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NYC Psychotherapy Blog: Are You Enabling Your Adult Children?
In prior articles, I've discussed the concepts of enabling and codependency (Relationships: Overcoming Codependency, Crossing the Line From Being Compassionate to Being Enabling, and Avoiding Codependency Between You and Your Children). In this article, I'm focusing on parents who enable their adult children.
Are You Enabling Your Adult Children?
Before I discuss parents enabling adult children, I want to stress that, of course, most parents want the best for their children and would never do anything intentionally that was harmful to them.
At the same time, good intentions can sometimes lead to bad outcomes. With regard to enabling adult children, this often means that these children don't learn the necessary skills to develop and grow psychologically.
What Does "Enabling" Mean?
Let's start by providing the negative definition of "enabling," which is how I'm using it in this article.
Basically, the concept of enabling developed in the recovery community to describe spouses, family members and others who intend to "help" someone with an addiction but who make the problems worse with their "help."
Common examples of enabling in this sense is the wife (or husband) who calls the spouse's employer to make an excuse for an absence when, in fact, the spouse is too drunk to go to work.
In this case, the intention is for the spouse with the drinking problem to keep his job. However, the unintended consequence is that the spouse with the drinking problem learns that he can continue to drink and his spouse will make sure that everything is taken care of with the boss.
Anyone with an ounce of compassion for the wife in this situation can understand why she's doing this. If her husband loses his job, she and the children will also suffer terrible consequences. But, at the same time, she is unknowingly and unintentionally making the situation worse because there are no consequences for the husband at home or at work--at least for a while.
Presumably, these excuses can't go on forever. In the meantime, since alcohol problems tend to be progressive, without help, the husband's drinking problem will get worse and can lead to serious health problems or death.
The same scenario often occurs when parents make excuses for their adult children to shield them from experiencing negative consequences.
Let's take a look at a fictionalized scenario to understand these consequences better:
Sandy and Ann:
Sandy's daughter, Ann, moved out with roommates when she was 25.
Initially, when Ann moved out, she wasn't making much money, so Sandy paid most of Ann's portion of the rent and helped to pay for Ann's phone.
As time went on, Ann got a promotion to a managerial position at her bank, and she was doing well enough so that she could afford her own apartment in Brooklyn. But she asked Sandy to continue to pay her rent for the next six months until she was settled in.
Sandy was happy to help her and continued to pay Ann's rent for the next six months--even though it came at a financial sacrifice. Sandy cancelled a vacation she planned to take. She also became a lot more careful about other everyday expenses so that she could continue to help Ann financially.
A few months later Ann approached Sandy for more money because she maxed out her credit cards and the balances were so high that she could no longer make minimum payments.
When Ann told Sandy what happened, Ann was shocked. She had no idea that Ann was running up her credit cards. She told Ann that she would pay off her credit cards, but she needed to be more careful in the future about how she spent money.
Are You Enabling Your Adult Children
Ann agreed to be more responsible, but this pattern continued for the next few years with Ann living above her means to have whatever she wanted and Sandy cutting back more and more on her own expenses to bail out Ann.
By the time Ann turned 30, she had a high paying executive position at her bank, but she continued to get deeper and deeper in debt.
As a single parent of an only child, Sandy was getting increasingly concerned about Ann's financial problems.
On some level, she knew that she had enabled this problem by constantly bailing Ann out, but she felt too guilty to refuse to help her.
Sandy offered to pay for Ann to see a financial advisor to help Ann develop better financial skills, but Ann wasn't interested.
Then, one day, Sandy came to visit Ann and told her that she had run up a new credit card and she couldn't pay it. When Ann told Sandy that the balance owed was $25,000, Sandy was shocked. At first, Sandy refused to help her.
At that point, Ann got enraged and said in a desperate tone, "What am I going to do!?! If the credit card company contacts my employer, I'll be so embarrassed! I'll lose my job! You have to help me."
Sandy wasn't sure what to do. She agonized about it for days. Then, she confided in her best friend, Meg, who was already familiar with this dynamic between Sandy and Ann.
Meg listened patiently, as she always did, and then she told Sandy that Ann was in deep trouble with her overspending habit. Then, she took a deep breath and told Sandy that Sandy was also in trouble for enabling Ann.
Sandy knew that Meg was right, but she felt too guilty and afraid not to help her daughter, so she took money from her savings account, gave it to Ann and told her that it was a loan and she had to pay her back.
Ann was taken aback when Sandy told her that she had to pay the money back, but she was feeling desperate, so she agreed to pay Ann back within the next two years.
After Sandy gave Ann the money, they each felt momentarily relieved and didn't talk about it again for a while. But as time went on and Ann didn't give Sandy any money to pay back the loan, Sandy got concerned.
Whenever Sandy brought up the loan, Ann got annoyed and reminded Ann that she had agreed to pay her back. Ann didn't like being "badgered" for the money and she said she would start paying back Sandy soon--as soon as she had the extra cash.
But time went on and Ann never brought up the loan and never gave Sandy any money. Instead, she got herself deep into debt again.
When Ann approached Sandy for more money, Sandy had run out of all options and, even though she felt very guilty, she told Ann that she would have to deal with this new debt on her own because she couldn't help her.
Ann panicked and approached her close friends for loans, but she had already borrowed money from all of them and still owed them so they told her they couldn't lend her any more money. Having no money to pay, Ann had to file for bankruptcy, which ruined her credit and caused her to lose her position with her bank.
Ann had no choice but to give up her apartment and move back home, which she resented. Then, instead of looking for another job, she spent all day sleeping in her former childhood room. She refused to speak to Sandy because she was unable to help her.
Faced with this increasingly difficult situation, Sandy began therapy because the situation was overwhelming to her.
In therapy, Sandy learned about the concept of enabling and how she had unwittingly contributed to her daughter's problems by constantly bailing her out.
Sandy knew that she had to make changes, but it was extremely difficult for her to say "no" to her daughter. But she took responsibility for her part in Ann's problems, and she began to deal with the underlying reasons that caused her to enable her daughter.
Sandy discussed in therapy that she had always felt guilty about separating from Ann's father, who was a gambler who gambled away their savings when Ann was a child.
After they separated, Ann's father disappeared from Ann's life and Ann blamed Sandy for this. As a child, Ann was too young to understand her father's gambling problems and Sandy never explained it to Ann--even when Ann became an adult. In fact, they never discussed it.
Ann realized in therapy that, due to her guilty feelings about the marital separation and Ann's father's abandoning Ann, Sandy felt she could never say "no" to Ann.
Ann had to work through her guilt in therapy before she could accept that she did what was best for herself and Ann when she left Ann's father, and it was time for her to stop trying to overcompensate for it by constantly bailing out Ann.
She also knew that she needed to speak with Ann about it and set limits with her, including how long she could continue to live with her and expect that Sandy would support her.
It was one of the most difficult conversations that Sandy had ever had in her life, but by the end of their conversation, Sandy explained why she separated from Ann's father. She also gave Ann six months to find a job or she would have to move out.
It took a few days before Ann came back to Sandy and apologized for her behavior.
She told Ann that after her father left, she blamed Sandy, but she also blamed herself. She felt that she must have been unlovable and that the only thing that made her feel good about herself was spending money (see my article: Learn to Stop Overspending as a Way to Avoid Uncomfortable Feelings).
She understood now that she had placed herself in an untenable situation and it was always headed for disaster, but she couldn't face it until now.
Ann began attending Debtor's Anonymous 12 Step meetings to deal with her overspending habit and she started putting her life together again.
Although it was very hard for Sandy to set limits with her daughter, she realized that neither she nor her daughter would have made any changes if she had not confronted her own underlying reasons for enabling Sandy. This, in turn, led to Ann facing her own problems.
Confronting your enabling behavior toward an adult child can be one of the most difficult things you do in your life.
You can find many rationalizations and excuses for your behavior, but until you face it and make a commitment to overcome it, neither you nor your child will be able to change this dysfunctional behavior.
In the scenario above, the enabling behavior was about giving an adult child money, but it can be about anything--enabling substance abuse, enabling overeating, enabling workaholism, and so on.
Parents often find it increasingly difficult to stop enabling their children on their own.
It's easy to continue to bargain with yourself, "I'll just do it this one time and then I'll stop," in much the same way that your adult child can bargain with him or herself to continue to engage in dysfunctional behavior.
Getting help in therapy can provide you with the tools you need to take yourself out of the cycle of ongoing enabling.
Parents are often surprised that once they take a stand and stop enabling their adult children's dysfunctional behavior, their children will have no choice but to confront their own behavior and make changes.
The first step, picking up the phone and setting up an appointment with a therapist for a consultation, can be the hardest step, but it's often the first step to making positive changes.
I have helped many clients to overcome negative enabling behavior so they could lead more fulfilling lives.
Labels: adult children, codependency, enabling, family, guilt, money, overspending, parents, psychotherapist, psychotherapy, setting limits, therapist, therapy
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redpajama_set_name: RedPajamaGithub


Vinita Gupta took over the reins of Lupin in September 2013 and is now working towards achieving the Mumbai pharma company's projected revenue target of $5 billion by 2018.
She took over the reins of Lupin in September 2013 and is now working towards achieving the Mumbai pharma company 's projected revenue target of $5 billion by 2018.
"My lens changed from regional to a global one from the business perspective," says Vinita Gupta, who has played a major role in establishing her company's presence in the US, the world's largest market for medicines. Besides building a good team, she is also focusing on strategic expansion in Europe and Latin America.
The two regions fall much behind the US, India and Japan, Lupin's primary markets. Talking about the company's acquisitions in the Netherlands and Mexico, Gupta says: "It was a lot of execution on a strategic plan." She has now brought in an executive who earlier worked with Israel's Teva Pharma and holds expertise in handling the European region.
She also hired an expert on mergers and acquisitions, another major focus area for the company. Gupta credits her father Desh Bandhu Gupta, Lupin's Founder and Chairman, for the entrepreneurial spirit that she so strongly demonstrates. "That is what drove me to move to the US and build the business in a region that had turned out to be very lucrative for the company," she says.
With siblings Kavita and Anuja (top left and right) and Richa and Nilesh Gupta (to her left and right) during a family function in Mumbai in the year 2008.
With son and husband at a family function in Mumbai in the year 2008.
A weekend, with husband Brij Sharma, and son Krish in Baltimore, US around 2009.


