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Prove that there are infinitely many primes $P_iequiv1pmod6$

The prime number matrix sieveAre there infinitely many primes of the form $p_1cdot p_2cdot…p_n+1$?Infinitely many primes of the form $6cdot k+1$ , where $k$ is an odd number?Proof that there are infinitely many primes of the form $6k+1$. Proof verificationRelevance of prime being divisble by $4k+1$ in proof that 'There are infinitely many primes of the shape $4k+3$'I conjecture that there are infinitely many linear relations $p_n + p_n + 3 = 2 p_n + 2$ in the sequence of primes!Infinitely many primes of the form $3k+2$Prove that there exists infinitely many primes of Digital root $2,5$ or $8$There are infinitely many primes congruent to 9 mod 10Proof that there are infinitely many primes (Euclid)Question on a proof that there are infinitely many primes

5

5

$begingroup$

Proving that there are infinitely many primes is fairly simple:

• Assume that there is a finite number of primes.


• Let $G$ be the set of all primes $P_1,P_2,ldots,P_n$.


• Compute $K = P_1 times P_2 times cdots times P_n + 1$.


• If $K$ is prime, then it is obviously not in $G$.


• Otherwise, none of its prime factors are in $G$.


• Conclusion: $G$ is not the set of all primes.

I thought I could use a similar method in order to prove:

• There are infinitely many primes $P_iequiv1pmod6$


• There are infinitely many primes $P_iequiv5pmod6$

But it doesn't appear to be that simple... any ideas?

number-theory prime-numbers

share|cite|improve this question

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, for simplicity, if $P_iequiv 1pmod 6$ and $P_iequiv 5pmod 6$ then $P_iequiv pm 1pmod 6$.
  $endgroup$
  – user477343
  Mar 5 '18 at 7:47
  

  
  
  
  

add a comment | 

5

5

$begingroup$

Proving that there are infinitely many primes is fairly simple:

• Assume that there is a finite number of primes.


• Let $G$ be the set of all primes $P_1,P_2,ldots,P_n$.


• Compute $K = P_1 times P_2 times cdots times P_n + 1$.


• If $K$ is prime, then it is obviously not in $G$.


• Otherwise, none of its prime factors are in $G$.


• Conclusion: $G$ is not the set of all primes.

I thought I could use a similar method in order to prove:

• There are infinitely many primes $P_iequiv1pmod6$


• There are infinitely many primes $P_iequiv5pmod6$

But it doesn't appear to be that simple... any ideas?

number-theory prime-numbers

share|cite|improve this question

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, for simplicity, if $P_iequiv 1pmod 6$ and $P_iequiv 5pmod 6$ then $P_iequiv pm 1pmod 6$.
  $endgroup$
  – user477343
  Mar 5 '18 at 7:47
  

  
  
  
  

add a comment | 

$begingroup$

Proving that there are infinitely many primes is fairly simple:

• Assume that there is a finite number of primes.


• Let $G$ be the set of all primes $P_1,P_2,ldots,P_n$.


• Compute $K = P_1 times P_2 times cdots times P_n + 1$.


• If $K$ is prime, then it is obviously not in $G$.


• Otherwise, none of its prime factors are in $G$.


• Conclusion: $G$ is not the set of all primes.

I thought I could use a similar method in order to prove:

• There are infinitely many primes $P_iequiv1pmod6$


• There are infinitely many primes $P_iequiv5pmod6$

But it doesn't appear to be that simple... any ideas?

number-theory prime-numbers

share|cite|improve this question

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

$endgroup$

share|cite|improve this question

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

share|cite|improve this question

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

edited Aug 29 '14 at 3:14

Michael Hardy

1

edited Aug 29 '14 at 3:14

Michael Hardy

1

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

asked Aug 27 '14 at 22:21

barak manosbarak manos

38k742103

4 Answers
4

active

oldest

votes

5

$begingroup$

There is an old argument (don't know the correct attribution) to prove that there are infinitely-many primes $=1 mod N$: let $f$ be the $N$-th cyclotomic polynomial. Note that $pmid f(n)$ implies that $n$ is a primitive $N$-th root of unity mod $p$, so $p=1bmod N$. Given a finite collection $p_1,ldots,p_k$ of primes $=1$ mod $N$, for sufficiently large integer $ell$, $f(ellcdot p_1ldots p_k)>1$, so has some prime factor...

share|cite|improve this answer

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

$endgroup$

add a comment | 

5

$begingroup$

The case of $5$ is easy: an integer $equiv 5 bmod 6$ must be divisible by at least one prime $equiv 5 bmod 6$.
I don't think the case of $1$ is so simple. In general, Dirichlet's theorem is not trivial.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is a simple method for $1$. See my hint.
  $endgroup$
  – Thomas Andrews
  Aug 27 '14 at 23:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those integers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are only finitely many primes $p_1, ldots, p_n equiv 5 mod 6$, consider $K = p_1 ldots p_n + 6$ (if $n$ is odd) or $p_1^2 ldots p_n + 6$ (if $n$ is even). Then $K equiv 5 mod 6$, and is not divisible by any of $p_1, ldots, p_n$.
  $endgroup$
  – Robert Israel
  Aug 28 '14 at 6:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Easier to just do $6p_1p_2dots p_n+5$, @RobertIsrael
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or rather $6p_1p_2dots p_n-1$, I suppose $6p_1dots p_n+5$ could be a power of $5$. :)
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 12:34
  

  
  
  
  

add a comment | 

4

$begingroup$

Hint: $-3$ is a square modulo a prime $p>3$ if and only if $pequiv 1pmod 3$.

