Solution of this Diophantine Equation The Next CEO of Stack OverflowA Diophantine equation solved when N is not a square?Find all integer solutions to $x^2-2y^2=1$Methods for quartic diophantine equationsolving this equation using prime numbersHas anyone solved this general Diophantine Equation?Generalization of a Diophantine Equation ProblemConjecture about linear diophantine equationsDiophantine equations for polynomialsFactor proofs problemWhy $n=2$ should be a prime number however it is even integer and is not similar with other primes?Does this qualify as a prime-representing Diophantine equation?What are the properties of abundancy numbers?
Is it okay to store user locations?
How to be diplomatic in refusing to write code that breaches the privacy of our users
Was a professor correct to chastise me for writing "Prof. X" rather than "Professor X"?
How can I open an app using Terminal?
What does "Its cash flow is deeply negative" mean?
Why were Madagascar and New Zealand discovered so late?
Apart from "berlinern", do any other German dialects have a corresponding verb?
Horror movie/show or scene where a horse creature opens its mouth really wide and devours a man in a stables
Text adventure game code
Science fiction (dystopian) short story set after WWIII
Anatomically Correct Mesopelagic Aves
How to count occurrences of text in a file?
% symbol leads to superlong (forever?) compilations
Why do professional authors make "consistency" mistakes? And how to avoid them?
What makes a siege story/plot interesting?
Grabbing quick drinks
How to make a software documentation "officially" citable?
MAZDA 3 2006 (UK) - poor acceleration then takes off at 3250 revs
How do I go from 300 unfinished/half written blog posts, to published posts?
Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis
Error when running sfdx update to 7.1.3 then sfdx push errors
How do I construct this japanese bowl?
Opposite of a diet
Why did we only see the N-1 starfighters in one film?
Solution of this Diophantine Equation
The Next CEO of Stack OverflowA Diophantine equation solved when N is not a square?Find all integer solutions to $x^2-2y^2=1$Methods for quartic diophantine equationsolving this equation using prime numbersHas anyone solved this general Diophantine Equation?Generalization of a Diophantine Equation ProblemConjecture about linear diophantine equationsDiophantine equations for polynomialsFactor proofs problemWhy $n=2$ should be a prime number however it is even integer and is not similar with other primes?Does this qualify as a prime-representing Diophantine equation?What are the properties of abundancy numbers?
$begingroup$
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$implies (x+sqrt2y)(x-sqrt2y)=1$
$implies (x+sqrt2y)=1$ and $(x-sqrt2y)=1$
$implies x=1$ 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
asked 8 hours ago
MrAPMrAP
1,25621432
$endgroup$
add a comment |
$begingroup$
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$implies (x+sqrt2y)(x-sqrt2y)=1$
$implies (x+sqrt2y)=1$ and $(x-sqrt2y)=1$
$implies x=1$ 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
asked 8 hours ago
MrAPMrAP
1,25621432
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 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
8 hours ago
add a comment |
$begingroup$
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$implies (x+sqrt2y)(x-sqrt2y)=1$
$implies (x+sqrt2y)=1$ and $(x-sqrt2y)=1$
$implies x=1$ 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
asked 8 hours ago
MrAPMrAP
1,25621432
$endgroup$
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$implies (x+sqrt2y)(x-sqrt2y)=1$
$implies (x+sqrt2y)=1$ and $(x-sqrt2y)=1$
$implies x=1$ 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
asked 8 hours ago
MrAPMrAP
1,25621432
asked 8 hours ago
MrAPMrAP
1,25621432
asked 8 hours ago
MrAPMrAP
1,25621432
asked 8 hours ago
MrAPMrAP
1,25621432
asked 8 hours ago
MrAPMrAP
1,25621432
1,25621432
1
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 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
8 hours ago
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 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
8 hours ago
1
1
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 hours ago
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 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
8 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
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What about
beginalign*&x^2-2y^2=1tag1\iff & x^2-1=(x+1)(x-1)=2y^2endalign*
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, $colorredy=2$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $colorblue(3, 2)$.
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 $$a·b=0implies a=0;text or ;b=0$$
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
$endgroup$
1
$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
add a comment |
$begingroup$
The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+sqrt2 *2)(3-sqrt2*2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.
answered 8 hours ago
MannMann
2,0851725
$endgroup$
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
8 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
8 hours ago
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
add a comment |
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
What about
beginalign*&x^2-2y^2=1tag1\iff & x^2-1=(x+1)(x-1)=2y^2endalign*
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, $colorredy=2$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $colorblue(3, 2)$.
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 $$a·b=0implies a=0;text or ;b=0$$
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
$endgroup$
1
$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
add a comment |
$begingroup$
What about
beginalign*&x^2-2y^2=1tag1\iff & x^2-1=(x+1)(x-1)=2y^2endalign*
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, $colorredy=2$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $colorblue(3, 2)$.
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 $$a·b=0implies a=0;text or ;b=0$$
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
$endgroup$
1
$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
add a comment |
$begingroup$
What about
beginalign*&x^2-2y^2=1tag1\iff & x^2-1=(x+1)(x-1)=2y^2endalign*
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, $colorredy=2$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $colorblue(3, 2)$.
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 $$a·b=0implies a=0;text or ;b=0$$
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
$endgroup$
What about
beginalign*&x^2-2y^2=1tag1\iff & x^2-1=(x+1)(x-1)=2y^2endalign*
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, $colorredy=2$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $colorblue(3, 2)$.
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 $$a·b=0implies a=0;text or ;b=0$$
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
edited 7 hours ago
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
answered 8 hours ago
Dr. MathvaDr. Mathva
3,010527
answered 8 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
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
add a comment |
1
$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 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
8 hours ago
$begingroup$
+1 for the correct solution but you did not answer my question. What is wrong with my method?
$endgroup$
– MrAP
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
7 hours ago
add a comment |
$begingroup$
The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+sqrt2 *2)(3-sqrt2*2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.
answered 8 hours ago
MannMann
2,0851725
$endgroup$
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
8 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
8 hours ago
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
add a comment |
$begingroup$
The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+sqrt2 *2)(3-sqrt2*2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.
answered 8 hours ago
MannMann
2,0851725
$endgroup$
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
8 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
8 hours ago
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
add a comment |
$begingroup$
The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+sqrt2 *2)(3-sqrt2*2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.
answered 8 hours ago
MannMann
2,0851725
$endgroup$
The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+sqrt2 *2)(3-sqrt2*2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.
answered 8 hours ago
MannMann
2,0851725
answered 8 hours ago
MannMann
2,0851725
answered 8 hours ago
MannMann
2,0851725
answered 8 hours ago
MannMann
2,0851725
2,0851725
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
8 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
8 hours ago
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
add a comment |
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
8 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
8 hours ago
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 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
8 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
8 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
8 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
8 hours ago
2
2
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
$begingroup$
Please read the question. I answered exactly what has been asked.
$endgroup$
– Mann
8 hours ago
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
8 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
8 hours ago