Hi Everyone, my name is Suganya and I am from the Windows Devices and Deployment Team. I would like to discuss one of the common issues customers face today; "Customizing the recovery partition after upgrading from Windows 8.1 to Windows 10".
What is Windows RE and why is it used?
Now we have two recovery partitions, but we would like to only have one and customize the partition based on our requirements.
Delete both of the current recovery partitions on the drive and follow these steps to create a new recovery partition and customize it.
Create the folders "Recovery" and "WindowsRE" on the R:\ drive.
Note: This is an example as drive letters may differ in your environment. Verify drive letters in WinRE with Diskpart.
Copy the Winre.wim to the "R:\Recovery\WindowsRE" folder.
Now, to configure the Windows® Recovery Environment, run the following command.
Now if you run the command "reagentc /info" from a command prompt it shows the Status as Disabled. When you enable it using the command "reagentc /enable", the Status will still show as Disabled.
This can happen if the Windows Boot Loader is not updated with the correct device information. You will have to edit the recovery.xml which is located in C:\WINDOWS\SYSTEM32\RECOVERY as it contains the older WinRE and OS image configurations. You will have to give everyone full control for the XML file before copying the following lines to the file.
When you run the command "reagent /info", you will still see the Status as Disabled. When you run the command "reagentc /enable", the Windows RE Status will now show as Enabled.
Also when you check the output of the command "bcdedit /enum all", you can see that the windows boot loader is updated with the correct device information.
It appears that the winre.wim is missing from the 1511 ISO.
Hi Will. The doing an "attrib winre.wim" shows it as having System and Hidden attributes. Do an "attrib winre.wim -s -h" and that should do it.
Do a "dir -force" on that directory… I suspect the file is marked "system readonly", which isn't shown by default.
Very thorough work you've done here. I upgraded to W10 during beta and since then, my external and internal auto-backup systems have had a rather tenacious bug. Surely there will be a user friendly fix for this in a forthcoming W10 update… right?
Question: Once you realign the proper WinRE, will subsequent Windows major version updates also update this partition, or is this going to have to be done with each major release of WinX?
Didn't work, it still shows up as disabled. However, there are boot entries that point to c:\Recovery, revealed by bcdedit /enum all.
Is there any other way of invoking the recovery environment by pressing a custom bootkey or programmatically?
it gives error:2 = the system cannot find the file specified.
I am currently converting the esm to wim is there any better way?
I am using Windows Server 2012 R2 Essentials to manage all my computer backups. The clients are Windows 10 Pro (1703) based computers with MBR (BIOS) or GPT (UEFI) partitions, depending on the computer hardware.
As a result, it seems that WS 2012 R2 Essentials cannot restore the Windows 10 installation in a way that all the features are working.
I was testing a scenario of total hard drive failure, and did the bare metal restore. Luckily I was able to boot the computers after restoring the partitions. But I got into the situation that I lost all the Reset My PC features, and this is due to partition ID and boot manager ID changes. The only differences I noticed that Windows Server 2012 R2 Essentials client restore service does not give the correct IDs to the Recovery Partitions, and I needed to set up the partition IDs by using diskpart. One time I had to manually give BCDBOOT C:\Windows on an UEFI-based computer to make Windows bootable again. Additionally, The restore service cannot restore the partitions at all on an empty GPT disk without any partitions. By creating the default partition layout the service can restore the Windows and recovery partition contents correctly.
I took all the steps described above to restore the WinRE, and managed to get it work. I continued the tests, and performed the total Reset My PC successfully.
But, there is something in the instructions above that are different to Windows installation process. The steps above require that I have a drive letter assigned to the recovery partition (R:). If I remove the drive letter assignment — as this is the case for the Recovery Partitions — the reagentc /info shows the Windows RE is disabled, and I am unable to reactivate the RE.
Does Microsoft have any plans to fix the Client restore process in WS2012R2Essentials for the Windows 10 clients so that the partitions are created on UEFI/BIOS computers and the RE is activated automatically?


text:
George Sherwin Simonds (March 12, 1874 – November 1, 1938) was a U.S. Army officer with the rank of major general.

Early life
He was born in Cresco, Iowa on March 12, 1874, the son of son of William O. Simonds and Ellen Sherwin Simonds .

Military career

He graduated from the United States Military Academy in 1899. After graduation, he was assigned to the 22d Infantry. He served in the Philippines and China. He returned to West Point in 1904 as an instructor in the Department of Law. In addition from 1915 to 1917, he also taught in the Department of Tactics.

When World War I began, he was sent to France as an observer.  After returning to America, he was assigned to the General Staff of the General Headquarters of the American Expeditionary Force and returned to France again in 1917. Lieutenant Colonel George S. Simonds, Infantry, was Chief of Staff of the 2nd Corps, American Expeditionary Force. He received a temporary promotion to colonel in June 1918 and to brigadier general in October 1918. Simonds received the Army Army Distinguished Service Medal for service in the war. The medal's citation reads:

After the war, Simonds graduated from the Army War College in 1920 and was promoted to colonel in July 1921. He was promoted to brigadier general in November 1924.  Simonds commanded the tank center until 1925, and was succeeded by James K. Parsons.

Brigadier General Simonds served as Chief of the War Plans Division of the War Department from September 1, 1927 to September 1, 1931.

In 1932, he was assigned to run the Army War College. Previously he was Douglas MacArthur's chief war planner, and his appointment was made to improve the planning ability in the Army War College.

Simonds was promoted to major general in 1933.

Simonds served as Deputy Chief of Staff in the War Department from February 2, 1935 to May 28, 1936.

In June 1936, Simonds took command of the Fourth United States Army and Ninth Corps; the headquarters was at the Presidio of San Francisco.

Simonds retired from the Army at San Francisco in 1938.

Death and legacy
He died on November 1, 1938 and was buried in the West Point Cemetery. The ship USAT General George S. Simonds was named for him.

References

External links

1874 births
1938 deaths
People from Cresco, Iowa
United States Military Academy alumni
Military personnel from Iowa
American military personnel of the Philippine–American War
United States Military Academy faculty
United States Army generals of World War I
Recipients of the Distinguished Service Medal (US Army)
United States Army War College alumni
United States Army generals
People from San Francisco
Burials at West Point Cemetery

redpajama_set_name: RedPajamaWikipedia


These guys rock! Every time I have been there I am greeted with a smile and helped with everything I need. The store is very neat and has tons of stuff for every budget.
We have 137 A Buck Or Two locations with hours of operation and phone number.


On our early morning walks recently along the Fleet we have frequently seen Roe Deer feeding in the fields. The numbers spotted at any one time vary between solitary specimens and as many as five in a group!
A family of four badgers have been seen crossing Camp Road in single file from the caravan park to the Tented Camp, these same badgers can also be seen scampering along Camp Road late at night.
A number of Pipistrel bats are currently around giving us an acrobatic demonstration as they swoop and dive catching low flying insects.


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Do not miss this heart patterned tile if you want to have a sweet and lovely kitchen space. This 2″ black white porcelain triangle tile with a single heart pattern looks gorgeous when used on the splashback, it's matte finish and wear-resistant, non-slip. Easy to clean and install, affordable price for you.


text:
#import <UIKit/UIControl.h>

// Not exported
@interface UIRemoveControlMinusButton : UIControl
{
    unsigned int _rotated:1;
    unsigned int _rotating:1;
    unsigned int _hiding:1;
    unsigned int _showAsPlus:1;
    unsigned int _reserved:28;
    float _verticalOffset;
}

+ (id)plusImage;
+ (id)minusImage;
+ (float)defaultWidth;
- (void)_toggleRotateAnimationDidStop:(id)arg1 finished:(_Bool)arg2;
- (_Bool)isRotating;
- (_Bool)isRotated;
- (void)toggleRotate:(_Bool)arg1;
- (void)animator:(id)arg1 stopAnimation:(id)arg2;
- (void)drawRect:(struct CGRect)arg1;
- (_Bool)isHiding;
- (void)setHiding:(_Bool)arg1;
- (void)dealloc;
- (id)initWithRemoveControl:(id)arg1;

@end


text:
\section{Introduction}
    Throughout this article, $K$ and $p$ denote a totally real field and a rational prime, respectively. The study of non-trivial solutions to the Diophantine equations is one of the exciting and most interesting areas in Mathematics. A prominent and most interesting example is the Fermat equation $x^p+y^p=z^p$ with exponent $p$.

    In~\cite[Theorem 0.5]{W95}, Wiles used the modularity of elliptic curves to show that $\Z$-integral
    solutions of $x^p+y^p=z^p$ are trivial. In~\cite[Theorem 1.3]{JM04}, Jarvis and Meekin showed
    the same result that continues to hold even over $\Z[\sqrt{2}]$. A similar study over $K$ has been initiated by Freitas and Siksek in~\cite[Theorem 3]{FS15} and showed that an asymptotic version of these results continues to hold over $K$, by employing some explicit bounds on the solutions of $S$-unit equation. In~\cite{D16}, Deconinck generalized the work of~\cite{FS15} to $Ax^p+By^p=Cz^p$ with $2 \nmid ABC$.

In~\cite[Theorem 3]{R97}, Ribet showed that $x^p+2^ry^p+z^p=0$ with exponent $p$ has no non-trivial
    $\Z$-solution for $1\leq r <p$. Over $K$, in~\cite[Theorem 3.2]{KS22}, we show that the equation $x^p+y^p=2^rz^p\ (r \in \N)$ has no asymptotic solution in $W_K$ (cf. \cite[Definition 3.2]{KS22} for $W_K$). Furthermore, we show that the equation $x^p+y^p=2^rz^p$ has no asymptotic solution in $\mcO_K^3$ for $r=2,3$ (cf. \cite[Theorem 3.3]{KS22}). The proofs of our results also depend on certain explicit bounds on the solutions of the $S$-unit equation.

    Similarly, Ivorra in~\cite{I03} studied $\Z$-solutions of $x^2=y^p+2^rz^p$ and $2x^2=y^p+2^rz^p$ for $0 \leq r \leq p$. In ~\cite[Theorem 1]{S03}, Siksek established that the only non-trivial primitive $\Z$-solutions of $x^2=y^p+2^rz^p$ are $r=3,\ x=\pm3,\ y=z=1$, where $p \in \mathbb{P}$ is arbitrary.

    Let us now make sense of what ``asymptotic'' would actually mean.

     \begin{dfn}
    	We say a Diophantine equation $Ax^2= By^p+Cz^p$  with exponent $p$ has no asymptotic solution in a set $S \subseteq \mcO_K^3$, if there exists a constant $V_{K,A,B,C}>0$ (depending on $K,A,B,C$) such that for primes $p>V_{K,A,B,C}$, the equation $Ax^2= By^p+Cz^p$  with exponent $p$ has no non-trivial primitive solution in $S$.
    \end{dfn}

    In~\cite{DM97}, Darmon and Merel showed that the equation $x^n+y^n=z^2$ with exponent $n \geq 4$ has no non-trivial primitive $\Z$-solutions.  In~\cite[Theorem 1.1]{IKO20}, I\c{s}ik, Kara, and Ozman proved that the ``asymptotic'' FLT holds for $x^p+y^p=z^2$ with exponent $p$ of certain type over $K$, whenever the narrow class number $h_K^+=1$ and there exists a $\mathfrak{P} \in P$ lying above $2$ with residual degree $1$. The proof of ~\cite[Theorem 1.1]{IKO20} relies on certain explicit bounds on the solutions of the $S$-unit equation. ~\cite[Theorem 1.1]{IKO20} was extended in two different directions as follows.
    \begin{itemize}
     \item In ~\cite[Theorem 5.2]{KS22}, we relaxed the assumptions in ~\cite[Theorem 1.1]{IKO20} and proved that $x^p+y^p=z^2$ has no asymptotic solution in $W_K^\prime$ (cf. \cite[Definition 5.2]{KS22} for $W_K^\prime$).

    \item  In ~\cite[Theorem 3]{M22}, Mocanu generalized \cite[Theorem 1.1]{IKO20} by replacing the assumption from $h_K^+=1$ to $\Cl_{S_K}(K)=1$ and its proof depends upon some explicit bounds on the solutions of $\alpha +\beta =\gamma^2$, with $\alpha, \beta \in \mcO_{S_K}^\ast$ and $\gamma \in  \mcO_{S_K}$ (cf. \S \ref{section for main result of x^2=By^p+2^rz^p} for the definitions of $\Cl_{S_K}(K)$, $\mcO_{S_K}$ and $\mcO_{S_K}^\ast$).
     \end{itemize}
     In the first part, we extend~\cite[Theorem 3]{M22} to  $x^2= By^p+Cz^p$ with exponent $p$ over $K$, where $B$ is an odd integer and $C$ is either an odd integer or $C=2^r$ for $r \in \N$. In Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}, we prove that the equation $x^2= By^p+Cz^p$ with exponent $p$ has no asymptotic solution in $W_K$ (cf. Definition~\ref{def for W_K} for $W_K$). For $r=1,2,4,5$, we also prove that $x^2= By^p+2^rz^p$ with exponent $p$ has no asymptotic solution in $\mcO_K^3 \setminus S_r$ for some finite set $S_r$ (cf. Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}). In Proposition~\ref{cor to main result2 for x^2=By^p+2^rz^p over K}, we provide situations where $S_r= \phi$.

