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Solution of this Diophantine Equation

3

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?

elementary-number-theory prime-numbers diophantine-equations

share|cite|improve this question

asked 2 hours ago

MrAPMrAP

1,24621432

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Pell%27s_equation
  $endgroup$
  – Sil
  2 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
  2 hours ago
  

  
  
  
  

add a comment | 

3

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?

elementary-number-theory prime-numbers diophantine-equations

share|cite|improve this question

asked 2 hours ago

MrAPMrAP

1,24621432

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Pell%27s_equation
  $endgroup$
  – Sil
  2 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
  2 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

share|cite|improve this question

asked 2 hours ago

MrAPMrAP

1,24621432

$endgroup$

share|cite|improve this question

asked 2 hours ago

MrAPMrAP

1,24621432

share|cite|improve this question

asked 2 hours ago

MrAPMrAP

1,24621432

asked 2 hours ago

MrAPMrAP

1,24621432

asked 2 hours ago

MrAPMrAP

1,24621432

2 Answers
2

active

oldest

votes

5

$begingroup$

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

edited 1 hour ago

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem with your method is the assumption that $a·b=1$ with $a,b$ irrational. This does not implie that $a=1, b=1$... In fact this only woks for $acdot b=0$ where you either have $a=0$ or $b=0$
  $endgroup$
  – Dr. Mathva
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you put that in your answer.
  $endgroup$
  – MrAP
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course! Give me a minute
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

MannMann

2,0751724

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment | 

Your Answer

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5

$begingroup$

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

edited 1 hour ago

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem with your method is the assumption that $a·b=1$ with $a,b$ irrational. This does not implie that $a=1, b=1$... In fact this only woks for $acdot b=0$ where you either have $a=0$ or $b=0$
  $endgroup$
  – Dr. Mathva
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you put that in your answer.
  $endgroup$
  – MrAP
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course! Give me a minute
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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

edited 1 hour ago

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem with your method is the assumption that $a·b=1$ with $a,b$ irrational. This does not implie that $a=1, b=1$... In fact this only woks for $acdot b=0$ where you either have $a=0$ or $b=0$
  $endgroup$
  – Dr. Mathva
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you put that in your answer.
  $endgroup$
  – MrAP
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course! Give me a minute
  $endgroup$
  – Dr. Mathva
  1 hour 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 $$a·b=0implies a=0;text or ;b=0$$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

answered 2 hours ago

Dr. MathvaDr. Mathva

2,982527

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

answered 2 hours ago

MannMann

2,0751724

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

MannMann

2,0751724

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

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.

share|cite|improve this answer

answered 2 hours ago

MannMann

2,0751724

$endgroup$

share|cite|improve this answer

answered 2 hours ago

MannMann

2,0751724

answered 2 hours ago

MannMann

2,0751724

answered 2 hours ago

MannMann

2,0751724