How to prove the existence of a prime number with elementary method?

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Suppose $A>5$ is an integer and there exists a prime number $p$ such that $A-2leq p^2<A$.



Show that there exists at least one prime number $q$ such that $p<q<A$.



This seems to be intuitive to me since $A$ and $p$ has large gap. How can we prove it?










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    up vote
    1
    down vote

    favorite












    Suppose $A>5$ is an integer and there exists a prime number $p$ such that $A-2leq p^2<A$.



    Show that there exists at least one prime number $q$ such that $p<q<A$.



    This seems to be intuitive to me since $A$ and $p$ has large gap. How can we prove it?










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose $A>5$ is an integer and there exists a prime number $p$ such that $A-2leq p^2<A$.



      Show that there exists at least one prime number $q$ such that $p<q<A$.



      This seems to be intuitive to me since $A$ and $p$ has large gap. How can we prove it?










      share|cite|improve this question













      Suppose $A>5$ is an integer and there exists a prime number $p$ such that $A-2leq p^2<A$.



      Show that there exists at least one prime number $q$ such that $p<q<A$.



      This seems to be intuitive to me since $A$ and $p$ has large gap. How can we prove it?







      number-theory prime-numbers






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      asked Dec 4 at 3:56









      user3813057

      25418




      25418




















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          Using one version of Bertrand's Postulate, if $q$ is the next prime after $p$ then
          $$p<q<2ple p^2<A .$$






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            5
            down vote













            Using one version of Bertrand's Postulate, if $q$ is the next prime after $p$ then
            $$p<q<2ple p^2<A .$$






            share|cite|improve this answer
























              up vote
              5
              down vote













              Using one version of Bertrand's Postulate, if $q$ is the next prime after $p$ then
              $$p<q<2ple p^2<A .$$






              share|cite|improve this answer






















                up vote
                5
                down vote










                up vote
                5
                down vote









                Using one version of Bertrand's Postulate, if $q$ is the next prime after $p$ then
                $$p<q<2ple p^2<A .$$






                share|cite|improve this answer












                Using one version of Bertrand's Postulate, if $q$ is the next prime after $p$ then
                $$p<q<2ple p^2<A .$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 at 4:03









                David

                67.4k663126




                67.4k663126



























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