    In the second part, we perform a similar analysis to the equation $2x^2= By^p+2^rz^p  \ (r \in \N)$ with exponent $p$ and show that it has no asymptotic solution in $\mcO_K^3$ (cf. Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}).
 The proofs of these results rely on certain explicit bounds on the solutions of the equation $\alpha +\beta =\gamma^2$, with $\alpha, \beta \in \mcO_{S_K^{\prime}}^\ast, \gamma \in \mcO_{S_K^{\prime}}$ (cf. \S \ref{notations section for x^2=By^p+Cz^p} for the definition of $S_K^{\prime}$).

  In the last part, we provide several local criteria of $K$ for Theorems~\ref{main result1 for x^2=By^p+Cz^p Type I}, ~\ref{main result1 for x^2=By^p+2^rz^p over K}, ~\ref{main result1 for 2x^2=By^p+Cz^p Type I}, which follow from Propositions~\ref{main result2 for x^2=By^p+Cz^p Type I}, ~\ref{main result2 for  x^2=By^p+2^rz^p over K}, ~\ref{main result2 for 2x^2=By^p+Cz^p Type I}, respectively, when $S_K^{\prime}=S_K$. Though these propositions can be thought of as an analog of ~\cite[Theorem 5]{M22} but they differ technically in the following sense.
    \begin{itemize}
    	\item The proof of~\cite[Theorem 5]{M22} uses the assumption ``$2$ is inert in $K$'', while we relax this assumption to ``$\mfP \in P$ is principal for all $\mfP|2$''.
    	\item In the final steps of the proof of ~\cite[Theorem 5]{M22}, the author needs to study the ramification behavior of a certain extension $L/K$. In our proofs, we need to invoke~\cite[Theorem 9(b)]{FKS20} in Proposition~\ref{main result2 for  x^2=By^p+2^rz^p over K} to study the extension $L/K$, here the discriminant of $L/K$ involves a power of $2$.
 \end{itemize}

	\section{Preliminaries}
	\label{section for preliminary}
	 We use the notations $\mcO_K$, $P$, and $\mfn$ to represent the ring of integers, $\mrm{Spec}(\mcO_K)$, and an ideal of $K$, respectively.  Let $E/K$ be an elliptic curve of conductor $\mfn$. For any $\mfq \in P$, let $\Delta_\mfq$ be the minimal discriminant of $E$ at $\mfq$. Let

	\begin{equation}
		\label{conductor of elliptic curve}
		\mfm_p:= \prod_{ p|v_\mfq(\Delta_\mfq), \ \mfq ||\mfn} \mfq \text{ and } \mfn_p:=\frac{\mfn}{\mfm_p}.
	\end{equation}

	We state a conjecture, which is an extension of the Eichler-Shimura theorem over $\Q$.
	\begin{conj}[Eichler-Shimura]
		\label{ES conj}
		Let $f$ be a Hilbert modular newform over $K$ of parallel weight $2$, level $\mfn$ and with coefficient field $\Q_f= \Q$. Then, there exists an elliptic curve $E_f /K$ with conductor $\mfn$ having same $L$-function as $f$.
	\end{conj}
    	In~\cite[Theorem 7.7]{D04}, Darmon showed that Conjecture~\ref{ES conj} holds over $K$, if  either
    $[K: \Q] $ is odd or there exists some $\mfq \in P$ such that $v_\mfq(\mfn) = 1$.
    In \cite[Corollary 2.2]{FS15}, Freitas and Siksek provided a partial answer to Conjecture~\ref{ES conj} in terms of mod $p$  Galois representations attached to $E$.

	\section{Solutions of the Diophantine equation $x^2=By^p+Cz^p$ over $W_K$}
	\label{notations section for x^2=By^p+Cz^p}
	In this section, we study the solutions of the following Diophantine equation
	\begin{equation}
		\label{x^2=By^p+Cz^p}
		x^2 =By^p+Cz^p
	\end{equation}
	with prime exponent $p\geq 3$ and $B,C \in \Z$. Throughout, we assume that $B$ is odd.
	\begin{itemize}
		\item We say the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ is of Type I, if $C$ is odd.
		\item We say the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ is of Type II, if $C=2^r$ for some $r\in \N$.
	\end{itemize}
	For $n \in \Z$, define $S_K(n):= \{ \mfP \in P :\ \mfP|2n \}$.
	Let $S_K:=S_K(1)$, $S_K^{\prime}:= S_K(BC)$ and $U_K:=\{ \mfP \in S_K: (3, v_\mfP(2))=1 \}$.

	\begin{dfn}[Trivial solution]
		We call a solution $(a, b, c)\in \mcO_K^3$ to the equation \eqref{x^2=By^p+Cz^p} with exponent $p$ is trivial, if $abc=0$,
		otherwise non-trivial.
		We say $(a, b, c)\in \mcO_K^3$ is primitive if $a,b,c$ are pairwise coprime.
	\end{dfn}
	\begin{dfn}
		\label{def for W_K}
			\label{remark for W_K}
	Let $W_K$ be the set of all non-trivial primitive solutions $(a, b, c)\in  \mcO_K^3$ to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type I or II with $\mfP |bc$ for every $\mfP \in S_K$. Note that, for any $\mfP \in S_K$ and $(a, b, c)\in W_K$,
	 $\mfP$ divides exactly one of $b$ and $c$.
\end{dfn}

	\subsection{Main result}
	\label{section for main result of x^2=By^p+2^rz^p}
	For any set $S \subseteq P$, let $\mcO_{S}:=\{\alpha \in K : v_\mfP(\alpha)\geq 0 \text{ for all } \mfP \in P \setminus S\}$ be the ring of $S$-integers in $K$ and $\mcO_{S}^*$ be the $S$-units of $\mcO_{S}$.
    Let $\Cl_S(K):= \mrm{Cl}(K)/\langle [\mfP]\rangle_{\mfP\in S}$ and $\Cl_S(K)[n]$ be its $n$-torsion points.
	We now show that the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type I or II has no asymptotic solution in $W_K$. More precisely;
	\begin{thm}
		\label{main result1 for x^2=By^p+Cz^p Type I}
		\label{main result1 for x^2=By^p+Cz^p Type II}
		Let $K$ be a totally real field with $\Cl_{S_K^{\prime}}(K)[2]=1$. Suppose for every solution $(\alpha, \beta, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to  $\alpha+\beta=\gamma^2$,
        there exists $\mfP \in S_K$ that satisfies
		\begin{equation}
			\label{assumption for main result1 for x^2=By^p+Cz^p Type I}
	\left| v_\mfP \left(\alpha \beta^{-1}\right) \right| \leq 6v_\mfP(2).
		\end{equation}
		Then, the Diophantine equation $x^2 =By^p+Cz^p$ with exponent $p$ of Type I or II has no asymptotic solution in $W_K$.
\end{thm}
By~\cite[Theorem 39]{M22}, for any finite set $S \subseteq P$, the equation $\alpha+\beta=\gamma^2$ with  $(\alpha, \beta, \gamma) \in \mcO_{S}^\ast \times \mcO_{S}^\ast \times \mcO_{S}$ has only finitely many solutions. The following proposition, which is a consequence of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I} will be relevant in \S\ref{section for local criteria for Diophantine equations over $W_K$}. We say that $S \subseteq P$ is principal if $\mfP$ is principal for every $\mfP \in S$.
\begin{prop}
	\label{main result2 for x^2=By^p+Cz^p Type I}
	\label{main result2 for x^2=By^p+Cz^p Type II}
	Let $K$ be a field such that $S_K^{\prime}=S_K$ is principal and $2 \nmid h_K$. Suppose for every solution $(\alpha, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to  $\alpha+1=\gamma^2$,

	there exists $\mfP \in S_K$ that satisfies
	\begin{equation}
	\label{assumption for main result2 for x^2=By^p+Cz^p Type I}
	|v_\mfP(\alpha)|\leq 6v_\mfP(2).
	\end{equation}
Then, the Diophantine equation $x^2 =By^p+Cz^p$ with exponent $p$ of Type I  or II has no asymptotic solution in $W_K$.
\end{prop}

	\subsection{Steps to prove Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}}
	For any non-trivial and primitive solution $(a, b, c)\in \mcO_K^3$ to the equation \eqref{x^2=By^p+Cz^p} with exponent $p$, the Frey curve $E:=E_{a,b,c}$ is given by
		\begin{equation}
			\label{Frey curve for x^2=By^p+Cz^p of Type I}
			E:=E_{a,b,c} : Y^2 = X(X^2+2aX+Bb^p),
		\end{equation}
    	with $c_4=2^4(Bb^p+4Cc^p),\ \Delta_E=2^{6}(B^2C)(b^2c)^{p}$ and $ j_E=2^{6} \frac{(Bb^p+4Cc^p)^3}{B^2C(b^2c)^{p}}$, where $j_E$ (resp., $\Delta_E$) denote the $j$-invariant (resp., discriminant) of $E$. We now prove the modularity of the Frey curve $E:=E_{a,b,c}$ in ~\eqref{Frey curve for x^2=By^p+Cz^p of Type I} associated to  $(a,b,c)\in W_K$.

	\begin{thm}
		\label{modularity of Frey curve x^2=By^p+Cz^p over W_K}
		Let $(a,b,c)\in W_K$, and let $E:=E_{a,b,c}$ be the Frey curve attached to $(a,b,c)$ as in~\eqref{Frey curve for x^2=By^p+Cz^p of Type I}. Then, there exists a constant $A=A_{K,B,C}$ (depending on $K,B,C$) such that for primes $p >A$, $E/K$ is modular.
	\end{thm}

	\begin{proof}
        By ~\cite[Theorem 5]{FLBS15},
        there exists finitely many elliptic curves over $K$ (up to $\bar{K}$-isomorphism) which are not modular.  Let $j_1,\ldots,j_t \in K$ be the $j$-invariants of these.
		The $j$-invariant $ j_E=2^{6} \frac{(Bb^p+4Cc^p)^3}{B^2C(b^2c)^{p}}=2^{6}\frac{(4-\lambda(E))^3}{\lambda(E)^2}$ for $\lambda(E)= -\frac{Bb^p}{Cc^p}$.
		For $i=1,2,\ldots,t$, the equation $j_E=j_i$ has at most three solutions in $K$. Hence, there exists
		$\lambda_1, \lambda_2, ..., \lambda_m \in K$ with $m\leq 3t$ such that $E$ is modular for all $\lambda(E) \notin\{\lambda_1, \lambda_2, ..., \lambda_m\}$.
		If $\lambda(E)= \lambda_k$ for some $k \in \{1, 2, \ldots, m \}$, then $\left(\frac{b}{c} \right)^p=\frac{-C\lambda_k}{B}$.
		This equation determines $p$ uniquely, denote it by $p_k$. Suppose $p \neq q$ are primes such that $\left(\frac{b}{c} \right)^p=\left(\frac{b}{c} \right)^q$, which means $\left(\frac{b}{c}\right)$ is a root of unity. Since $K$ is totally real, we get $b=\pm c$. For $\mfP \in S_K$, $\mfP \mid bc$ implies that $\mfP |a$,
		contradicts $(a,b,c) \in W_K$. Now, the proof of the theorem follows by taking $A=\max \{p_1,...,p_m\}$.
	\end{proof}

	\subsection{Reduction type}

	The following lemma characterizes the type of reduction of the Frey curve $E:= E_{a,b,c}$
	at primes $\mfq$ away from $S_K^{\prime}$.

	\begin{lem}
		\label{reduction away from S}
		Let $(a,b,c) \in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type I or II, and let $E$ be the associated Frey curve. Then at all primes  $\mfq \notin S_K^{\prime}$, $E$ is minimal, semi-stable at $\mfq$ and satisfies $p | v_\mfq(\Delta_E)$. Let $\mfn$ be the conductor of $E$ and $\mfn_p$ be as in \eqref{conductor of elliptic curve}. Then,
			\begin{equation}
				\label{conductor of E and E' x^2=By^p+Cz^p Type I}
				\mfn=\prod_{\mfP \in S_K^{\prime}}\mfP^{r_\mfP} \prod_{\mfq|bc,\ \mfq \notin S_K^{\prime}}\mfq,\ \mfn_p=\prod_{\mfP \in S_K^{\prime}}\mfP^{r_\mfP^{\prime}},
			\end{equation}
			where $0 \leq r_\mfP^{\prime} \leq r_\mfP $ with $r_\mfP
			\leq 2+6v_\mfP(2)$ for $\mfP |2$ and $ r_\mfP\leq 2+3v_\mfP(3)$ for $\mfP \nmid 2$.
\end{lem}

	\begin{proof}
	Let $\mfq \in P \setminus S_K^\prime$. If $\mfq \not|\Delta_E$, then $E$ has good reduction at $\mfq$
	and $p | v_\mfq(\Delta_E)=0$.

