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Infinite number of primes congruent to 5 mod 6

Is there an infinite number of primes constructed as in Euclid's proof?Relevance of prime being divisble by $4k+1$ in proof that 'There are infinitely many primes of the shape $4k+3$'Prove that there exists infinitely many primes of Digital root $2,5$ or $8$There are infinitely many primes congruent to 9 mod 10Lower bound related to Goldbach conjectureproof that there are infinitely many primes congruent to $1$ mod $3$Proof of the infinitude of primes by contradiction using polynomialsinfinitude of primes that are $11 bmod 12$Proof of an infinite number of primes of the form 8k+3.I don't understand the following proof involving infinitely many primes.

0

$begingroup$

1. In the proof that there are an infinite number of primes equivalent to 5 mod 6, we suppose by way of contradiction that $5, p_1, . . . , p_r$ is the complete list, and we consider the integer $m=6(p_1 ·p_2···p_r)+5$.

b. Show that at least one prime factor $q$ of $m$ is congruent to $5 bmod 6$.

I don't understand why $p_j$ cannot be congruent to $1 bmod 6$.

number-theory

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edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

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0

$begingroup$

1. In the proof that there are an infinite number of primes equivalent to 5 mod 6, we suppose by way of contradiction that $5, p_1, . . . , p_r$ is the complete list, and we consider the integer $m=6(p_1 ·p_2···p_r)+5$.

b. Show that at least one prime factor $q$ of $m$ is congruent to $5 bmod 6$.

I don't understand why $p_j$ cannot be congruent to $1 bmod 6$.

number-theory

share|cite|improve this question

edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

1. In the proof that there are an infinite number of primes equivalent to 5 mod 6, we suppose by way of contradiction that $5, p_1, . . . , p_r$ is the complete list, and we consider the integer $m=6(p_1 ·p_2···p_r)+5$.

b. Show that at least one prime factor $q$ of $m$ is congruent to $5 bmod 6$.

I don't understand why $p_j$ cannot be congruent to $1 bmod 6$.

number-theory

share|cite|improve this question

edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 17 hours ago

Bernard

122k741116

edited 17 hours ago

Bernard

122k741116

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

New contributor

Maitreyee Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 17 hours ago

Maitreyee JoshiMaitreyee Joshi

1

2 Answers
2

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2

$begingroup$

The product of numbers that are $1$ modulo $6$ is again $1pmod6$. If $m$ consisted of primes of this form only, then their product, $m$, would also be of this form. But in fact, $m$ isn't of this form, $$m=6(p_1cdots p_r)+5=5pmod6.$$ Hence there must be at least one prime factor of type $5pmod6$.

(To complete the proof, one should add that this prime factor, call it $q$, is not one of the $p_i$'s. Otherwise $q$ would divide $m$ and also $6p_1cdots p_r$, so must divide $5$.)

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

ChrystomathChrystomath

1,388512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  maybe this is a stupid clarification but I am unable to understand why 1 mod 6 is introduced in this proof and what do you mean by "But it isn't".
  $endgroup$
  – Maitreyee Joshi
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mai If all prime factors of $m$ are $equiv 1pmod!6$ then so too is their product $m$. But $,mequiv -1pmod!6$ so $m$ must have some prime factor $notequiv 1pmod!6$ (hence $equiv -1pmod!6,,$ by $m$ coprime to $2,3,,$ i.e. prime $p> 3$ implies $,pequiv pm1pmod!6) $
  $endgroup$
  – Bill Dubuque
  13 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

We want to show that there are infinitely many primes congruent to $5$ modulo $6$. In order to prove that, we assume for contradiction that there are only finitely many of them. The list $5, p_q, p_2, ldots, p_r$ is the complete list of all primes congruent to $5$ modulo $6$. An element $p_j$ in this list cannot be congruent to $1$ modulo $6$ because if that were the case then $p_j$ wouldn't be in this list.

share|cite|improve this answer

answered 16 hours ago

ArthurArthur

117k7116200

$endgroup$

add a comment | 

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2

$begingroup$

The product of numbers that are $1$ modulo $6$ is again $1pmod6$. If $m$ consisted of primes of this form only, then their product, $m$, would also be of this form. But in fact, $m$ isn't of this form, $$m=6(p_1cdots p_r)+5=5pmod6.$$ Hence there must be at least one prime factor of type $5pmod6$.

(To complete the proof, one should add that this prime factor, call it $q$, is not one of the $p_i$'s. Otherwise $q$ would divide $m$ and also $6p_1cdots p_r$, so must divide $5$.)

