcd(a,b)=1$. The
argument below is inspired by [10, Theorem 2.1] and it will show by
contradiction that these latter solutions do not exist for $p$ in a set of
primes of density $>0$.

Suppose there is a non-trivial primitive solution $(a,b,c)$ to (1.1) with
$a,b$ coprime111In the paper [6] where our Frey curve $F$ was introduced, a
solution $(a,b,c)$ is called primitive when $\gcd(a,b)=1$. Here we decided to
use the condition $\gcd(a,b,c)=1$ instead because this is more standard in the
context of the generalized Fermat equation $Ax^{r}+By^{q}=Cz^{p}$..

By Proposition 2.6, for sufficiently large $p$, we have
$\bar{\rho}_{F,p}\simeq\bar{\rho}_{f,\mathfrak{p}}$, where $f$ is a Hilbert
newform of level
$2^{s}\cdot\mathfrak{p}_{r}^{t}\cdot\mbox{Rad}_{2r}(d_{0}d_{1}^{\prime})$,
where $\mathfrak{p}$ is a prime in the field of coefficients $K_{f}$ of $f$.
Since the field of coefficients of $F$ is $\mathbb{Q}$, by enlarging $p$ if
needed, we can assume that $K_{f}=\mathbb{Q}$. Moreover, since $d\neq 1$,
there is at least one prime $\mathfrak{q}$ in $K$ dividing
$\mbox{Rad}_{2r}(d_{0}d_{1}^{\prime})$, which is a Steinberg prime of $f$.
Therefore, by the Eichler-Shimura correspondence for $f$, there is an isogeny
class of elliptic curves defined over $K$ corresponding to $f$. Let $E$ denote
an elliptic curve in that isogeny class. Since $F$ has full $2$-torsion over
$K$, by further enlarging $p$ if necessary, we can also assume that $E$ has
full $2$-torsion; see [9, §2.4 and §4] for the previous claims.

From the above, in particular, we have an isomorphism of $G_{K}$-modules
$\phi:F[p]\rightarrow E[p]$.

(1) Suppose $2\mid a+b$. Let $\mathfrak{q}_{2}\mid 2$ and $\mathfrak{q}\mid d$
be primes in $K$.

Recall that $d=d_{0}d_{1}$ where a prime $q\mid d_{1}$ if and only if $q\equiv
1\pmod{r}$.

(1a) Assume that $\mathfrak{q}\mid d_{0}$. Then the curve $F$ has
multiplicative reduction at $\mathfrak{q}_{2}$ and $\mathfrak{q}$ by
Proposition 2.2; we will apply Lemma 3.1.

Note that $a+b=2^{sp}\cdot r^{k}\cdot d_{0}\cdot c_{0}^{p}$ with $s\geq 1$,
therefore from (2.5) and (2.7) we have,

$\upsilon_{\mathfrak{q}_{2}}(\Delta(F))\equiv-8\pmod{p}\qquad\text{and}\qquad\upsilon_{\mathfrak{q}}(\Delta(F))\equiv
4\upsilon_{\mathfrak{q}}(d_{0})\not\equiv 0\pmod{p}.$

Therefore, by Lemma 3.1 we must have

$-2\upsilon_{\mathfrak{q}}(d_{0})=\upsilon_{\mathfrak{q}_{2}}(\Delta(E))\upsilon_{\mathfrak{q}}(\Delta(E))\text{
in }(\mathbb{F}_{p})^{*}/(\mathbb{F}_{p}^{*})^{2}.$

Let $k$ be the number of isogeny classes of elliptic curves over $K$ with full
$2$-torsion and conductor $2\cdot\mbox{Rad}_{2r}(d)\cdot\mathfrak{p}_{r}^{t}$.
For $i=1,\ldots,k$, we let $E_{i}$ be a representative of each isogeny class
and set

$n_{i}:=-2\upsilon_{\mathfrak{q}}(d_{0})\upsilon_{\mathfrak{q}_{2}}(\Delta(E_{i}))\upsilon_{\mathfrak{q}}(\Delta(E_{i}))$

which are negative integers as the valuations used in their definition are
positive.

Observe that our result now follows if, for a positive density of primes $p$,
we have that $n_{i}$ is not a square modulo $p$ for all $i$. We claim that the
set of such primes is non-empty. Then the Dirichlet density theorem guarantees
that this set has density $\geq(1/2)^{k}.$

To finish the proof we now prove the claim. Choose $p\equiv 7\pmod{8}$ such
that for all odd primes $q\mid\upsilon_{\mathfrak{q}}(d_{0})\prod_{i}n_{i}$
the condition $\left(\frac{q}{p}\right)=1$ is satisfied. Such primes exist by
the Dirichlet density theorem. Let $n_{i}=-2^{s}q_{1}^{e_{1}}\ldots
q_{j}^{e_{j}}$ be the prime factorisation. Then

$\left(\frac{n_{i}}{p}\right)=\left(\frac{-1\cdot 2^{s}\cdot
q_{1}^{e_{1}}\ldots
q_{j}^{e_{j}}}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)^{s}\cdot
1\cdot\ldots 1=-1,$

that is, $n_{i}$ is not a square mod $p$ for all $i$, as desired.