        If $\mfq|\Delta_E=2^{6}B^2C(b^2c)^{p}$, then $\mfq$ divides precisely one of $b$ and $c$, since $(a,b,c)$ is primitive and $\mfq \nmid 2BC$. This implies that $\mfq\nmid c_4=2^4(Bb^p+4Cc^p)$, hence $E$ is minimal and has multiplicative reduction at $\mfq$.  Since $v_\mfq(\Delta_E)=p v_\mfq(b^2c) $, $p | v_\mfq(\Delta_E)$.

        By the definition of $\mfn_p$ in~\eqref{conductor of elliptic curve}, we get $\mfq \nmid \mfn_p$ for all  $\mfq \notin S_K^{\prime}$. Finally, for $\mfP \in S_K^{\prime}$, the bounds on $r_\mfP$ follow from \cite[Theorem IV.10.4]{S94}.
	\end{proof}

	\subsubsection{Type of reduction with image of inertia}
    For any elliptic curve $E/K$, let $\bar{\rho}_{E,p} : G_K \rightarrow \mathrm{Aut}(E[p]) \simeq \GL_2(\F_p)$ be the residual Galois representation of $G_K$, induced by the action of $G_K$ on  $E[p]$, the $p$-torsion of $E$.
    We first recall ~\cite[Lemmas 3.4, 3.6]{FS15}, which will be useful for the types of the reduction of the Frey curve at  $\mfP \in P$.
	\begin{lem}
		\label{criteria for potentially multiplicative reduction}
		Let $E/K$ be an elliptic curve and $p>5$ be a prime. For $\mfq \in P$ with $\mfq \nmid p$, $E$ has potentially multiplicative reduction at $\mfq$ and $p \nmid v_\mfq(j_E)$ if and only if $p | \# \bar{\rho}_{E,p}(I_\mfq)$.
	\end{lem}
	\begin{lem}
		\label{3 divides discriminant}
		Let $E/K$ be an elliptic curve and $p\geq 3$ be a prime.
		Suppose $E$ has potential good reduction at $\mfP$ for some $\mfP \in S_K$. Then, $3 \nmid v_\mfP(\Delta_E)$ if and only if $3 | \#\bar{\rho}_{E,p}(I_\mfP)$.
	\end{lem}

   The following lemma determines the type of reduction of $E_{a,b,c}$ at primes $\mfq \nmid 2pBC$.
	\begin{lem}
		\label{Type of reduction at q away from 2,p,B x^2=By^p+Cz^p}
		Let $(a,b,c)\in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p>5$ of Type I or II, and let $E$ be the associated Frey curve. Suppose $\mfq \in P$ with $\mfq\nmid 2pBC$. Then $p \nmid \#\bar{\rho}_{E,p}(I_\mfq)$.
 	\end{lem}

	\begin{proof}
		By Lemma~\ref{criteria for potentially multiplicative reduction}, it is enough to show that

		either $v_\mfq(j_E) \geq 0$ or $p | v_\mfq(j_E)$. Recall that $\Delta_E=2^{6}(B^2C)(b^2c)^{p}$ and $c_4=2^4(Bb^p+4Cc^p)$.
		If $\mfq \nmid \Delta_E$, then $E$ has good reduction at $\mfq$, and hence $v_\mfq(j_E)\geq 0$. If $\mfq | \Delta_E$ then $\mfq |bc$,
		and hence $\mfq$ divides exactly one of $b$ and $c$. Therefore, $\mfq \nmid c_4$ and $p |v_\mfq(j_E)=-p
		v_\mfq(b^2c)$, which completes the proof.
	\end{proof}

	We discuss the type of reduction of the Frey curve $E_{a,b,c}$ at  $\mfP \in S_K$ with $(a,b,c) \in W_K$.
	\begin{lem}
		\label{reduction on T and S}
		Let $(a,b,c)\in W_K$, and let $E:=E_{a,b,c}$ be the Frey curve attached to $(a,b,c)$ as in~\eqref{Frey curve for x^2=By^p+Cz^p of Type I}.
		For $\mfP \in S_K$, if $p > 6v_\mfP(2)+v_\mfP(C)$, then $\ v_\mfP(j_E) < 0$ and $p \nmid v_\mfP(j_E)$, equivalently $p | \#\bar{\rho}_{E,p}(I_\mfP)$.
    \end{lem}
    \begin{proof}
    Since $(a,b,c)\in W_K$, $\mfP$ divides exactly one of $b$ and $c$.

    Suppose the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$
    is of Type I. Recall that $j_E=2^{6} \frac{(Bb^p+4Cc^p)^3}{B^2C(b^2c)^{p}}$. If $\mfP |b$, then $\mfP \nmid c$.
    Since $p > 6v_\mfP(2)$, $v_\mfP(j_E)= 6v_\mfP(2)+6v_\mfP(2)-2pv_\mfP(b)= 2(6v_\mfP(2)-pv_\mfP(b))$, $v_\mfP(j_E) <0$ and $p \nmid v_\mfP(j_E)$. Similar proof works for $\mfP |c$ as well.

    Suppose the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$
    is of Type II, i.e., $C=2^r$ for some $r \in \N$. If $\mfP |b$, then $v_\mfP(j_E)=  (6-r)v_\mfP(2)+3(r+2)v_\mfP(2)-2pv_\mfP(b)= (12+2r)v_\mfP(2)-2pv_\mfP(b) <0$ and $p \nmid v_\mfP(j_E)$, since $p > (6+r)v_\mfP(2)$. If $\mfP |c$ then $v_\mfP(j_E)=  (6-r)v_\mfP(2)-pv_\mfP(c) <0$ since $p > (6+r)v_\mfP(2)$.
		 Since $p > (6+r)v_\mfP(2)> (6-r)v_\mfP(2) \geq0$ for $1 \leq r\leq 6$ and $-p< (-6-r)v_\mfP(2)<(6-r)v_\mfP(2)<0$ for $r> 6$, we get $p \nmid v_\mfP(j_E)$.
		Hence, by Lemma~\ref{criteria for potentially multiplicative reduction}, we get $p | \#\bar{\rho}_{E,p}(I_\mfP)$.

	\end{proof}

	\subsection{Proof of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}.}
The proof of this theorem depends on the following result.
	\begin{thm}
		\label{auxilary result x^2=By^p+Cz^p over W_K}
		Let $K$ be a totally real field. Then, there is a constant $V=V_{K,B,C}>0$ (depending on $K,B,C$) such that the following hold.
		Let $(a,b,c)\in W_K$ be a solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p>V$ of Type I or II, and let $E$ be the Frey curve as in \eqref{Frey curve for x^2=By^p+Cz^p of Type I}. Then, there exists an elliptic curve $E^\prime/K$ such that:
		\begin{enumerate}
			\item $E^\prime$ has good reduction away from $S_K^{\prime}$ and has a non-trivial $2$-torsion point;
			\item $\bar{\rho}_{E,p} \sim\bar{\rho}_{E^\prime,p}$, and  $v_\mfP(j_{E^\prime})<0$ for $\mfP \in S_K$.
		\end{enumerate}
	\end{thm}

    \begin{proof}[Proof of Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}]
		By Theorem~\ref{modularity of Frey curve x^2=By^p+Cz^p over W_K}, $E$ is modular for primes $p>A=A_{K,B,C}$ with $A \gg 0$. By Lemma~\ref{reduction away from S}, $E$ is semi-stable away from $S_K^{\prime}$. If necessary, we can take the Galois closure of $K$ to ensure that $\bar{\rho}_{E,p}$ is irreducible for $p \gg 0$ (cf.~\cite[Theorem 2]{FS15 Irred}).

			By ~\cite[Theorem 7]{FS15}, there exists a Hilbert
        modular newform $f$ of parallel weight $2$, level $\mfn_p$ and some prime $\omega$ of $\Q_f$ such that $\omega | p$ and $\bar{\rho}_{E,p} \sim \bar{\rho}_{f,\omega}$ for $p \gg 0$, where $\bar{\rho}_{f,\omega}$ denote the residual Galois representation attached to $f,\omega$. By allowing $p$ to be sufficiently large, we can assume  $\Q_f=\Q$ (cf. ~\cite[\S 4]{FS15} for more details).

		Let $\mfP \in S_K$. Then $E$ has potential multiplicative reduction at $\mfP$ and $p | \#\bar{\rho}_{E,p}(I_\mfP)$ for $p \gg 0$
		(cf. Lemma~\ref{reduction on T and S}). The existence of $E_f$ then follows from \cite[Corollary 2.2]{FS15} for all $p\gg 0$ after leaving primes $p$ with $p \mid \left( \text{Norm}(K/\Q)(\mfP) \pm 1 \right)$. Therefore, $\bar{\rho}_{E,p} \sim \bar{\rho}_{E_f,p}$ for some elliptic curve $E_f$ with conductor $\mfn_p$ for $p>V=V_{K,B,C}$, where $V_{K,B,C}$ is the maximum of all the above implicit/explicit lower bounds.

		\begin{itemize}
			\item Since the conductor of $E_f$ is $\mfn_p$ given in \eqref{conductor of E and E' x^2=By^p+Cz^p Type I}, $E_f$ has good reduction away from $S_K^{\prime}$. Now, arguing as in~\cite[page 1247]{M22}, we can enlarge the constant $V$ and by possibly replacing $E_f$ with an isogenous curve, say $E^\prime$, we get $E^\prime/ K$ has a non-trivial $2$-torsion point. Since $E_f \sim E^\prime$, $E^\prime$ has good reduction away from $S_K^{\prime}$.

			\item Since $E_f$ is isogenous to $E^\prime$ and $\bar{\rho}_{E,p} \sim \bar{\rho}_{E_f,p}$ implies $\bar{\rho}_{E,p} \sim \bar{\rho}_{E^\prime,p}$. As a result, we obtain $p | \# \bar{\rho}_{E,p}(I_\mfP)=\# \bar{\rho}_{E^\prime,p}(I_\mfP)$ for any $\mfP \in S_K$. Finally, by Lemma~\ref{criteria for potentially multiplicative reduction}, we have $v_\mfP(j_{E^\prime})<0$ for any $\mfP \in S_K$.
			\end{itemize}
		This completes the proof of the theorem.
	\end{proof}
	We now prove Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}, and its inspired from that of~\cite[Theorem 3]{M22}.
	\begin{proof}[Proof of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}]
		Suppose $(a,b,c)\in W_K$ is a solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p>V$ of Type I or Type II, where $V=V_{K,B,C}$ be the constant as in Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}. By Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}, there exists an elliptic curve $E^\prime/K$ having a non-trivial $2$-torsion point and good reduction away from $S_K^{\prime}$.
Then the elliptic curve $E^\prime/K$ has a model of the form
		\begin{equation}
			\label{j invariant of E'}
		E^\prime:y^2=x^3+cx^2+dx
		\end{equation}
	for some $c,d \in K$ with $j$-invariant $j_{E^\prime}=2^8 \frac{(c^2-3d)^3}{d^2(c^2-4d)}$. Since $E^\prime$ has good reduction away from $S_K^{\prime}$, $j_{E^\prime} \in \mcO_{S_K^{\prime}}$.

