If $p=x^2-6y^2$ is a prime then $pequiv 1 , 19 mod24$

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Let $p > 3$ be a prime for which there exists $x, y in mathbbZ$ such that $p = x^2 - 6y^2$.

Show that $p equiv 1 or 19 mod24$.

---

The fact that $p = x^2 - 6y^2$ implies that $left(frac6pright) = 1$ (Legendre symbol) and therefore $p equiv 1, 5, 19, 23 bmod24$, but I don't know how to continue.

elementary-number-theory prime-numbers modular-arithmetic

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edited Jul 16 at 15:10

Robert Soupe

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asked Jul 14 at 14:06

ned grekerzberg

436218

add a comment | 

up vote
2
down vote

favorite

Let $p > 3$ be a prime for which there exists $x, y in mathbbZ$ such that $p = x^2 - 6y^2$.

Show that $p equiv 1 or 19 mod24$.

---

The fact that $p = x^2 - 6y^2$ implies that $left(frac6pright) = 1$ (Legendre symbol) and therefore $p equiv 1, 5, 19, 23 bmod24$, but I don't know how to continue.

elementary-number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

asked Jul 14 at 14:06

ned grekerzberg

436218

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $p > 3$ be a prime for which there exists $x, y in mathbbZ$ such that $p = x^2 - 6y^2$.

Show that $p equiv 1 or 19 mod24$.

---

The fact that $p = x^2 - 6y^2$ implies that $left(frac6pright) = 1$ (Legendre symbol) and therefore $p equiv 1, 5, 19, 23 bmod24$, but I don't know how to continue.

elementary-number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

asked Jul 14 at 14:06

ned grekerzberg

436218

Let $p > 3$ be a prime for which there exists $x, y in mathbbZ$ such that $p = x^2 - 6y^2$.

Show that $p equiv 1 or 19 mod24$.

---

The fact that $p = x^2 - 6y^2$ implies that $left(frac6pright) = 1$ (Legendre symbol) and therefore $p equiv 1, 5, 19, 23 bmod24$, but I don't know how to continue.

elementary-number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

asked Jul 14 at 14:06

ned grekerzberg

436218

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

edited Jul 16 at 15:10

Robert Soupe

10.1k21947

10.1k21947

asked Jul 14 at 14:06

ned grekerzberg

436218

asked Jul 14 at 14:06

ned grekerzberg

436218

asked Jul 14 at 14:06

ned grekerzberg

436218

436218

add a comment | 

add a comment | 

1 Answer
1

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votes

up vote
3
down vote



accepted

If you know that $p equiv 1,5,19,23 pmod24$ are your only options you can exclude 5 and 23 by looking at the equation modulo 3. Since $3 nmid x$ you have $x equiv pm 1 pmod 3$ and thus $p equiv x^2 equiv 1 pmod 3$.

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answered Jul 14 at 14:14

L. Gitin

584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
  – ned grekerzberg
  Jul 14 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
  – L. Gitin
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
  – ned grekerzberg
  Jul 14 at 14:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i hope that's more clear now
  – ned grekerzberg
  Jul 14 at 14:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
  – Keith Backman
  Jul 14 at 17:07
  
  

  
  
  
  

 | 
show 3 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

If you know that $p equiv 1,5,19,23 pmod24$ are your only options you can exclude 5 and 23 by looking at the equation modulo 3. Since $3 nmid x$ you have $x equiv pm 1 pmod 3$ and thus $p equiv x^2 equiv 1 pmod 3$.

share|cite|improve this answer

answered Jul 14 at 14:14

L. Gitin

584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
  – ned grekerzberg
  Jul 14 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
  – L. Gitin
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
  – ned grekerzberg
  Jul 14 at 14:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i hope that's more clear now
  – ned grekerzberg
  Jul 14 at 14:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
  – Keith Backman
  Jul 14 at 17:07
  
  

  
  
  
  

 | 
show 3 more comments

up vote
3
down vote



accepted

If you know that $p equiv 1,5,19,23 pmod24$ are your only options you can exclude 5 and 23 by looking at the equation modulo 3. Since $3 nmid x$ you have $x equiv pm 1 pmod 3$ and thus $p equiv x^2 equiv 1 pmod 3$.

share|cite|improve this answer

answered Jul 14 at 14:14

L. Gitin

584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
  – ned grekerzberg
  Jul 14 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
  – L. Gitin
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
  – ned grekerzberg
  Jul 14 at 14:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i hope that's more clear now
  – ned grekerzberg
  Jul 14 at 14:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
  – Keith Backman
  Jul 14 at 17:07
  
  

  
  
  
  

 | 
show 3 more comments

up vote
3
down vote



accepted

up vote
3
down vote



accepted

If you know that $p equiv 1,5,19,23 pmod24$ are your only options you can exclude 5 and 23 by looking at the equation modulo 3. Since $3 nmid x$ you have $x equiv pm 1 pmod 3$ and thus $p equiv x^2 equiv 1 pmod 3$.

share|cite|improve this answer

answered Jul 14 at 14:14

L. Gitin

584

If you know that $p equiv 1,5,19,23 pmod24$ are your only options you can exclude 5 and 23 by looking at the equation modulo 3. Since $3 nmid x$ you have $x equiv pm 1 pmod 3$ and thus $p equiv x^2 equiv 1 pmod 3$.

share|cite|improve this answer

answered Jul 14 at 14:14

L. Gitin

584

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 14:14

L. Gitin

584

answered Jul 14 at 14:14

L. Gitin

584

answered Jul 14 at 14:14

L. Gitin

584

584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
  – ned grekerzberg
  Jul 14 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
  – L. Gitin
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
  – ned grekerzberg
  Jul 14 at 14:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i hope that's more clear now
  – ned grekerzberg
  Jul 14 at 14:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
  – Keith Backman
  Jul 14 at 17:07
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
  – ned grekerzberg
  Jul 14 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
  – L. Gitin
  Jul 14 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
  – ned grekerzberg
  Jul 14 at 14:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i hope that's more clear now
  – ned grekerzberg
  Jul 14 at 14:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
  – Keith Backman
  Jul 14 at 17:07
  
  

  
  
  
  

first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
– ned grekerzberg
Jul 14 at 14:23

first, welcome to the site and second, thank you very much (obviously i didn't thought about the question through) :-)
– ned grekerzberg
Jul 14 at 14:23

To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
– L. Gitin
Jul 14 at 14:25

To be honest, I don't see why $p equiv 1,5,19,23 pmod24$ are the only options though.
– L. Gitin
Jul 14 at 14:25

1

1

$p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
– ned grekerzberg
Jul 14 at 14:29

$p=x^2-6y^2$ therefor $x^2 equiv 6y^2 mod(p)$ therefor $( frac6y^2p) = ( frac6p) ( fracy^2p) = ( frac6p) = 1$ (i'm refering to Legendre symbol) and it's known that $( frac6p) = 1 iff p equiv 1,5,19,23 mod24$
– ned grekerzberg
Jul 14 at 14:29

i hope that's more clear now
– ned grekerzberg
Jul 14 at 14:30

i hope that's more clear now
– ned grekerzberg
Jul 14 at 14:30

1

1

Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
– Keith Backman
Jul 14 at 17:07

Just to flesh out the "it's known that" part of the reasoning: $left(frac2pright)=pm1$, so $left(frac4pright)=left(frac2pright)left(frac2pright)=1$, so $left(frac24pright)=left(frac4pright)left(frac6pright)=1$. The numbers whose squares are $1 mod24$ are $pm1,pm5$.
– Keith Backman
Jul 14 at 17:07

 | 
show 3 more comments

 

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