Density of the set of numbers of the form $x^2+2y^2$

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Is there a natural number $k$ such that for all intervals (of positive numbers) of length $k$, there is at least one element of the form $x^2+2y^2,$ where $x$ and $y$ are integers?

number-theory

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asked Jul 18 at 17:57

Danut Preda

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up vote
4
down vote

favorite

Is there a natural number $k$ such that for all intervals (of positive numbers) of length $k$, there is at least one element of the form $x^2+2y^2,$ where $x$ and $y$ are integers?

number-theory

share|cite|improve this question

asked Jul 18 at 17:57

Danut Preda

312

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

Is there a natural number $k$ such that for all intervals (of positive numbers) of length $k$, there is at least one element of the form $x^2+2y^2,$ where $x$ and $y$ are integers?

number-theory

share|cite|improve this question

asked Jul 18 at 17:57

Danut Preda

312

Is there a natural number $k$ such that for all intervals (of positive numbers) of length $k$, there is at least one element of the form $x^2+2y^2,$ where $x$ and $y$ are integers?

number-theory

share|cite|improve this question

asked Jul 18 at 17:57

Danut Preda

312

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 17:57

Danut Preda

312

asked Jul 18 at 17:57

Danut Preda

312

asked Jul 18 at 17:57

Danut Preda

312

312

add a comment | 

add a comment | 

1 Answer
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up vote
3
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Let $p$ be a prime congruent to $5$ or $7$ modulo $8$. Then $p$
can only divide $x^2+2y^2$ if $pmid x$ and $pmid y$, so that then
$p^2mid(x^2+2y^2)$. Therefore if $mequiv ppmodp^2$ then
$m$ cannot be of the form $x^2+2y^2$.

Let $p_1,ldots,p_k$ be $k$ distinct primes of this form
(there are infinitely many by Dirichlet). By the Chinese remainder theorem
there is some $zinBbb N$ with $zequiv p_j-jpmodp_j^2$
for all $j$. Then $z+j$ cannot be an $x^2+ny^2$. Then none
of $z+1,ldots,z+k$ is an $x^2+ny^2$.

share|cite|improve this answer

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
  – Daniel Fischer♦
  Jul 18 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
  – Lord Shark the Unknown
  Jul 19 at 5:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. $5$ or $7$ is enough for the purpose at hand.
  – Daniel Fischer♦
  Jul 19 at 8:10
  

  
  
  
  

add a comment | 

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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

Let $p$ be a prime congruent to $5$ or $7$ modulo $8$. Then $p$
can only divide $x^2+2y^2$ if $pmid x$ and $pmid y$, so that then
$p^2mid(x^2+2y^2)$. Therefore if $mequiv ppmodp^2$ then
$m$ cannot be of the form $x^2+2y^2$.

Let $p_1,ldots,p_k$ be $k$ distinct primes of this form
(there are infinitely many by Dirichlet). By the Chinese remainder theorem
there is some $zinBbb N$ with $zequiv p_j-jpmodp_j^2$
for all $j$. Then $z+j$ cannot be an $x^2+ny^2$. Then none
of $z+1,ldots,z+k$ is an $x^2+ny^2$.

share|cite|improve this answer

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
  – Daniel Fischer♦
  Jul 18 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
  – Lord Shark the Unknown
  Jul 19 at 5:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. $5$ or $7$ is enough for the purpose at hand.
  – Daniel Fischer♦
  Jul 19 at 8:10
  

  
  
  
  

add a comment | 

up vote
3
down vote

Let $p$ be a prime congruent to $5$ or $7$ modulo $8$. Then $p$
can only divide $x^2+2y^2$ if $pmid x$ and $pmid y$, so that then
$p^2mid(x^2+2y^2)$. Therefore if $mequiv ppmodp^2$ then
$m$ cannot be of the form $x^2+2y^2$.