	Take $\lambda := \frac{c^2}{d}$ and $\mu := \lambda-4 \in \mcO_{S_K^{\prime}}^\ast$
	(cf.~\cite[Lemma 16(i)]{M22}). By~\cite[Lemma-17(i)]{M22},
	we get $ \lambda  \mcO_K=I^2J$ for some fractional ideal $I$ and $S_K^{\prime}$-ideal $J$. Since $J$ is $S_K^{\prime}$-ideal,  $1= [I]^2 \in \Cl_{S_K^{\prime}}(K)$. By hypothesis $\Cl_{S_K^{\prime}}(K)[2]=1$ which gives $I=\gamma I_1$ for some
	$\gamma \in \mcO_K$ and $S_K^{\prime}$-ideal $I_1$. Thus, $\lambda \mcO_K=\gamma^2I_1^2J$ and hence $(\frac{\lambda}{\gamma^2}) \mcO_K$ is an $S_K^{\prime}$-ideal. Therefore, $u=\frac{\lambda}{\gamma^2} \in \mcO_{S_K^{\prime}}^\ast$.
	Now, divide the equation $\mu +4=\lambda$ by $u$ to obtain
	$\alpha +\beta =\gamma^2$, where $\alpha= \frac{\mu}{u} \in \mcO_{S_K^{\prime}}^\ast$ and $\beta =\frac{4}{u} \in \mcO_{S_K^{\prime}}^\ast$, which implies $ \alpha \beta^{-1}=\frac{\mu}{4}$.
	By~\eqref{assumption for main result1 for x^2=By^p+Cz^p Type I}, there exists $\mfP \in S_K$ with $|v_\mfP(\alpha \beta^{-1})|= |v_\mfP(\frac{\mu}{4} )| \leq 6 v_\mfP(2)$. This means
	\begin{equation}
		\label{inequality for valution of mu}
		-4v_\mfP(2) \leq v_\mfP(\mu) \leq 8 v_\mfP(2)
	\end{equation}
    We now show that the bounds on $v_\mfP(\mu)$ would imply that $v_\mfP(j_{E^\prime}) \geq 0$.
	Write $j_{E^\prime}$ in terms of $\mu$ yields $j_{E^\prime}= 2^8 \frac{(\mu+1)^3}{\mu}$,
	which means $v_\mfP(j_{E^\prime})=8v_\mfP(2)+3 v_\mfP(\mu +1)-v_\mfP(\mu)$.
\begin{itemize}
 \item If $v_\mfP(\mu) <0$, then $v_\mfP(\mu +1)=v_\mfP(\mu)$.
       By \eqref{inequality for valution of mu}, we get $v_\mfP(j_{E^\prime})\geq 0$.
 \item  If $v_\mfP(\mu)=0$, then $v_\mfP(\mu +1)\geq 0$, hence $v_\mfP(j_{E^\prime})\geq 8v_\mfP(2) \geq 0$.
 \item  If $v_\mfP(\mu) >0$, then $v_\mfP(\mu +1)=0$.
        By \eqref{inequality for valution of mu}, we get $v_\mfP(j_{E^\prime})=8v_\mfP(2)- v_\mfP(\mu) \geq 0$.
\end{itemize}
In all cases, we get $v_\mfP(j_{E^\prime}) \geq 0$, which is a contradiction to Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}. This completes the proof of the theorem.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{main result2 for x^2=By^p+Cz^p Type I}]
	By Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I}, it suffices to show that for every solution $(\alpha, \beta, \gamma)\in  \mcO_{S_K}^\ast \times \mcO_{S_K}^\ast \times \mcO_{S_K}$ to the equation $\alpha+\beta=\gamma^2$, there exists $\mfP\in S_K$ such that $|v_\mfP(\alpha\beta^{-1}) |\leq 6v_\mfP(2)$.
	Let $\mfP \in S_K$. If necessary, by scaling even powers of $\mfP$ and swapping $\alpha, \beta$, we can assume $0\leq v_\mfP(\beta)\leq v_\mfP(\alpha)$ with $v_\mfP(\beta)=0$ or $1$.
	\begin{enumerate}
		\item Suppose $v_\mfP(\beta)=1$ for some $\mfP \in S_K$. If $v_\mfP(\alpha)>6v_\mfP(2)>1$ then $v_\mfP(\gamma^2)=v_\mfP(\alpha+\beta)=v_\mfP(\beta)=1$, which cannot occur since $v_\mfP(\gamma^2)$ is even. The inequality $v_\mfP(\alpha)\leq 6v_\mfP(2)$ implies
		$\left|v_\mfP(\alpha \beta^{-1}) \right|\leq 6v_\mfP(2)-1 <6v_\mfP(2)$.

		\item Suppose $v_\mfq(\beta)=0$ for all $\mfq \in S_K$, i.e., $\beta$ is a unit in $K$.
		\begin{itemize}
		 \item If $\beta$ is a square, then divide the equation $\alpha +\beta= \gamma^2$ by $\beta$ to obtain an equation of the form $\alpha'+1=\gamma'^2$, where $\alpha'=  \alpha \beta^{-1} \in \mcO_{S_K^{\prime}}^\ast$ and $\gamma' \in \mcO_{S_K^{\prime}}$. By~\eqref{assumption for main result2 for x^2=By^p+Cz^p Type I}, we obtain $|v_\mfP(\alpha \beta^{-1}) |=|v_\mfP(\alpha')|\leq 6v_\mfP(2)$ for some $\mfP \in S_K$.

		 \item Suppose $\beta$ is not a square. If $v_\mfP(\alpha)\leq 6v_\mfP(2)$ for some $\mfP \in S_K$, then we are done. Otherwise, $v_\mfq(\alpha)> 6v_\mfq(2)>1$ for all $\mfq \in S_K$. This gives $\alpha \equiv 0 \pmod {2^6}$ and $\gamma^2=\alpha+\beta \equiv \beta\pmod {2^6}$.
		 Since $v_\mfq(\gamma^2)= v_\mfq(\alpha+\beta)=0$ for all $\mfq \in S_K$ and $S_K=S_K^\prime$, we get $\gamma \in \mcO_K$.

		 Now, consider the field $L=K(\theta)$, where $\theta:= \frac{\gamma - \sqrt{\beta}}{2}$. The minimal polynomial of $\theta$ is $m_\theta(x)= x^2-\gamma x+  \frac{\gamma^2 - \beta}{4}$. Then $m_\theta(x) \in \mcO_K[x]$ with discriminant $\beta$.
		 Therefore, $L$ is everywhere unramified extension of degree 2 over $K$ implying $2| h_K$, which contradicts our hypothesis that $2\nmid h_K$.
        \end{itemize}
    \end{enumerate}
This completes the proof of the proposition.
\end{proof}

\section{Solutions of the Diophantine equation $x^2=By^p+2^rz^p$ over $K$}
\label{section for $x^2=By^p+2^rz^p$ and $2x^2=By^p+2^rz^p$ over $K$}
In this section, we examine the solutions of the Diophantine equation \eqref{x^2=By^p+Cz^p} with exponent $p$ of Type II, i.e., $x^2 =By^p+2^rz^p$ over $K$. Here, $S_K^{\prime}=\{\mfP \in P : \mfP |2B \}$. Let $h_K^+$ be the narrow class number of $K$. We follow notations as in \S\ref{notations section for x^2=By^p+Cz^p}.

\subsection{Main result}
We write $(ES)$ for ``either $[K: \Q]\equiv 1 \pmod 2$ or Conjecture \ref{ES conj} holds for $K$".
For $r\in \N$, let $S_r:=\{ (\pm \sqrt{2^r+B},1,1), \ (\pm \sqrt{2^r-B},-1,1), \\
(\pm \sqrt{-2^r+B},1,-1),\ (\pm \sqrt{-2^r-B},1,1)\}$.
For $r\in \{1,2,4,5\}$, we show that the equation \eqref{x^2=By^p+Cz^p} with exponent $p$ of Type II has no asymptotic solution in $\mcO_K^3 \setminus S_r$. More precisely;

\begin{thm}
	\label{main result1 for x^2=By^p+2^rz^p over K}
	Let $K$ be a totally real field satisfying $(ES)$ with $\Cl_{S_K^{\prime}}(K)[2]=1$. Suppose for every solution $(\alpha, \beta, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to $\alpha+\beta=\gamma^2$,
	there exists $\mfP \in U_K$ that satisfies
	\begin{equation}
		\label{assumption for main result1 x^2=By^p+2^rz^p over K}
		\left|v_\mfP( \alpha \beta^{-1}) \right| \leq 6v_\mfP(2) \text{ and } v_\mfP \left(\alpha \beta^{-1} \right)\equiv 0 \pmod 3.
	\end{equation}
	Then for $r \in \{1,2,4,5\}$, the Diophantine equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type II has no asymptotic solution in $\mcO_K^3 \setminus S_r$.

\end{thm}
The following proposition is a consequence of Theorem~\ref{main result1 for  x^2=By^p+2^rz^p over K} and will be relevant in \S\ref{section for local criteria for Diophantine equations over $K$}.
\begin{prop}
	\label{main result2 for x^2=By^p+2^rz^p over K}
	\label{cor to main result2 for x^2=By^p+2^rz^p over K}
	Let $K$ be a field satisfying $(ES)$ with degree $n>1$ and $2 \nmid h_K^+$. Assume $B= \pm1$, and $2$ is inert in $K$. Suppose for every solution $(\alpha, \gamma) \in \mcO_{S_K}^\ast \times \mcO_{S_K}$ to $\alpha+1=\gamma^2$, there exists $\mfP \in U_K$ that satisfies
	\begin{equation}
		\label{assumption for main result2 x^2=By^p+2^rz^p over K}
		\left|v_\mfP(\alpha) \right| \leq 6v_\mfP(2) \text{ and } v_\mfP \left(\alpha \right)\equiv 0 \pmod 3.
	\end{equation}
	Then for $r \in \{1,2,4,5\}$, the Diophantine equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type II has no asymptotic solution in $\mcO_K^3 \setminus S_r$.
	In particular, if $[K:\Q]$ is odd, then $S_1=\{(\pm1, -1,1)\}$ (resp., $\{(\pm1, 1,1)\}$) for $B=1$ (resp., $B=-1$), and $S_r= \phi $ for $r=2,4,5$.
\end{prop}

\subsection{Steps to prove Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}.}

We now prove the modularity of the Frey curve $E:=E_{a,b,c}$ in \eqref{Frey curve for x^2=By^p+Cz^p of Type I} associated to any non-trivial primitive solution $(a,b,c)\in$ $\mcO_K^3 \setminus S_r$.
\begin{thm}
	\label{modularity of Frey curve of x^2=By^p+2^rz^p over K}
	Let $(a,b,c)\in \mcO_K^3 \setminus S_r$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p$ of Type II, and let $E:=E_{a,b,c}$ be the associated Frey curve. Then, there exists a constant $A=A_{K,B}$ (depending on $K,B$) such that for primes $p >A$, $E/K$ is modular.
\end{thm}

\begin{proof}
	Arguing as in the proof of Theorem~\ref{modularity of Frey curve x^2=By^p+Cz^p over W_K}, there exists $\lambda_k \in K$ with $1 \leq k \leq m$ such that $E/K$ is modular for all $\lambda(E) \notin\{\lambda_1, \lambda_2, ..., \lambda_m\}$.
	If $\lambda= \lambda_k$ for some $k \in \{1, 2, \ldots, m \}$, then $\left(\frac{b}{c} \right)^p= \frac{-\lambda_k}{B}$.
	The above equation determines $p$ uniquely, denote it $p_k$. Otherwise, we get $b=\pm c$.
	Since $a^2=Bb^p+2^rc^p$ and $(a,b,c)$ is primitive, we get $b=\pm1$ and $c=\pm1$, hence $(a,b,c) \in S_r$, which is not possible.

