Solution of this Diophantine Equation












4












$begingroup$



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
    6 hours ago










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    6 hours ago
















4












$begingroup$



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
    6 hours ago










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    6 hours ago














4












4








4


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 6 hours ago









MrAPMrAP

1,25121432




1,25121432








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    6 hours ago










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    6 hours ago














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    6 hours ago










  • $begingroup$
    See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
    $endgroup$
    – Sil
    6 hours ago








1




1




$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
6 hours ago




$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
6 hours ago












$begingroup$
See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
$endgroup$
– Sil
6 hours ago




$begingroup$
See A Diophantine equation solved when N is not a square? and Find all integer solutions to $x^2-2y^2=1$
$endgroup$
– Sil
6 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

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











$endgroup$









  • 1




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    5 hours ago










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    5 hours ago



















2












$begingroup$

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

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

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
    5 hours ago






  • 1




    $begingroup$
    "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
    5 hours ago






  • 2




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    5 hours ago














Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166201%2fsolution-of-this-diophantine-equation%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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











$endgroup$









  • 1




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    5 hours ago










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    5 hours ago
















6












$begingroup$

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











$endgroup$









  • 1




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    5 hours ago










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    5 hours ago














6












6








6





$begingroup$

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 5 hours ago

























answered 6 hours ago









Dr. MathvaDr. Mathva

3,010527




3,010527








  • 1




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    5 hours ago










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    5 hours ago














  • 1




    $begingroup$
    +1 for the correct solution but you did not answer my question. What is wrong with my method?
    $endgroup$
    – MrAP
    5 hours ago










  • $begingroup$
    Can you put that in your answer.
    $endgroup$
    – MrAP
    5 hours ago








1




1




$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
5 hours ago




$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
5 hours ago












$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
5 hours ago




$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
5 hours ago











2












$begingroup$

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

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

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
    5 hours ago






  • 1




    $begingroup$
    "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
    5 hours ago






  • 2




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    5 hours ago


















2












$begingroup$

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

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

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
    5 hours ago






  • 1




    $begingroup$
    "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
    5 hours ago






  • 2




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    5 hours ago
















2












2








2





$begingroup$

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

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

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, $(3+sqrt{2} *2)(3-sqrt{2}*2)=1$

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










answered 6 hours ago









MannMann

2,0851724




2,0851724








  • 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
    5 hours ago






  • 1




    $begingroup$
    "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
    5 hours ago






  • 2




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    5 hours ago
















  • 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
    5 hours ago






  • 1




    $begingroup$
    "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
    5 hours ago






  • 2




    $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    5 hours ago










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
5 hours ago




$begingroup$
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
5 hours ago




1




1




$begingroup$
"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
5 hours ago




$begingroup$
"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
5 hours ago




2




2




$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
5 hours ago






$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
5 hours ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166201%2fsolution-of-this-diophantine-equation%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Futebolista

Image
Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
Read more

Jornalista

Image
Uma repórter de televisão segurando um microfone na frente de um cinegrafista. Fotojornalistas no Campeonato Mundial de Atletismo 2013 Uma repórter entrevistando Boris Johnson, Prefeito de Londres. Jornalista é o profissional formado em Jornalismo. É a pessoa responsável pela apuração, investigação e apresentação de notícias, reportagens, entrevistas ou distribuição de notícias ou outra informação de interesse coletivo. O trabalho do jornalista é chamado jornalismo. Um jornalista pode trabalhar com questões gerais ou especializar-se em determinadas áreas. No entanto, a maioria dos jornalistas tendem a se especializar, e cooperando com outros jornalistas, produzir publicações que abrangem muitos tópicos. [ 1 ] Por exemplo, um jornalista esportivo cobre notícias dentro do mundo dos esportes, mas este jornalista pode ser uma parte de um jornal que cobre diversos temas. O exercício do Jornalismo é privativo de jornalista. Entre as áreas em que o jornalista trabalha estão o...
Read more

F# list compare

Image
0 I'm asking for help to solve a programming exercice in F#. I have to create a list where books and movies are listed in. All the books that have the same name as the movies sbould be listed in another list. I link what I've done until now and what the inputs are and what results i should get. Thank you in advance. type Movie = { movieName: string duration: Nat fileSize: Nat } type Book = { bookName: string pages: Nat } type Activity = | Watch of Movie | Read of Book let booksWithMovie(activities: Activity list): Book list = match activities with | -> | [Read book] -> match activities with | x :: xs -> match x with | Watch same -> if (same.bookName = same.movieName) then [same] else booksWithMovie(xs) Here are the inputs: Set.ofL...
Read more