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












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

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






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marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun 10 hours ago


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
















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











$endgroup$



marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun 10 hours ago


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
    13 hours ago














1












1








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











$endgroup$





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.





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







elementary-number-theory




elementary-number-theory






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share|cite|improve this question













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edited 13 hours ago









Daniele Tampieri

2,3972922




2,3972922










asked 13 hours ago









Darko DekanDarko Dekan

342113




342113




marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun 10 hours ago


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.









marked as duplicate by lulu, Servaes, Song, J. M. is not a mathematician, Shaun 10 hours ago


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
    13 hours ago














  • 1




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








1




1




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




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










1 Answer
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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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    1 Answer
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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.






    share|cite|improve this answer









    $endgroup$


















      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









      $endgroup$
















        13












        13








        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









        $endgroup$



        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered 13 hours ago









        J.G.J.G.

        30.1k23048




        30.1k23048















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