	Finally, the proof of the theorem follows by taking $A_K=\max \{p_1,...,p_m\}$.
\end{proof}

\subsubsection{Type of reduction with image of inertia.}
The following lemma specifies the type of reduction of the Frey curve $E:=E_{a,b,c}$ given in ~\eqref{Frey curve for x^2=By^p+Cz^p of Type I} at  $\mfP \in U_K$, when $(a,b,c)\in$ $\mcO_K^3$ and $r\in \{1,2,4,5\}$. More precisely;
\begin{lem}
	\label{reduction on T and S x^2=By^p+Cz^p Type II over K}
	Let $r\in \{1,2,4,5\}$. Let $(a,b,c)\in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p>(6+r)v_\mfP(2)$ of Type II, and let $E$ be the associated Frey curve. If $\mfP \in U_K$, then either $p | \#\bar{\rho}_{E,p}(I_\mfP)$ or $3 | \#\bar{\rho}_{E,p}(I_\mfP)$.

\end{lem}

\begin{proof}
Recall that $\Delta_E=2^{r+6}B^2(b^2c)^{p}$ and $j_E=2^{6-r} \frac{(Bb^p+2^{r+2}c^p)^3}{B^2(b^2c)^{p}}$. If $\mfP |bc$, then $p | \#\bar{\rho}_{E,p}(I_\mfP)$ by Lemma~\ref{reduction on T and S} and due to the fact that $p>(6+r)v_\mfP(2)$. If $\mfP \nmid bc$, then $v_\mfP(j_E)= (6-r)v_\mfP(2)$ and $v_\mfP(\Delta_E)=(6+r)v_\mfP(2)$. Since $\mfP \in U_K$ and $r\in \{1,2,4,5\}$, we get $v_\mfP(j_E)\geq 0$ and $3 \nmid v_\mfP(\Delta_E)$.
Hence, by Lemma~\ref{3 divides discriminant}, we get  $3 | \#\bar{\rho}_{E,p}(I_\mfP)$.
\end{proof}

\subsection{Proof of Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}}
The proof of this theorem depends on the following result.
\begin{thm}
\label{auxilary result x^2=By^p+2^rz^p over K}
Let $K$ be a totally real field satisfying $(ES)$, and let $r\in \{1,2,4,5\}$.
Then, there is a constant $V=V_{K,B}>0$ (depending on $K,B$) such that the following hold. Let $(a,b,c)\in \mcO_K^3 \setminus
S_r$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p >V$ of Type II, and let $E$ be the Frey curve as in ~\eqref{Frey curve for x^2=By^p+Cz^p of Type I}. Then there exists an elliptic curve $E^\prime/K$ such that:

\begin{enumerate}
	\item $E^\prime/K$ has good reduction away from $S_K^{\prime}$ and has a non-trivial $2$-torsion point, and
	$\bar{\rho}_{E,p} \sim\bar{\rho}_{E^\prime,p}$;
	\item  For $\mfP \in U_K$, either $v_\mfP(j_{E^\prime})<0$ or $3\nmid v_\mfP(j_{E^\prime})$.

\end{enumerate}
\end{thm}
\begin{proof}
Arguing as in the proof of Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}, the first part of Theorem~\ref{auxilary result x^2=By^p+2^rz^p over K} follows from \cite[Theorem 8]{FS15}, Theorem~\ref{conductor of E and E' x^2=By^p+Cz^p Type I}, Theorem~\ref{modularity of Frey curve of x^2=By^p+2^rz^p over K} and Lemma~\ref{reduction on T and S x^2=By^p+Cz^p Type II over K}. Let $\mfP \in U_K$. If $p | \# \bar{\rho}_{E,p}(I_\mfP)= \# \bar{\rho}_{E^\prime,p}(I_\mfP)$, then by Lemma~\ref{criteria for potentially multiplicative reduction}, we get $v_\mfP(j_{E^\prime})<0$. If $p \nmid \# \bar{\rho}_{E,p}(I_\mfP)$, then by Lemma~\ref{reduction on T and S x^2=By^p+Cz^p Type II over K}, we conclude that $3 | \# \bar{\rho}_{E,p}(I_\mfP)=\# \bar{\rho}_{E^\prime,p}(I_\mfP)$. If $v_\mfP(j_{E^\prime}) < 0$, then we are done. If $v_\mfP(j_{E^\prime}) \geq 0$, then  by Lemma~\ref{3 divides discriminant}, we have $3\nmid v_\mfP(\Delta_{E^\prime})$. Since $j_{E^\prime} = \frac{c_4^3}{\Delta_{E'}}$ and $3\nmid v_\mfP(\Delta_{E^\prime})$, we get $3\nmid v_\mfP(j_{E^\prime})$. This completes the proof of the theorem.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}]
Let $r\in \{1,2,4,5\}$. Let $(a,b,c)\in \mcO_K^3 \setminus S_r$ be a non-trivial primitive solution to the equation~\eqref{x^2=By^p+Cz^p} with exponent $p >V$ of Type II, where $V=V_{K,B}$ be the constant as in Theorem~\ref{auxilary result x^2=By^p+2^rz^p over K}. By Theorem~\ref{auxilary result x^2=By^p+2^rz^p over K}, there exists an elliptic curve $E^\prime/K$ having a non-trivial $2$-torsion point and good reduction away from $S_K^{\prime}$.
By ~\eqref{assumption for main result1 x^2=By^p+2^rz^p over K}, there exists some $\mfP \in U_K$ that satisfies $|v_\mfP(\alpha\beta^{-1}) |\leq 6v_\mfP(2)$ and $v_\mfP(\alpha\beta^{-1})\equiv 0 \pmod 3$.

Now, arguing as in the proof of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type II}, we find $v_\mfP(j_{E^\prime}) \geq 0$ by using $|v_\mfP(\alpha\beta^{-1}) |\leq 6v_\mfP(2)$.
Recall that $j_{E^\prime} = 2^8\frac{(1+\mu)^3}{\mu}$ with $\frac{\mu}{4}= \alpha\beta^{-1}$. This implies $v_\mfP(j_{E^\prime}) \equiv 2v_\mfP(2)-v_\mfP(\mu)= -v_\mfP(\alpha\beta^{-1}) \pmod3 $.
Since $v_\mfP(\alpha\beta^{-1})\equiv 0 \pmod 3$, $v_\mfP(j_{E^\prime})\equiv 0 \pmod 3$ and hence $3 | v_\mfP(j_{E^\prime})$.
Therefore, $v_\mfP(j_{E^\prime}) \geq0$ and $3 | v_\mfP(j_{E^\prime})$, which contradicts Theorem~\ref{auxilary result x^2=By^p+2^rz^p over K}. This completes the proof of the theorem.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{main result2 for x^2=By^p+2^rz^p over K}]
Since $B=\pm1$, and $2$ is inert in $K$, we find $S_K^\prime =S_K$ is principal. Let $\mfP \in U_K$ be the unique prime lying above $2$. Let $(\alpha, \beta, \gamma)\in  \mcO_{S_K}^\ast \times \mcO_{S_K}^\ast \times \mcO_{S_K}$ be a solution to $\alpha+\beta=\gamma^2$. According to Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}, it suffices to show that $|v_\mfP(\alpha\beta^{-1}) |\leq 6v_\mfP(2)$ and $v_\mfP(\alpha\beta^{-1}) \equiv 0 \pmod3$. If necessary by scaling even powers of $\mfP$ and swapping $\alpha, \beta$, we can assume $0\leq v_\mfP(\beta)\leq v_\mfP(\alpha)$ with $v_\mfP(\beta)=0$ or $1$.
\begin{enumerate}
\item Suppose $v_\mfP(\beta)=1$. If $v_\mfP(\alpha)>1$, then $v_\mfP(\gamma^2)=v_\mfP(\alpha+\beta)=v_\mfP(\beta)=1$, which cannot occur because $v_\mfP(\gamma^2)$ is even. As a result, $v_\mfP(\alpha)=1$. Thus $|v_\mfP(\alpha\beta^{-1})|=0<6v_\mfP(2)$ and $v_\mfP(\alpha\beta^{-1}) \equiv 0 \pmod 3$.

\item Suppose $v_\mfP(\beta)=0$, i.e., $\beta$ is a unit in $K$.
\begin{itemize}
	\item Assume $\beta$ is a square. Then divide the equation $\alpha +\beta= \gamma^2$ by $\beta$ to obtain an equation of the form $\alpha'+1=\gamma'^2$, where $\alpha'= \alpha\beta^{-1} \in \mcO_{S_K}^\ast$, $\gamma' \in \mcO_{S_K}$. Using~\eqref{assumption for main result2 x^2=By^p+2^rz^p over K}, we get $|v_\mfP(\alpha\beta^{-1}) |\leq 6v_\mfP(2)$ and $v_\mfP(\alpha\beta^{-1}) \equiv 0 \pmod 3$.

	\item Assume $\beta$ is not a square. Consider the field $L=K(\sqrt{\beta})$. The minimal polynomial of $\beta$ is $m_\beta (x)=x^2-\beta$. Then $m_\beta (x)\in \mcO_K[x]$ with discriminant $4\beta$. Hence $L$ is unramified away from $2$. By~\cite[Theorem 9(b)]{FKS20}, we conclude that $2$ totally ramified in $K$, which contradicts $2$ being inert in $K$.
\end{itemize}
\end{enumerate}
This completes the proof of the proposition.
\end{proof}

\section{Solutions of the Diophantine equation $2x^2=By^p+2^rz^p$ over $K$}
\label{notations section for 2x^2=By^p+Cz^p}
In this section, we study the solutions of the following Diophantine equation
\begin{equation}
	\label{2x^2=By^p+Cz^p}
	2x^2 =By^p+2^rz^p,
\end{equation}
with exponent $p\geq3$, $B$ is an odd integer and $r \in \N$. Here, $S_K^{\prime}=\{\mfP \in P : \mfP |2B \}$.

\begin{dfn}[Trivial solution]
	We call a solution $(a, b, c) \in \mcO_K^3$ to the equation \eqref{2x^2=By^p+Cz^p} with exponent $p$ is trivial, if $abc=0$,
	otherwise non-trivial.

\end{dfn}

\subsection{Main result}
\label{section for main result of 2x^2=By^p+2^rz^p}
We now show that the equation~\eqref{2x^2=By^p+Cz^p} with exponent $p$ has no asymptotic solution in $\mcO_K^3$. More specifically;
\begin{thm}
	\label{main result1 for 2x^2=By^p+Cz^p Type I}
	Let $K$ be a totally real field with $\Cl_{S_K^{\prime}}(K)[2]=1$. Suppose for every solution $(\alpha, \beta, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to $\alpha+\beta=\gamma^2$,
	there exists $\mfP \in S_K$ that satisfies
	\begin{equation}
		\label{assumption for main result1 for 2x^2=By^p+Cz^p Type I}
		\left| v_\mfP \left( \alpha \beta^{-1} \right) \right| \leq 6v_\mfP(2).
	\end{equation}
Then, the Diophantine equation~\eqref{2x^2=By^p+Cz^p} with exponent $p$ has no asymptotic solution in $\mcO_K^3$.
\end{thm}

The following proposition is a consequence of Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}, which will be useful in \S\ref{section for local criteria for Diophantine equations over $W_K$}.
\begin{prop}
	\label{main result2 for 2x^2=By^p+Cz^p Type I}
	\label{main result2 for 2x^2=By^p+Cz^p Type II}
	Let $K$ be a field with $S_K^{\prime}=S_K$ is principal and $2 \nmid h_K$. Suppose for every solution $(\alpha, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to $\alpha+1=\gamma^2$,

	there exists $\mfP \in S_K$ that satisfies
	\begin{equation}
		\label{assumption for main result2 for 2x^2=By^p+Cz^p Type I}
		|v_\mfP(\alpha)|\leq 6v_\mfP(2).
	\end{equation}
Then, the Diophantine equation~\eqref{2x^2=By^p+Cz^p} with exponent $p$ has no asymptotic solution in $\mcO_K^3$.
\end{prop}

\begin{remark}
	\label{def for W_K'}
	\label{remark for W_K'}
	There are no non-trivial primitive solutions $(a, b, c)\in  \mcO_K^3$ to the equation $2x^2=By^p+Cz^p$ with exponent $p > [K:\Q]$ with $\mfP |bc$ for every $\mfP \in S_K$, where $B,C$ are odd integers. If not, let $(a, b, c)$	be such a solution such that  $\mfP |bc$ for every $\mfP \in S_K$. Since $B,C$ are odd, $\mfP $ divides both $b$ and $c$, which implies $\mfP^p | Bb^p+Cc^p=2a^2$. Since $p>[K:\Q]$, we have $\mfP |a$, which is a contradiction to $(a,b,c)$ is primitive.
\end{remark}

\subsection{Steps to prove Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}.}
	For any non-trivial and primitive solution $(a, b, c)\in \mcO_K^3$ to the equation \eqref{2x^2=By^p+Cz^p} with exponent $p$, the Frey curve $E:=E_{a,b,c}$ is given by

		\begin{equation}
			\label{Frey curve for 2x^2=By^p+Cz^p of Type I}
			E=E_{a,b,c} : Y^2 = X(X^2-4aX+2Bb^p),
		\end{equation}
		with $c_4=2^5(Bb^p+2^{r+2}c^p),\ \Delta_E=2^{9+r}B^2(b^2c)^{p}$ and $ j_E=2^{6-r} \frac{(Bb^p+2^{r+2}c^p)^3}{B^2(b^2c)^{p}}$.