Let $p_1,ldots,p_k$ be $k$ distinct primes of this form
(there are infinitely many by Dirichlet). By the Chinese remainder theorem
there is some $zinBbb N$ with $zequiv p_j-jpmodp_j^2$
for all $j$. Then $z+j$ cannot be an $x^2+ny^2$. Then none
of $z+1,ldots,z+k$ is an $x^2+ny^2$.

share|cite|improve this answer

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
  – Daniel Fischer♦
  Jul 18 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
  – Lord Shark the Unknown
  Jul 19 at 5:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. $5$ or $7$ is enough for the purpose at hand.
  – Daniel Fischer♦
  Jul 19 at 8:10
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Let $p$ be a prime congruent to $5$ or $7$ modulo $8$. Then $p$
can only divide $x^2+2y^2$ if $pmid x$ and $pmid y$, so that then
$p^2mid(x^2+2y^2)$. Therefore if $mequiv ppmodp^2$ then
$m$ cannot be of the form $x^2+2y^2$.

Let $p_1,ldots,p_k$ be $k$ distinct primes of this form
(there are infinitely many by Dirichlet). By the Chinese remainder theorem
there is some $zinBbb N$ with $zequiv p_j-jpmodp_j^2$
for all $j$. Then $z+j$ cannot be an $x^2+ny^2$. Then none
of $z+1,ldots,z+k$ is an $x^2+ny^2$.

share|cite|improve this answer

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

Let $p$ be a prime congruent to $5$ or $7$ modulo $8$. Then $p$
can only divide $x^2+2y^2$ if $pmid x$ and $pmid y$, so that then
$p^2mid(x^2+2y^2)$. Therefore if $mequiv ppmodp^2$ then
$m$ cannot be of the form $x^2+2y^2$.

Let $p_1,ldots,p_k$ be $k$ distinct primes of this form
(there are infinitely many by Dirichlet). By the Chinese remainder theorem
there is some $zinBbb N$ with $zequiv p_j-jpmodp_j^2$
for all $j$. Then $z+j$ cannot be an $x^2+ny^2$. Then none
of $z+1,ldots,z+k$ is an $x^2+ny^2$.

share|cite|improve this answer

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 18:24

edited Jul 18 at 18:24

edited Jul 18 at 18:24

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

answered Jul 18 at 18:21

Lord Shark the Unknown

85.5k951112

85.5k951112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
  – Daniel Fischer♦
  Jul 18 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
  – Lord Shark the Unknown
  Jul 19 at 5:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. $5$ or $7$ is enough for the purpose at hand.
  – Daniel Fischer♦
  Jul 19 at 8:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
  – Daniel Fischer♦
  Jul 18 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
  – Lord Shark the Unknown
  Jul 19 at 5:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. $5$ or $7$ is enough for the purpose at hand.
  – Daniel Fischer♦
  Jul 19 at 8:10
  

  
  
  
  

1

1

Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
– Daniel Fischer♦
Jul 18 at 18:27

Just for the record, since Dirichlet's theorem is rather advanced: One can elementarily show that there are infinitely many primes $equiv 5,7 pmod8$. If $p_1,dotsc,p_k$ are odd primes, then $(p_1cdot dotsc cdot p_k)^2 + 4$ has a prime factor $equiv 5 pmod8$, and that is different from the $p_kappa$. And for $p equiv 7 pmod8$ one considers $(p_1cdot dotsc cdot p_k)^2 - 2$.
– Daniel Fischer♦
Jul 18 at 18:27

@DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
– Lord Shark the Unknown
Jul 19 at 5:06

@DanielFischer One can be even more elementary: consider the prime factorisation of $8p_1cdots p_k-1$. It has a prime factor $qequiv 5$ or $7pmod 8$ distinct from the $p_j$.
– Lord Shark the Unknown
Jul 19 at 5:06

Good point. $5$ or $7$ is enough for the purpose at hand.
– Daniel Fischer♦
Jul 19 at 8:10

Good point. $5$ or $7$ is enough for the purpose at hand.
– Daniel Fischer♦
Jul 19 at 8:10

add a comment | 

 

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