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How many prime numbers are there that can't be written as a sum of two composite numbers? [duplicate]

Prove by contradiction that every integer greater than 11 is a sum of two composite numbersThere are infinitely many triangular numbers that are the sum of two other such numberscan all triangle numbers that are squares be expressed as sum of squaresProve that there are infinitely many natural numbers that can't be written as $a^2+p$Show that there are infinitely many prime numbers ending in 3 or 7 (when written in decimal)Proof that there are infinitely many prime numbersUse this sequence to prove that there are infinitely many prime numbers.There are infinitely many numbers that can't be written as a sum of a prime and a triangular numberProve that there are infinitely many natural numbers that can't be written as $6xy+x+y$How can I show that there are infinitely many prime numbers such that $p equiv 3 pmod4$?Prove that there are infinitely many composite number $m$ such that $a^m-1 - 1$ is divisible by $m$

1

$begingroup$

This question already has an answer here:

• 
  Prove by contradiction that every integer greater than 11 is a sum of two composite numbers
  
  4 answers
  
  

Obviously $2,3,5,ldots$ but I'm not sure for what other numbers does it hold, or if there are infinitely many.

elementary-number-theory

share|cite|improve this question

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

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marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you talking about only Natural numbers? Otherwise $2=12+(-10)$ or like that.
  $endgroup$
  – Love Invariants
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  Prove by contradiction that every integer greater than 11 is a sum of two composite numbers
  
  4 answers
  
  

Obviously $2,3,5,ldots$ but I'm not sure for what other numbers does it hold, or if there are infinitely many.

elementary-number-theory

share|cite|improve this question

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

$endgroup$

marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you talking about only Natural numbers? Otherwise $2=12+(-10)$ or like that.
  $endgroup$
  – Love Invariants
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Prove by contradiction that every integer greater than 11 is a sum of two composite numbers
  
  4 answers
  
  

Obviously $2,3,5,ldots$ but I'm not sure for what other numbers does it hold, or if there are infinitely many.

elementary-number-theory

share|cite|improve this question

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

$endgroup$

share|cite|improve this question

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

share|cite|improve this question

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

edited yesterday

Daniele Tampieri

2,5222922

edited yesterday

Daniele Tampieri

2,5222922

asked yesterday

Darko DekanDarko Dekan

342113

asked yesterday

Darko DekanDarko Dekan

342113

1 Answer
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oldest

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13

$begingroup$

Any $nge 13$, prime or otherwise, is either $8$ or $9$ more than an even number $ge 4$. Therefore, any such $n$ is a sum of two composite numbers. We can exhaustively check the only primes lacking such an expression are the primes from $2$ to $11$ inclusive, i.e. $5$ of them.

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answered yesterday

J.G.J.G.

30.2k23148

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add a comment | 

13

$begingroup$

Any $nge 13$, prime or otherwise, is either $8$ or $9$ more than an even number $ge 4$. Therefore, any such $n$ is a sum of two composite numbers. We can exhaustively check the only primes lacking such an expression are the primes from $2$ to $11$ inclusive, i.e. $5$ of them.

share|cite|improve this answer

answered yesterday

J.G.J.G.

30.2k23148

$endgroup$

add a comment | 

13

$begingroup$

Any $nge 13$, prime or otherwise, is either $8$ or $9$ more than an even number $ge 4$. Therefore, any such $n$ is a sum of two composite numbers. We can exhaustively check the only primes lacking such an expression are the primes from $2$ to $11$ inclusive, i.e. $5$ of them.

share|cite|improve this answer

answered yesterday

J.G.J.G.

30.2k23148

$endgroup$

add a comment | 

$begingroup$

Any $nge 13$, prime or otherwise, is either $8$ or $9$ more than an even number $ge 4$. Therefore, any such $n$ is a sum of two composite numbers. We can exhaustively check the only primes lacking such an expression are the primes from $2$ to $11$ inclusive, i.e. $5$ of them.

share|cite|improve this answer

answered yesterday

J.G.J.G.

30.2k23148

$endgroup$

share|cite|improve this answer

answered yesterday

J.G.J.G.

30.2k23148

answered yesterday

J.G.J.G.

30.2k23148

answered yesterday

J.G.J.G.

30.2k23148