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?
$begingroup$
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
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 |
Your Answer
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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
8 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
8 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
8 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
8 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
8 hours ago
$begingroup$
Can you put that in your answer.
$endgroup$
– MrAP
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
$endgroup$
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 |
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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