(1b) Assume now $\mathfrak{q}\mid d_{1}$. Thus $\mathfrak{q}\nmid a+b$ and,
after replacing $\mathfrak{q}$ by a conjugate if needed, we can assume that
$\mathfrak{q}\mid\mathfrak{c}$ where $\mathfrak{c}$ is given by Proposition
2.2. Thus $F$ has multiplicative reduction at $\mathfrak{q}_{2}$ and
$\mathfrak{q}$. From (2.8) we also know that
$\upsilon_{\mathfrak{q}}(\Delta(F))\equiv
2\upsilon_{\mathfrak{q}}(\Delta(d_{1}^{\prime}))\not\equiv 0\pmod{p}$. The
conclusion now follows similar to the previous case where the integers $n_{i}$
are instead defined by
$n_{i}:=-\upsilon_{\mathfrak{q}}(d_{1}^{\prime})\upsilon_{\mathfrak{q}_{2}}(\Delta(E_{i}))\upsilon_{\mathfrak{q}}(\Delta(E_{i}))$.

(2) Suppose $r\mid a+b$. Now the curve $F$ has multiplicative reduction at
$\mathfrak{p}_{r}$ and $\mathfrak{q}$ by Proposition 2.2. Because
$\upsilon_{r}(\Phi_{r}(a,b))=1$ when $r\mid a+b$, we have $a+b=2^{sp}\cdot
r^{kp-1}\cdot d\cdot c_{0}^{p}$. Therefore it follows from the proof of
Proposition 2.2 that a minimal discriminant at $\mathfrak{p}_{r}$ for $F$
satisfies

$\upsilon_{\mathfrak{p}_{r}}(\Delta)=10+4\upsilon_{\mathfrak{p}_{r}}(a+b)-12=-2+4(sp-1)\frac{r-1}{2}\equiv-2r\pmod{p}$

We now apply Lemma 3.1 with primes $\mathfrak{p}_{r}$ and $\mathfrak{q}$.
Recall that $\upsilon_{\mathfrak{q}}(\Delta(F))\equiv
4\upsilon_{\mathfrak{q}}(d)\pmod{p}$, leading to

$\upsilon_{\mathfrak{q}}(\Delta(F))\upsilon_{\mathfrak{p}_{r}}(\Delta(F))=-2\upsilon_{\mathfrak{q}}(d)r\in(\mathbb{F}_{p})^{*}/(\mathbb{F}_{p}^{*})^{2}$

and

$n_{i}=-2\upsilon_{\mathfrak{q}}(d)r\cdot\upsilon_{\mathfrak{q}_{2}}(\Delta(E_{i}))\upsilon_{\mathfrak{p}_{r}}(\Delta(E_{i})).$

Since the $n_{i}$ are again all negative we complete the proof analogously to
case (1). ∎

######  Remark 3.2.

The argument above succeeds due to the negative sign in the definition of the
integers $n_{i}$. More precisely, this sign arises due to the congruence
$\upsilon_{\mathfrak{q}_{2}}(\Delta(F))\equiv-8\pmod{p}$ in case (1) and
$\upsilon_{\mathfrak{q}_{r}}(\Delta(F))\equiv-2r\pmod{p}$ in case (2). We
observe these congruences hold only because $d$ is odd and not divisible by
$r$.

##  4\. Proof of Theorem 1.3

As in Theorem 1.2 we apply the modular method with the Frey curve $F$.
Moreover, the simplifications at the start of the proof of Theorem 1.2 also
apply.

Namely, let $(a,b,c)$ be a non-trivial solution to (1.1) such that $p\mid a+b$
and $\gcd(a,b)=1$. For primes $p$ sufficiently large, we obtain
$\bar{\rho}_{F,p}\simeq\bar{\rho}_{E_{f},p}$, where $E_{f}$ is an elliptic
curve with full $2$-torsion associated with a Hilbert newform $f$ of level
$2^{s}\cdot\mathfrak{p}_{r}^{t}\cdot\mbox{Rad}_{2r}(d_{0}d_{1}^{\prime})$.