	\subsection{Modularity of the Frey curve}
			We now prove the modularity of the Frey curve $E:=E_{a,b,c}$ in \eqref{Frey curve for 2x^2=By^p+Cz^p of Type I}, which is analogous to that of Theorem~\ref{modularity of Frey curve x^2=By^p+Cz^p over W_K}.

	\begin{thm}
		\label{modularity of Frey curve 2x^2=By^p+Cz^p over W_K}
		Let $(a,b,c) \in \mcO_K^3$ be a non-trivial primitive solution to the equation $\eqref{2x^2=By^p+Cz^p}$ with exponent $p > [K: \Q]$, and let $E:=E_{a,b,c}$ be the associated Frey curve. Then, there exists a constant $A=A_{K,B,r}$ (depending on $K,B,r$) such that for primes $p >A$, $E/K$ is modular.
	\end{thm}
	\begin{proof}
  Arguing as in the proof of Theorem~\ref{modularity of Frey curve x^2=By^p+Cz^p over W_K}, there exists $\lambda_k \in K$ with $1 \leq k \leq m$ such that $E/K$ is modular for all $\lambda(E) \notin\{\lambda_1, \lambda_2, ..., \lambda_m\}$.
  If $\lambda(E)= \lambda_k$ for some $k \in \{1, 2, \ldots, m \}$, then $\left(\frac{b}{c} \right)^p= \frac{-2^r\lambda_k}{B}$.
  The above equation determines $p$ uniquely, denote it $p_k$. If not, since $K$ is totally real,
  we get $b=\pm c$.

Since $2a^2= Bb^p+2^rc^p$ for some $r \in \N$ and $B$ is odd, we get $\mfP|b$ for any $\mfP \in S_K$.
Since $b=\pm c$, $\mfP^p |Bb^p+2^rc^p=2a^2$. As $p > [K: \Q]$, we get $\mfP | a$, which contradicts $(a,b,c)$ is primitive. Finally, the proof of the theorem follows by taking $A=\max \{p_1,...,p_m, [K: \Q]+1\}$.
\end{proof}

	\subsection{Reduction type.}
		The following lemma describes the type of reduction of the Frey curve $E:= E_{a,b,c}$ in~\eqref{Frey curve for 2x^2=By^p+Cz^p of Type I} at primes $\mfq$ away from $S_K^{\prime}$.

	\begin{lem}
		\label{reduction away from S 2x^2=By^p+Cz^p}
		Let $(a,b,c) \in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{2x^2=By^p+Cz^p} with exponent $p$, and let $E$ be the associated Frey curve. Then at all primes  $\mfq \notin S_K^{\prime}$, $E$ is minimal, semi-stable at $\mfq$ and satisfies $p | v_\mfq(\Delta_E)$. Let $\mfn$ be the conductor of $E$ and $\mfn_p$ be as in \eqref{conductor of elliptic curve}. Then,
			\begin{equation}
				\label{conductor of E and E' 2x^2=By^p+Cz^p Type I}
				\mfn=\prod_{\mfP \in S_K^{\prime}}\mfP^{r_\mfP} \prod_{\mfq|bc,\ \mfq \notin S_K^{\prime}}\mfq,\ \mfn_p=\prod_{\mfP \in S_K^{\prime}}\mfP^{r_\mfP^{\prime}},
			\end{equation}
			where $0 \leq r_\mfP^{\prime} \leq r_\mfP $ with $r_\mfP
			\leq 2+6v_\mfP(2)$ for $\mfP |2$, and $ r_\mfP\leq 2+3v_\mfP(3)$ for $\mfP \nmid 2$.
	\end{lem}

	\begin{proof}
	The proof of Lemma~\ref{reduction away from S 2x^2=By^p+Cz^p} is similar to that of Lemma~\ref{reduction away from S}, except that here $c_4=2^5(Bb^p+2^{r+2}c^p)$ and $ \Delta_E=2^{9+r}B^2(b^2c)^{p}$.
	\end{proof}

	\subsubsection{Type of reduction with image of inertia.}
	The following lemma, whose proof is similar to that of Lemma~\ref{Type of reduction at q away from 2,p,B x^2=By^p+Cz^p}, specifies the type of reduction of $E_{a,b,c}$ at primes $\mfq \nmid 2pB$.
	\begin{lem}
		\label{Type of reduction at q away from 2,p,B 2x^2=By^p+Cz^p}
		Let $(a,b,c)\in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{2x^2=By^p+Cz^p} with exponent $p>5$, and let $E$ be the associated Frey curve. Suppose $\mfq\in P$ with $\mfq\nmid 2pB$. Then $p \nmid \#\bar{\rho}_{E,p}(I_\mfq)$.
	\end{lem}

	We will now discuss the type of reduction of the Frey curve $E_{a,b,c}$ at $\mfP \in S_K$.
	\begin{lem}
		\label{reduction on T and S 2x^2=By^p+Cz^p}
		Let $(a,b,c)\in \mcO_K^3$ be a non-trivial primitive solution to the equation~\eqref{2x^2=By^p+Cz^p} with exponent $p > \max \{ (6+r)v_\mfP(2),\ [K : \Q]\}$, and let $E$ be the associated Frey curve.
		For $\mfP \in S_K$, we have $\ v_\mfP(j_E) < 0$ and $p \nmid v_\mfP(j_E)$, equivalently $p | \#\bar{\rho}_{E,p}(I_\mfP)$.
	\end{lem}

	\begin{proof}

		Since $(a,b,c)$ is a solution, we get $\mfP |b$. As observed earlier, since $p>[K: \Q] $,   $\mfP \nmid c$. Recall $j_E=2^{6-r} \frac{(Bb^p+2^{r+2}c^p)^3}{B^2(b^2c)^{p}}$. Since $p > (6+r)v_\mfP(2)$, $v_\mfP(j_E)= (12+2r)v_\mfP(2)-2pv_\mfP(b) <0$ and $p \nmid v_\mfP(j_E)$.
Hence, by Lemma~\ref{criteria for potentially multiplicative reduction}, we get $p | \#\bar{\rho}_{E,p}(I_\mfP)$.

	\end{proof}

	\subsection{Proof of Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}.}
	The proof of this theorem depends on the following result.
	\begin{thm}

		\label{auxilary result 2x^2=By^p+Cz^p over W_K}
		Let $K$ be a totally real field. Then, there is a constant $V=V_{K,B,r}>0$ (depending on $K,B,r$) such that the following hold.
		Let $(a,b,c)\in \mcO_K^3$ be a non-trivial primitive solution to the equation ~\eqref{2x^2=By^p+Cz^p} with exponent $p>V$, and let $E$ be the Frey curve as in \eqref{Frey curve for x^2=By^p+Cz^p of Type I}. Then, there exists an elliptic curve $E^\prime/K$ such that:
		\begin{enumerate}
			\item $E^\prime/K$ has good reduction away from $S_K^{\prime}$ and has a non-trivial $2$-torsion point;
			\item $\bar{\rho}_{E,p} \sim\bar{\rho}_{E^\prime,p}$, and  $v_\mfP(j_{E^\prime})<0$ for $\mfP \in S_K$.
		\end{enumerate}
	\end{thm}

	\begin{proof}
		Arguing as in the proof of Theorem~\ref{auxilary result x^2=By^p+Cz^p over W_K}, the proof of Theorem~\ref{auxilary result 2x^2=By^p+Cz^p over W_K} follows from Theorem~\ref{modularity of Frey curve 2x^2=By^p+Cz^p over W_K}, Lemma~\ref{reduction away from S 2x^2=By^p+Cz^p}, and Lemma~\ref{reduction on T and S 2x^2=By^p+Cz^p}.
	\end{proof}

	\begin{proof}[Proof of Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}]
		Suppose $(a,b,c) \in \mcO_K^3$ is a non-trivial primitive solution to the equation $\eqref{2x^2=By^p+Cz^p}$ with exponent $p>V$, where $V=V_{K,B,r}$ be the constant as in Theorem~\ref{auxilary result 2x^2=By^p+Cz^p over W_K}.
		By Theorem~\ref{auxilary result 2x^2=By^p+Cz^p over W_K}, there exists an elliptic curve $E^\prime/K$ having a non-trivial $2$-torsion point and good reduction away from $S_K^{\prime}$. Hence $j_{E^\prime} \in \mcO_{S_K^{\prime}}$.
		Arguing as in the proof of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I} and by \eqref{assumption for main result1 for 2x^2=By^p+Cz^p Type I}, we obtain $v_\mfP(j_{E'}) \geq 0$ for some $\mfP \in S_K$, which contradicts Theorem~\ref{auxilary result 2x^2=By^p+Cz^p over W_K}. This completes the proof of the theorem.
	\end{proof}

	\begin{proof}[Proof of Proposition~\ref{main result2 for 2x^2=By^p+Cz^p Type I}]
	Arguing as in the proof of Proposition~\ref{main result2 for x^2=By^p+Cz^p Type I}, we can show that hypothesis~\eqref{assumption for main result2 for 2x^2=By^p+Cz^p Type I} of Proposition~\ref{main result2 for 2x^2=By^p+Cz^p Type I} implies hypothesis \eqref{assumption for main result1 for 2x^2=By^p+Cz^p Type I} of Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}. Hence, the proof of the proposition follows from Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}.
\end{proof}

\section{Local criteria for the solutions of Diophantine equations}
\label{section for loc criteria}
In this section, we present several local criteria of $K$ which imply Theorems~\ref{main result1 for x^2=By^p+Cz^p Type I},~\ref{main result1 for  x^2=By^p+2^rz^p over K}, ~\ref{main result1 for 2x^2=By^p+Cz^p Type I}.
We start this discussion with a lemma.

