Solution of this Diophantine Equation












6












$begingroup$



If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.




My attempt:



$x^2-2y^2=1$



$implies (x+sqrt{2}y)(x-sqrt{2}y)=1$



$implies (x+sqrt{2}y)=1$ and $(x-sqrt{2}y)=1$



$implies x=1$ and $y=0$



Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?






elementary-number-theory prime-numbers diophantine-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    Mar 28 at 18:04










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    Mar 28 at 18:06
















6












$begingroup$



If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.




My attempt:



$x^2-2y^2=1$



$implies (x+sqrt{2}y)(x-sqrt{2}y)=1$



$implies (x+sqrt{2}y)=1$ and $(x-sqrt{2}y)=1$



$implies x=1$ and $y=0$



Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?






elementary-number-theory prime-numbers diophantine-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    Mar 28 at 18:04










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    Mar 28 at 18:06














6












6








6


1



$begingroup$



If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.




My attempt:



$x^2-2y^2=1$



$implies (x+sqrt{2}y)(x-sqrt{2}y)=1$



$implies (x+sqrt{2}y)=1$ and $(x-sqrt{2}y)=1$



$implies x=1$ and $y=0$



Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?






elementary-number-theory prime-numbers diophantine-equations







share|cite|improve this question









$endgroup$





If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.




My attempt:



$x^2-2y^2=1$



$implies (x+sqrt{2}y)(x-sqrt{2}y)=1$



$implies (x+sqrt{2}y)=1$ and $(x-sqrt{2}y)=1$



$implies x=1$ and $y=0$



Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?






elementary-number-theory prime-numbers diophantine-equations




elementary-number-theory prime-numbers diophantine-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 28 at 17:55









MrAPMrAP

1,29321432




1,29321432








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    Mar 28 at 18:04










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    Mar 28 at 18:06














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    Mar 28 at 18:04










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    Mar 28 at 18:06








1




1




$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
Mar 28 at 18:04




$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
Mar 28 at 18:04












$begingroup$
See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
$endgroup$
– Sil
Mar 28 at 18:06




$begingroup$
See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
$endgroup$
– Sil
Mar 28 at 18:06










2 Answers
2






active

oldest

votes


















11












$begingroup$

What about




begin{align*}&x^2-2y^2=1tag{1}\iff & x^2-1=(x+1)(x-1)=2y^2end{align*}




Since $2mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4mid 2y^2implies 2mid y^2implies 2mid y$$ and since $y$ is prime, $color{red}{y=2}$. Can you end it now?



From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $color{blue}{(3, 2)}$.




Addendum



The problem with your method is that for $a,binmathbb R$



$$a·b=1notRightarrow a=1;text{ and };b=1$$



In fact, this only works if $$a·b=0implies a=0;text{ or };b=0$$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    Mar 28 at 18:33










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    Mar 28 at 18:35



















22












$begingroup$

The fault is that irrationals can also produce the product to $1$.

Consider $x=3$ and $y=2$ then we get, $(3+sqrt{2} cdot 2)(3-sqrt{2}cdot 2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt{2}$ in that step.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
    $endgroup$
    – B. Goddard
    Mar 28 at 18:17






  • 3




    $begingroup$
    "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
    $endgroup$
    – Mann
    Mar 28 at 18:18






  • 18




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    Mar 28 at 18:18








  • 12




    $begingroup$
    @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
    $endgroup$
    – Martin Argerami
    Mar 29 at 18:29












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

What about




begin{align*}&x^2-2y^2=1tag{1}\iff & x^2-1=(x+1)(x-1)=2y^2end{align*}




Since $2mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4mid 2y^2implies 2mid y^2implies 2mid y$$ and since $y$ is prime, $color{red}{y=2}$. Can you end it now?



From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $color{blue}{(3, 2)}$.




Addendum



The problem with your method is that for $a,binmathbb R$



$$a·b=1notRightarrow a=1;text{ and };b=1$$



In fact, this only works if $$a·b=0implies a=0;text{ or };b=0$$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    Mar 28 at 18:33










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    Mar 28 at 18:35
















11












$begingroup$

What about




begin{align*}&x^2-2y^2=1tag{1}\iff & x^2-1=(x+1)(x-1)=2y^2end{align*}




Since $2mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4mid 2y^2implies 2mid y^2implies 2mid y$$ and since $y$ is prime, $color{red}{y=2}$. Can you end it now?



From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $color{blue}{(3, 2)}$.




Addendum



The problem with your method is that for $a,binmathbb R$



$$a·b=1notRightarrow a=1;text{ and };b=1$$



In fact, this only works if $$a·b=0implies a=0;text{ or };b=0$$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    Mar 28 at 18:33










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    Mar 28 at 18:35














11












11








11





$begingroup$

What about




begin{align*}&x^2-2y^2=1tag{1}\iff & x^2-1=(x+1)(x-1)=2y^2end{align*}




Since $2mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4mid 2y^2implies 2mid y^2implies 2mid y$$ and since $y$ is prime, $color{red}{y=2}$. Can you end it now?