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

ChrystomathChrystomath

1,388512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  maybe this is a stupid clarification but I am unable to understand why 1 mod 6 is introduced in this proof and what do you mean by "But it isn't".
  $endgroup$
  – Maitreyee Joshi
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mai If all prime factors of $m$ are $equiv 1pmod!6$ then so too is their product $m$. But $,mequiv -1pmod!6$ so $m$ must have some prime factor $notequiv 1pmod!6$ (hence $equiv -1pmod!6,,$ by $m$ coprime to $2,3,,$ i.e. prime $p> 3$ implies $,pequiv pm1pmod!6) $
  $endgroup$
  – Bill Dubuque
  13 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The product of numbers that are $1$ modulo $6$ is again $1pmod6$. If $m$ consisted of primes of this form only, then their product, $m$, would also be of this form. But in fact, $m$ isn't of this form, $$m=6(p_1cdots p_r)+5=5pmod6.$$ Hence there must be at least one prime factor of type $5pmod6$.

(To complete the proof, one should add that this prime factor, call it $q$, is not one of the $p_i$'s. Otherwise $q$ would divide $m$ and also $6p_1cdots p_r$, so must divide $5$.)

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

ChrystomathChrystomath

1,388512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  maybe this is a stupid clarification but I am unable to understand why 1 mod 6 is introduced in this proof and what do you mean by "But it isn't".
  $endgroup$
  – Maitreyee Joshi
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mai If all prime factors of $m$ are $equiv 1pmod!6$ then so too is their product $m$. But $,mequiv -1pmod!6$ so $m$ must have some prime factor $notequiv 1pmod!6$ (hence $equiv -1pmod!6,,$ by $m$ coprime to $2,3,,$ i.e. prime $p> 3$ implies $,pequiv pm1pmod!6) $
  $endgroup$
  – Bill Dubuque
  13 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The product of numbers that are $1$ modulo $6$ is again $1pmod6$. If $m$ consisted of primes of this form only, then their product, $m$, would also be of this form. But in fact, $m$ isn't of this form, $$m=6(p_1cdots p_r)+5=5pmod6.$$ Hence there must be at least one prime factor of type $5pmod6$.

(To complete the proof, one should add that this prime factor, call it $q$, is not one of the $p_i$'s. Otherwise $q$ would divide $m$ and also $6p_1cdots p_r$, so must divide $5$.)

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

ChrystomathChrystomath

1,388512

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

ChrystomathChrystomath

1,388512

answered 16 hours ago

ChrystomathChrystomath

1,388512

answered 16 hours ago

ChrystomathChrystomath

1,388512

0

$begingroup$

We want to show that there are infinitely many primes congruent to $5$ modulo $6$. In order to prove that, we assume for contradiction that there are only finitely many of them. The list $5, p_q, p_2, ldots, p_r$ is the complete list of all primes congruent to $5$ modulo $6$. An element $p_j$ in this list cannot be congruent to $1$ modulo $6$ because if that were the case then $p_j$ wouldn't be in this list.

share|cite|improve this answer

answered 16 hours ago

ArthurArthur

117k7116200

$endgroup$

add a comment | 

0

$begingroup$

We want to show that there are infinitely many primes congruent to $5$ modulo $6$. In order to prove that, we assume for contradiction that there are only finitely many of them. The list $5, p_q, p_2, ldots, p_r$ is the complete list of all primes congruent to $5$ modulo $6$. An element $p_j$ in this list cannot be congruent to $1$ modulo $6$ because if that were the case then $p_j$ wouldn't be in this list.

share|cite|improve this answer

answered 16 hours ago

ArthurArthur

117k7116200

$endgroup$

add a comment | 

$begingroup$

We want to show that there are infinitely many primes congruent to $5$ modulo $6$. In order to prove that, we assume for contradiction that there are only finitely many of them. The list $5, p_q, p_2, ldots, p_r$ is the complete list of all primes congruent to $5$ modulo $6$. An element $p_j$ in this list cannot be congruent to $1$ modulo $6$ because if that were the case then $p_j$ wouldn't be in this list.

share|cite|improve this answer

answered 16 hours ago

ArthurArthur

117k7116200

$endgroup$

share|cite|improve this answer

answered 16 hours ago

ArthurArthur

117k7116200

answered 16 hours ago

ArthurArthur

117k7116200

answered 16 hours ago

ArthurArthur

117k7116200