  \begin{lem}
  	\label{lemma to loc crit1 for x^2=By^p+2^rz^p over W_K}
 	\label{lem for loc crit1 for x^2=By^p+2^rz^p over K}
 	Suppose the $S_K^{\prime}$-unit equation $\lambda+\mu=1$ with $\lambda, \mu \in \mcO_{S_K^{\prime}}^\ast$ has only solutions $(-1,2), (2,-1)$ and $(\frac{1}{2}, \frac{1}{2})$. Then every solution $(\alpha, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to the equation $\alpha+1=\gamma^2$ satisfies $|v_\mfP(\alpha)|\leq 3v_\mfP(2)$ and $v_\mfP(\alpha) \equiv 0 \pmod 3$ for all $\mfP \in S_K$.
 \end{lem}
 \begin{proof}
The solution $(\alpha, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to the equation $\alpha+1=\gamma^2$ gives rise to a solution of $\lambda +\mu=1$ with $\lambda, \mu \in \mcO_{S_K^{\prime}}$ as follows.
Take $\lambda :=\frac{\gamma+1}{2}, \mu :=\frac{1-\gamma}{2}$. Since $\gamma \in \mcO_{S_K^{\prime}}$,
then so is $\lambda, \mu$. Since $\alpha =-4 \lambda \mu$ with $\alpha \in \mcO_{S_K^{\prime}}^\ast$, we get $\lambda, \mu \in \mcO_{S_K^{\prime}}^\ast$ with $\lambda +\mu=1$. The choices of $(\lambda, \mu) \in \{(-1,2), (2,-1), (\frac{1}{2}, \frac{1}{2}) \}$ implies $\alpha= -1$ or $8$. Therefore, $|v_\mfP(\alpha)|\leq 3v_\mfP(2)$ and $v_\mfP(\alpha) \equiv 0 \pmod 3$ for all $\mfP\in S_K$.
 \end{proof}

\subsection{Local criteria for Theorems~\ref{main result1 for x^2=By^p+Cz^p Type I},~\ref{main result1 for 2x^2=By^p+Cz^p Type I}:}
\label{section for local criteria for Diophantine equations over $W_K$}
In this section, we give local criteria of $K$ which imply Theorems~\ref{main result1 for x^2=By^p+Cz^p Type I}, ~\ref{main result1 for 2x^2=By^p+Cz^p Type I}. Throughout this section, we assume $B= \pm1$ and $C= \pm1$ or $2^r$ for some $r \in \N$ to get  $S_K^{\prime}=S_K$.

\begin{prop}[Quadratic]
	\label{loc crit for real quadratic field over W_K}
Let $K=\Q(\sqrt{d})$ for some prime $d$ with $d \equiv 5 \pmod8$. Then the conclusion of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I} (resp., Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}) holds over $K$.
\end{prop}

\begin{proof}
	Since $d \equiv 5 \pmod 8$, $K$ has discriminant $d$,  $2$ is inert in $K$, and hence
	$S_K$ is principal. The assumptions on $B,C$ give $S_K^{\prime}= S_K$.
		By~\cite[Proposition 1.3.2]{Z21}(or~\cite[\S 3.8]{M22}), we have $2 \nmid h_K$.
 	By~\cite[Table 1, \S 6]{FS15}, the $S_K$-unit equation $\lambda+\mu=1$ with $\lambda, \mu \in \mcO_{S_K}^\ast$ has only solutions $(-1,2), (2,-1)$ and $(\frac{1}{2}, \frac{1}{2})$.
  Now the proof of the proposition follows from Lemma~\ref{lemma to loc crit1 for x^2=By^p+2^rz^p over W_K} and Proposition~\ref{main result2 for x^2=By^p+Cz^p Type I} (resp., Proposition~\ref{main result2 for 2x^2=By^p+Cz^p Type I}).
\end{proof}

\begin{prop}[Odd degree]
	\label{loc crit for odd degree field over W_K}
	Let $K$ be a field of degree $n$ with $2 \nmid h_K$. Suppose
	\begin{enumerate}
        \item $q \geq5$ be a prime with $\gcd(n, q-1)=1$, totally ramifies in $K$;
		\item $2$ is either inert or $2=\mfP ^n $ for some principal ideal $\mfP \in P$.
	\end{enumerate}
Then the conclusion of Theorem~\ref{main result1 for x^2=By^p+Cz^p Type I} (resp., Theorem~\ref{main result1 for 2x^2=By^p+Cz^p Type I}) holds over $K$.
\end{prop}

\begin{proof}
Let $\mfP \in S_K$ be the unique prime ideal lying above $2$. Arguing as in the proof of Lemma~\ref{lemma to loc crit1 for x^2=By^p+2^rz^p over W_K},
every solution $(\alpha, \gamma) \in \mcO_{S_K^{\prime}}^\ast \times \mcO_{S_K^{\prime}}$ to the equation $\alpha+1=\gamma^2$ gives rise to a solution $\lambda, \mu \in \mcO_{S_K^{\prime}}^\ast$ with  $\lambda +\mu=1$,
where $\lambda=\frac{\gamma+1}{2}$, $\mu=\frac{1-\gamma}{2}$, and a relation $\alpha=-4\lambda \mu$.
Since $S_K^{\prime}=S_K$, by~\cite[Lemma 4.1]{FKS21}, we get $ \max\{|v_\mfP(\lambda)|, |v_\mfP(\mu)|\} < 2v_\mfP(2)$. In particular, $|v_\mfP(\alpha)|= |2v_\mfP(2)+ v_\mfP(\lambda)+ v_\mfP(\mu)| < 6v_\mfP(2)$. Now, the proof of the proposition follows from Proposition~\ref{main result2 for x^2=By^p+Cz^p Type I}
(resp, Proposition~\ref{main result2 for 2x^2=By^p+Cz^p Type I}).
\end{proof}

\subsection{Local criteria for Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}:}
\label{section for local criteria for Diophantine equations over $K$}
In this section, we provide local criteria of $K$ which imply Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K}. Throughout this section, we assume that $B=\pm1$ and $C=2^r$ with $r=1,2,4,5$ to get $S_K^{\prime}=S_K$.

\begin{prop}[Quadratic]
	\label{loc crit for real quadratic field over K}
Assume that Conjecture~\ref{ES conj} holds over $K=\Q(\sqrt{d})$ for some prime $d$ with $d \equiv 5 \pmod8$.
Then the conclusion of Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K} holds over $K$.
\end{prop}

\begin{proof}
 Since $d$ is a prime with $d\equiv 5 \pmod 8$, $2 \nmid h_K^+$, $2$ is inert in $K$ and hence $S_K^{\prime}=S_K$ is principal. Arguing as in the proof of Proposition~\ref{loc crit for real quadratic field over W_K},
 we find that the $S_K$-unit equation $\lambda+\mu=1$ with $\lambda, \mu \in \mcO_{S_K}^\ast$ has only solutions $(\frac{1}{2},\frac{1}{2} ), (2,-1)$ and $ (-1,2)$. Now the proof of the proposition follows from Lemma~\ref{lem for loc crit1 for x^2=By^p+2^rz^p over K} and  Proposition~\ref{main result2 for x^2=By^p+2^rz^p over K}.
\end{proof}

\begin{prop}[Odd degree]
	\label{loc crit2 for odd degree field over K}
Let $K$ be a field of degree $n>1$ with $2 \nmid h_K^+$. Suppose
	\begin{enumerate}
 \item $q \geq5$ be a prime with $\gcd(n, q-1)=1$ and totally ramifies in $K$;
		\item $2$ is inert in $K$.
\end{enumerate}
 Then the conclusion of Theorem~\ref{main result1 for x^2=By^p+2^rz^p over K} holds over $K$.
\end{prop}

Define $\Lambda_{S_K}:=\{(\lambda, \mu) : \lambda+\mu=1, \lambda,\mu \in \mcO_{S_K}^\ast\}$.
By the discussion in \cite[\S 5]{FS15},
the action of the symmetric group $\mfS_3$ on $\mbP^1(K) \setminus \{0,1, \infty\}$ induces an action on $\Lambda_{S_K}$ as $(\lambda, \mu)^\sigma :=(\lambda^\sigma, \mu^\sigma)$ with $\sigma \in \mfS_3$. For any $(\lambda, \mu) \in \Lambda_{S_K}$ and $\mfP \in S_K$, define $m_{\lambda, \mu}(\mfP) := \max \{|v_\mfP(\lambda)|, |v_\mfP(\mu)|\}$. If $\mfP \in S_K$ is unique, then we write $m_{\lambda, \mu}$ for $m_{\lambda, \mu}(\mfP)$.
\begin{lem}
	\label{integral reduction of lambda, mu}
	Suppose $2$ is inert in $K$ and let $\mfP \in S_K$ be the unique prime lying over $2$. Then, for any $(\lambda, \mu) \in \Lambda_{S_K}$, there exists $(\lambda^\prime, \mu^\prime) \in \Lambda_{S_K}$ with $\lambda^\prime \in \mcO_K$, $\mu^\prime \in \mcO_K^\ast$ and $\sigma \in \mfS_3$ such that $(\lambda^\prime, \mu^\prime)= (\lambda, \mu)^\sigma$ and $m_{\lambda, \mu}= m_{\lambda^\prime, \mu^\prime}$.
\end{lem}

\begin{proof}
	If $v_\mfP(\lambda)= v_\mfP(\mu)=0$, or $v_\mfP(\lambda)>0$ (in this case $v_\mfP(\mu)=0$),
	then take  $\lambda^\prime=\lambda,\ \mu^\prime=\mu$ and $\sigma (\lambda)= \lambda$.
	If $v_\mfP(\mu)>0$, then $v_\mfP(\lambda)=0$ and take $\lambda^\prime=\mu,\ \mu^\prime=\lambda$ and $\sigma (\lambda)=1-\lambda$. If $v_\mfP(\lambda)<0$ then $v_\mfP(\mu)= v_\mfP(\lambda)= -m_{\lambda, \mu}<0$ and take $\lambda^\prime= \frac{1}{\lambda}$, $\mu^\prime=1-\frac{1}{\lambda}$ and  $\sigma (\lambda)= \frac{1}{\lambda}$. In all cases, we can choose $\lambda^\prime \in \mcO_K$, $\mu^\prime \in \mcO_K^\ast$ with $m_{\lambda^\prime, \mu^\prime}=m_{\lambda, \mu}.$
\end{proof}
\begin{proof}[Proof of Proposition~\ref{loc crit2 for odd degree field over K}]
	Let $\mfP\in U_K$ be the unique prime lying above $2$.
By~\cite[Lemma 4.1]{FKS21}, we have $m_{\lambda, \mu} <2v_\mfP(2)=2$ for all $(\lambda, \mu) \in \Lambda_{S_K}$. By Lemma~\ref{integral reduction of lambda, mu}, there exists $\lambda^\prime \in \mcO_K$, $\mu^\prime \in \mcO_K^\ast$ such that
$m_{\lambda^\prime, \mu^\prime}<2$. If $m_{\lambda^\prime, \mu^\prime}=0$ then $\lambda^\prime, \mu^\prime \in \mcO_K^\ast$, contradicts~\cite[Theorem 4]{FKS21}. So, $m_{\lambda^\prime, \mu^\prime}=1$. Since $\lambda^\prime \in \mcO_K, \mu^\prime \in \mcO_K^\ast$, we get
$v_\mfP(\lambda^\prime)=1$ and hence $v_\mfP(\lambda^\prime \mu^\prime)=1=v_\mfP(2)$. By~\cite[Lemma 6.2(ii)]{FS15}, we have
	\begin{equation}
		\label{valuation of lambda mu}
		v_\mfP(\lambda \mu)\equiv v_\mfP(2) \pmod3.
	\end{equation}
    We now show that the proof of Proposition~\ref{loc crit2 for odd degree field over K} follows from Proposition~\ref{cor to main result2 for x^2=By^p+2^rz^p over K}. Now, arguing as in the proof of Lemma~\ref{lemma to loc crit1 for x^2=By^p+2^rz^p over W_K},
    every solution $(\alpha, \gamma) \in \mcO_{S_K}^\ast \times \mcO_{S_K}$  to the equation $\alpha+1=\gamma^2$
gives rise to an element $(\lambda,\mu)\in \Lambda_{S_K}$ with a relation $\alpha=-4\lambda \mu$.
Since $m_{\lambda, \mu} <2$, $|v_\mfP(\alpha)| \leq 2 + |v_\mfP(\lambda )|+ |v_\mfP(\mu)|<6=6v_\mfP(2)$. By \eqref{valuation of lambda mu}, we get $ v_\mfP(\alpha)\equiv 0 \pmod3$. Note that, here $n$ is odd, hence $K$ satisfies $(ES)$.
\end{proof}

redpajama_set_name: RedPajamaArXiv