From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $color{blue}{(3, 2)}$.




Addendum



The problem with your method is that for $a,binmathbb R$



$$a·b=1notRightarrow a=1;text{ and };b=1$$



In fact, this only works if $$a·b=0implies a=0;text{ or };b=0$$






share|cite|improve this answer











$endgroup$



What about




begin{align*}&x^2-2y^2=1tag{1}\iff & x^2-1=(x+1)(x-1)=2y^2end{align*}




Since $2mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4mid 2y^2implies 2mid y^2implies 2mid y$$ and since $y$ is prime, $color{red}{y=2}$. Can you end it now?



From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $color{blue}{(3, 2)}$.




Addendum



The problem with your method is that for $a,binmathbb R$



$$a·b=1notRightarrow a=1;text{ and };b=1$$



In fact, this only works if $$a·b=0implies a=0;text{ or };b=0$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 18:40

























answered Mar 28 at 17:58









Dr. MathvaDr. Mathva

3,428630




3,428630








  • 4




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    Mar 28 at 18:33










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    Mar 28 at 18:35














  • 4




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    Mar 28 at 18:33










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    Mar 28 at 18:35








4




4




$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
Mar 28 at 18:33




$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
Mar 28 at 18:33












$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
Mar 28 at 18:35




$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
Mar 28 at 18:35











22












$begingroup$

The fault is that irrationals can also produce the product to $1$.

Consider $x=3$ and $y=2$ then we get, $(3+sqrt{2} cdot 2)(3-sqrt{2}cdot 2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt{2}$ in that step.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
    $endgroup$
    – B. Goddard
    Mar 28 at 18:17






  • 3




    $begingroup$
    "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
    $endgroup$
    – Mann
    Mar 28 at 18:18






  • 18




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    Mar 28 at 18:18








  • 12




    $begingroup$
    @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
    $endgroup$
    – Martin Argerami
    Mar 29 at 18:29
















22












$begingroup$

The fault is that irrationals can also produce the product to $1$.

Consider $x=3$ and $y=2$ then we get, $(3+sqrt{2} cdot 2)(3-sqrt{2}cdot 2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt{2}$ in that step.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
    $endgroup$
    – B. Goddard
    Mar 28 at 18:17






  • 3




    $begingroup$
    "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
    $endgroup$
    – Mann
    Mar 28 at 18:18






  • 18




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    Mar 28 at 18:18








  • 12




    $begingroup$
    @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
    $endgroup$
    – Martin Argerami
    Mar 29 at 18:29














22












22








22





$begingroup$

The fault is that irrationals can also produce the product to $1$.

Consider $x=3$ and $y=2$ then we get, $(3+sqrt{2} cdot 2)(3-sqrt{2}cdot 2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt{2}$ in that step.






share|cite|improve this answer











$endgroup$



The fault is that irrationals can also produce the product to $1$.

Consider $x=3$ and $y=2$ then we get, $(3+sqrt{2} cdot 2)(3-sqrt{2}cdot 2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt{2}$ in that step.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 29 at 21:59









Xander Henderson

15.1k103556




15.1k103556










answered Mar 28 at 18:06









MannMann

2,3501727




2,3501727








  • 1




    $begingroup$
    This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
    $endgroup$
    – B. Goddard
    Mar 28 at 18:17






  • 3




    $begingroup$
    "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
    $endgroup$
    – Mann
    Mar 28 at 18:18






  • 18




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    Mar 28 at 18:18








  • 12




    $begingroup$
    @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
    $endgroup$
    – Martin Argerami
    Mar 29 at 18:29














  • 1




    $begingroup$
    This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
    $endgroup$
    – B. Goddard
    Mar 28 at 18:17






  • 3




    $begingroup$
    "Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
    $endgroup$
    – Mann
    Mar 28 at 18:18






  • 18




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    Mar 28 at 18:18








  • 12




    $begingroup$
    @B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
    $endgroup$
    – Martin Argerami
    Mar 29 at 18:29








1




1




$begingroup$
This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
$endgroup$
– B. Goddard
Mar 28 at 18:17




$begingroup$
This is not an answer to the question. If it's a comment on the other answer, then it goes under the comments section there.
$endgroup$
– B. Goddard
Mar 28 at 18:17




3




3




$begingroup$
"Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
$endgroup$
– Mann
Mar 28 at 18:18




$begingroup$
"Clearly x and y are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?"
$endgroup$
– Mann
Mar 28 at 18:18




18




18




$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
Mar 28 at 18:18






$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
Mar 28 at 18:18






12




12




$begingroup$
@B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
$endgroup$
– Martin Argerami
Mar 29 at 18:29




$begingroup$
@B.Goddard: this answer addresses precisely what the OP asked. Why would you think it should be a comment and not an answer?
$endgroup$
– Martin Argerami
Mar 29 at 18:29


















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