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?

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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?










4












$begingroup$



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




My attempt:



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



$implies (x+sqrt2y)(x-sqrt2y)=1$



$implies (x+sqrt2y)=1$ 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?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    3 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
    3 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:



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



$implies (x+sqrt2y)(x-sqrt2y)=1$



$implies (x+sqrt2y)=1$ 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?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    3 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
    3 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:



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



$implies (x+sqrt2y)(x-sqrt2y)=1$



$implies (x+sqrt2y)=1$ 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?










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:



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



$implies (x+sqrt2y)(x-sqrt2y)=1$



$implies (x+sqrt2y)=1$ 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









MrAPMrAP

1,25121432




1,25121432







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    3 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
    3 hours ago












  • 1




    $begingroup$
    en.wikipedia.org/wiki/Pell%27s_equation
    $endgroup$
    – Sil
    3 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
    3 hours ago







1




1




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




$begingroup$
en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– Sil
3 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
3 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
3 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

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 $$a·b=0implies a=0;text or ;b=0$$






share|cite|improve this answer











$endgroup$












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










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


















1












$begingroup$

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.






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
    2 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
    2 hours ago










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












Your Answer





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Post as a guest















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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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 $$a·b=0implies a=0;text or ;b=0$$






share|cite|improve this answer











$endgroup$












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










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















6












$begingroup$

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 $$a·b=0implies a=0;text or ;b=0$$






share|cite|improve this answer











$endgroup$












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










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













6












6








6





$begingroup$

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 $$a·b=0implies a=0;text or ;b=0$$






share|cite|improve this answer











$endgroup$



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 $$a·b=0implies a=0;text or ;b=0$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









Dr. MathvaDr. Mathva

2,992527




2,992527











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










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
















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










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















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




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












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




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











1












$begingroup$

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.






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
    2 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
    2 hours ago










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
















1












$begingroup$

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.






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
    2 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
    2 hours ago










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














1












1








1





$begingroup$

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.






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, $(3+sqrt2 *2)(3-sqrt2*2)=1$

Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+bsqrt2$ in that step.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









MannMann

2,0751724




2,0751724







  • 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
    2 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
    2 hours ago










  • $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    2 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
    2 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
    2 hours ago










  • $begingroup$
    Please read the question. I answered exactly what has been asked.
    $endgroup$
    – Mann
    2 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
2 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
2 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
2 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
2 hours ago












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





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


















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ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result