Furthermore, $E_{f}$ has good reduction at all primes $\mathfrak{p}\mid p$ in
$K$ as $p\nmid 2rd$, and $F$ has multiplicative reduction at all
$\mathfrak{p}\mid p$ as $p\mid a+b$ by Proposition 2.2.

The assumption $p\equiv\pm 1\pmod{r}$ implies that $p$ splits completely in
$K=\mathbb{Q}(\zeta_{r})^{+}$.

In particular, locally at any $\mathfrak{p}\mid p$ the curves $F$ and $E_{f}$
become curves over $\mathbb{Q}_{p}$; therefore, by restricting
$\bar{\rho}_{F,p}\simeq\bar{\rho}_{E_{f},p}$ to decomposition subgroups at
$\mathfrak{p}$, we also have the isomorphism

(4.1)
$\bar{\rho}_{F,p}|_{\operatorname{Gal}({\overline{\mathbb{Q}}}_{p}/\mathbb{Q}_{p})}\simeq\bar{\rho}_{E_{f},p}|_{\operatorname{Gal}({\overline{\mathbb{Q}}}_{p}/\mathbb{Q}_{p})},$

where the representation on the right is the reduction of a crystaline
representation and the one on the left is the reduction of a semistable non-
crystaline representation. More precisely, the $p$-adic representation
$\rho_{F,p}$ when restricted to
$D_{\mathfrak{p}}\simeq\operatorname{Gal}({\overline{\mathbb{Q}}}_{p}/\mathbb{Q}_{p})$
is given by the first case in [14, §2.2.3] with $e=1$ whilst $\rho_{E_{f},p}$
restricted to $D_{\mathfrak{p}}$ is given by the second case also with $e=1$.
In particular, their semisimplifications are given by

(4.2)
$(\rho_{E_{f},p}|_{D_{\mathfrak{p}}})^{ss}\simeq\eta_{u}^{-1}\chi\oplus\eta_{u}\quad\text{
and
}\quad(\rho_{F,p}|_{D_{\mathfrak{p}}})^{ss}\simeq\eta_{-1}^{a}\chi\oplus\eta_{-1}^{a},$

where the characters $\eta_{u}$, $\eta_{-1}$ are unramified, $\eta_{-1}$ is
quadratic, $a\in\\{0,1\\}$, $\chi$ is the $p$-adic cyclotomic character and
$a_{p}(E_{f}/\mathbb{Q}_{p})=u^{-1}p+u$ with
$u=\eta_{u}(\operatorname{Frob})\in\mathbb{Z}_{p}^{\times}$. By putting
together the reduction mod $p$ of the representations in (4.2) with the
semisimplification of (4.1) we conclude that either

$(\eta_{u}^{-1}\chi,\eta_{u})\equiv(\eta_{-1}^{a}\chi,\eta_{-1}^{a})\pmod{p}\quad\text{
or
}\quad(\eta_{u}^{-1}\chi,\eta_{u})\equiv(\eta_{-1}^{a},\eta_{-1}^{a}\chi)\pmod{p}.$

Since $\chi\pmod{p}$ is the mod $p$ cyclotomic character which is ramified and
$\eta_{u},\eta_{-1}$ are unramified we must be in the first case. Therefore,

$a_{p}(E_{f}/\mathbb{Q}_{p})=u^{-1}p+u\equiv
u=\eta_{u}(\operatorname{Frob})\equiv\eta_{-1}^{a}(\operatorname{Frob})=\pm
1\pmod{p}.$

Since $a_{\mathfrak{p}}(E_{f})=a_{p}(E_{f}/\mathbb{Q}_{p})$ we have

$a_{\mathfrak{p}}(E_{f})\equiv\pm 1\pmod{p}\implies
a_{\mathfrak{p}}(E_{f})=\pm 1+kp,\quad k\in\mathbb{Z}.$

Furthermore, from the Weil bound, we get

$|a_{\mathfrak{p}}(E_{f})|\leq
2\sqrt{\operatorname{Norm}(\mathfrak{p})}\implies(\pm 1+kp)^{2}\leq 4p$

since $\operatorname{Norm}(\mathfrak{p})=p$ because $p$ splits completely in
$K$. The previous inequality does not hold for large $p$ unless $k=0$, that is
$a_{\mathfrak{p}}(E_{f})=\pm 1$.

On the other hand, since $E_{f}$ has full $2$-torsion over $\mathbb{Q}$ and
reduction modulo a rational prime $q$ of good reduction is injective on
$E_{f}(\mathbb{Q})_{tors}$, it follows that $a_{\ell}(E_{f})$ is even for all
primes $\ell$ in $K$ of good reduction; in particular,
$a_{\mathfrak{p}}(E_{f})$ is even, a contradiction. ∎

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