So any number of the form $X^2+3$ with $X$ even and relatively prime to $3$ is only divisibly by primes $p$ of the form $6k+1$.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those numbers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's why it was a hint. Given a finite number of such primes, pick $X$ carefully, similar to Euclid's proof that there are infinitely many primes.
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:40
  

  
  
  
  

add a comment | 

0

$begingroup$

If we cross out from sequence of positive integers all numbers divisible by $2$ and all numbers divisible by $3$, then all remaining numbers will be in one of two forms:

$S1(n)=6n−1=5,11,17,...$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$.

share|cite|improve this answer

edited Mar 16 at 7:45

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This shows that $S1(n) cup S2(n)$ is infinite but does not show that $S2(n)$ is infinite.
  $endgroup$
  – DanielWainfleet
  Mar 14 at 9:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see [link](math.stackexchange.com/questions/3037819/…)
  $endgroup$
  – Boris Sklyar
  Mar 14 at 16:50
  

  
  
  
  

add a comment | 

Your Answer

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5

$begingroup$

There is an old argument (don't know the correct attribution) to prove that there are infinitely-many primes $=1 mod N$: let $f$ be the $N$-th cyclotomic polynomial. Note that $pmid f(n)$ implies that $n$ is a primitive $N$-th root of unity mod $p$, so $p=1bmod N$. Given a finite collection $p_1,ldots,p_k$ of primes $=1$ mod $N$, for sufficiently large integer $ell$, $f(ellcdot p_1ldots p_k)>1$, so has some prime factor...

share|cite|improve this answer

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

$endgroup$

add a comment | 

5

$begingroup$

There is an old argument (don't know the correct attribution) to prove that there are infinitely-many primes $=1 mod N$: let $f$ be the $N$-th cyclotomic polynomial. Note that $pmid f(n)$ implies that $n$ is a primitive $N$-th root of unity mod $p$, so $p=1bmod N$. Given a finite collection $p_1,ldots,p_k$ of primes $=1$ mod $N$, for sufficiently large integer $ell$, $f(ellcdot p_1ldots p_k)>1$, so has some prime factor...

share|cite|improve this answer

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

$endgroup$

add a comment | 

$begingroup$

There is an old argument (don't know the correct attribution) to prove that there are infinitely-many primes $=1 mod N$: let $f$ be the $N$-th cyclotomic polynomial. Note that $pmid f(n)$ implies that $n$ is a primitive $N$-th root of unity mod $p$, so $p=1bmod N$. Given a finite collection $p_1,ldots,p_k$ of primes $=1$ mod $N$, for sufficiently large integer $ell$, $f(ellcdot p_1ldots p_k)>1$, so has some prime factor...

share|cite|improve this answer

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

$endgroup$

share|cite|improve this answer

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

edited Aug 29 '14 at 3:11

Michael Hardy

1

edited Aug 29 '14 at 3:11

Michael Hardy

1

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

answered Aug 27 '14 at 22:41

paul garrettpaul garrett

32.1k362119

5

$begingroup$

The case of $5$ is easy: an integer $equiv 5 bmod 6$ must be divisible by at least one prime $equiv 5 bmod 6$.
I don't think the case of $1$ is so simple. In general, Dirichlet's theorem is not trivial.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is a simple method for $1$. See my hint.
  $endgroup$
  – Thomas Andrews
  Aug 27 '14 at 23:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those integers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are only finitely many primes $p_1, ldots, p_n equiv 5 mod 6$, consider $K = p_1 ldots p_n + 6$ (if $n$ is odd) or $p_1^2 ldots p_n + 6$ (if $n$ is even). Then $K equiv 5 mod 6$, and is not divisible by any of $p_1, ldots, p_n$.
  $endgroup$
  – Robert Israel
  Aug 28 '14 at 6:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Easier to just do $6p_1p_2dots p_n+5$, @RobertIsrael
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or rather $6p_1p_2dots p_n-1$, I suppose $6p_1dots p_n+5$ could be a power of $5$. :)
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 12:34
  

  
  
