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The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$. (UPDATE: As the user Gerry Myerson mentioned last night, $x_{n}$ and $y_{n}$ are the natural numbers defined by $x_{n}+y_{n}\sqrt{3} = (2+\sqrt{3})^{n}$; if after reading the previous remark it is not apparent how it is that the equality allows one to define both $x_{n}$ and $y_{n}$, please take a look at any text on number theory that discusses the Pell equation.)

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armored?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$. (UPDATE: As the user Gerry Myerson mentioned last night, $x_{n}$ and $y_{n}$ are the natural numbers defined by $x_{n}+y_{n}\sqrt{3} = (2+\sqrt{3})^{n}$; if after reading the previous remark it is not apparent how it is that the equality allows one to define both $x_{n}$ and $y_{n}$, please take a look at any text on number theory that discusses the Pell equation.)

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armored?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$. (UPDATE: As the user Gerry Myerson mentioned last night, $x_{n}$ and $y_{n}$ are the natural numbers defined by $x_{n}+y_{n}\sqrt{3} = (2+\sqrt{3})^{n}$; if after reading the previous remark it is not apparent how that equality allows one to define both $x_{n}$ and $y_{n}$, please take a look at any text on number theory that discusses the Pell equation.)

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armoured?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$. (UPDATE: As the user Gerry Myerson mentioned last night, $x_{n}$ and $y_{n}$ are the natural numbers defined by $x_{n}+y_{n}\sqrt{3} = (2+\sqrt{3})^{n}$; if after reading the previous remark it is not apparent how it is that the equality allows one to define both $x_{n}$ and $y_{n}$, please take a look at any text on number theory that discusses the Pell equation.)

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armored?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$.

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armoured?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$. (UPDATE: As the user Gerry Myerson mentioned last night, $x_{n}$ and $y_{n}$ are the natural numbers defined by $x_{n}+y_{n}\sqrt{3} = (2+\sqrt{3})^{n}$; if after reading the previous remark it is not apparent how that equality allows one to define both $x_{n}$ and $y_{n}$, please take a look at any text on number theory that discusses the Pell equation.)

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armoured?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.