  
  

add a comment | 

5

$begingroup$

The case of $5$ is easy: an integer $equiv 5 bmod 6$ must be divisible by at least one prime $equiv 5 bmod 6$.
I don't think the case of $1$ is so simple. In general, Dirichlet's theorem is not trivial.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is a simple method for $1$. See my hint.
  $endgroup$
  – Thomas Andrews
  Aug 27 '14 at 23:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those integers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are only finitely many primes $p_1, ldots, p_n equiv 5 mod 6$, consider $K = p_1 ldots p_n + 6$ (if $n$ is odd) or $p_1^2 ldots p_n + 6$ (if $n$ is even). Then $K equiv 5 mod 6$, and is not divisible by any of $p_1, ldots, p_n$.
  $endgroup$
  – Robert Israel
  Aug 28 '14 at 6:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Easier to just do $6p_1p_2dots p_n+5$, @RobertIsrael
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or rather $6p_1p_2dots p_n-1$, I suppose $6p_1dots p_n+5$ could be a power of $5$. :)
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 12:34
  

  
  
  
  

add a comment | 

$begingroup$

The case of $5$ is easy: an integer $equiv 5 bmod 6$ must be divisible by at least one prime $equiv 5 bmod 6$.
I don't think the case of $1$ is so simple. In general, Dirichlet's theorem is not trivial.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

$endgroup$

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

edited Aug 29 '14 at 3:12

Michael Hardy

1

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

answered Aug 27 '14 at 22:36

Robert IsraelRobert Israel

328k23216469

4

$begingroup$

Hint: $-3$ is a square modulo a prime $p>3$ if and only if $pequiv 1pmod 3$.

So any number of the form $X^2+3$ with $X$ even and relatively prime to $3$ is only divisibly by primes $p$ of the form $6k+1$.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those numbers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's why it was a hint. Given a finite number of such primes, pick $X$ carefully, similar to Euclid's proof that there are infinitely many primes.
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:40
  

  
  
  
  

add a comment | 

4

$begingroup$

Hint: $-3$ is a square modulo a prime $p>3$ if and only if $pequiv 1pmod 3$.

So any number of the form $X^2+3$ with $X$ even and relatively prime to $3$ is only divisibly by primes $p$ of the form $6k+1$.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a silly question, but how do you prove that there isn't a finite amount of such primes that divide all those numbers?
  $endgroup$
  – barak manos
  Aug 28 '14 at 5:11
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's why it was a hint. Given a finite number of such primes, pick $X$ carefully, similar to Euclid's proof that there are infinitely many primes.
  $endgroup$
  – Thomas Andrews
  Aug 28 '14 at 11:40
  

  
  
  
  

add a comment | 

$begingroup$

Hint: $-3$ is a square modulo a prime $p>3$ if and only if $pequiv 1pmod 3$.

So any number of the form $X^2+3$ with $X$ even and relatively prime to $3$ is only divisibly by primes $p$ of the form $6k+1$.

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

$endgroup$

share|cite|improve this answer

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

edited Aug 29 '14 at 3:12

Michael Hardy

1

edited Aug 29 '14 at 3:12

Michael Hardy

1

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

answered Aug 27 '14 at 23:19

Thomas AndrewsThomas Andrews

130k12147298

0

$begingroup$

If we cross out from sequence of positive integers all numbers divisible by $2$ and all numbers divisible by $3$, then all remaining numbers will be in one of two forms:

$S1(n)=6n−1=5,11,17,...$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$.

share|cite|improve this answer

edited Mar 16 at 7:45

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This shows that $S1(n) cup S2(n)$ is infinite but does not show that $S2(n)$ is infinite.
  $endgroup$
  – DanielWainfleet
  Mar 14 at 9:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see [link](math.stackexchange.com/questions/3037819/…)
  $endgroup$
  – Boris Sklyar
  Mar 14 at 16:50
  

  
  
  
  

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If we cross out from sequence of positive integers all numbers divisible by $2$ and all numbers divisible by $3$, then all remaining numbers will be in one of two forms:

$S1(n)=6n−1=5,11,17,...$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$.

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edited Mar 16 at 7:45

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

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  This shows that $S1(n) cup S2(n)$ is infinite but does not show that $S2(n)$ is infinite.
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  – DanielWainfleet
  Mar 14 at 9:14
  

  
  
  
  


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  Please see [link](math.stackexchange.com/questions/3037819/…)
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  – Boris Sklyar
  Mar 14 at 16:50
  

  
  
  
  

add a comment | 

$begingroup$

If we cross out from sequence of positive integers all numbers divisible by $2$ and all numbers divisible by $3$, then all remaining numbers will be in one of two forms:

$S1(n)=6n−1=5,11,17,...$ or $S2(n)=6n+1=7,13,19,....n=1,2,3,...$ So all prime numbers also will be in one of these two forms and ratio 0f number of primes in the sequence $S1(n)$ to number of primes in the sequence $S2(n)$ tends to be $1$.

share|cite|improve this answer

edited Mar 16 at 7:45

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

$endgroup$

share|cite|improve this answer

edited Mar 16 at 7:45

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715

answered Mar 14 at 7:07

Boris SklyarBoris Sklyar

2715