Are there infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and odd $k$












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It is clear from Dirichlet's theorem on arithmetic progressions that for a fixed $n$, there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and $k=1,2,3,..$. However, what if we keep $k$ odd?



From Arturo's comment on Are there infinitely many primes of the form $kcdot 2^n +1$?, he writes:



And if you want $k$ odd, start with $a=2^n+1$, $b=2^ncdot 2$, and look at $a+bm$.



Can someone explain whether this is correct and why, and whether this proved there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and odd $k$.






number-theory elementary-number-theory dirichlet-series distribution-of-primes







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    $begingroup$


    It is clear from Dirichlet's theorem on arithmetic progressions that for a fixed $n$, there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and $k=1,2,3,..$. However, what if we keep $k$ odd?



    From Arturo's comment on Are there infinitely many primes of the form $kcdot 2^n +1$?, he writes:



    And if you want $k$ odd, start with $a=2^n+1$, $b=2^ncdot 2$, and look at $a+bm$.



    Can someone explain whether this is correct and why, and whether this proved there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and odd $k$.






    number-theory elementary-number-theory dirichlet-series distribution-of-primes







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      It is clear from Dirichlet's theorem on arithmetic progressions that for a fixed $n$, there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and $k=1,2,3,..$. However, what if we keep $k$ odd?



      From Arturo's comment on Are there infinitely many primes of the form $kcdot 2^n +1$?, he writes:



      And if you want $k$ odd, start with $a=2^n+1$, $b=2^ncdot 2$, and look at $a+bm$.



      Can someone explain whether this is correct and why, and whether this proved there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and odd $k$.






      number-theory elementary-number-theory dirichlet-series distribution-of-primes







      share|cite|improve this question











      $endgroup$




      It is clear from Dirichlet's theorem on arithmetic progressions that for a fixed $n$, there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and $k=1,2,3,..$. However, what if we keep $k$ odd?



      From Arturo's comment on Are there infinitely many primes of the form $kcdot 2^n +1$?, he writes:



      And if you want $k$ odd, start with $a=2^n+1$, $b=2^ncdot 2$, and look at $a+bm$.



      Can someone explain whether this is correct and why, and whether this proved there are infinitely many primes of the form $kcdot 2^n+1$ for a fixed $n$ and odd $k$.






      number-theory elementary-number-theory dirichlet-series distribution-of-primes




      number-theory elementary-number-theory dirichlet-series distribution-of-primes






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 0:32







      Tejas Rao

















      asked Dec 29 '18 at 23:06









      Tejas RaoTejas Rao

      31811




      31811






















          1 Answer
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          3












          $begingroup$

          That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            $gcd(2^n+1,2^{n+1}) = 1$ is important
            $endgroup$
            – mathworker21
            Dec 29 '18 at 23:09










          • $begingroup$
            Indeed it is. Added something about that.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:11










          • $begingroup$
            How does / does this prove my question, though?
            $endgroup$
            – Tejas Rao
            Dec 29 '18 at 23:18










          • $begingroup$
            The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:21










          • $begingroup$
            @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
            $endgroup$
            – Peter
            Dec 30 '18 at 10:02











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            $gcd(2^n+1,2^{n+1}) = 1$ is important
            $endgroup$
            – mathworker21
            Dec 29 '18 at 23:09










          • $begingroup$
            Indeed it is. Added something about that.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:11










          • $begingroup$
            How does / does this prove my question, though?
            $endgroup$
            – Tejas Rao
            Dec 29 '18 at 23:18










          • $begingroup$
            The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:21










          • $begingroup$
            @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
            $endgroup$
            – Peter
            Dec 30 '18 at 10:02
















          3












          $begingroup$

          That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            $gcd(2^n+1,2^{n+1}) = 1$ is important
            $endgroup$
            – mathworker21
            Dec 29 '18 at 23:09










          • $begingroup$
            Indeed it is. Added something about that.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:11










          • $begingroup$
            How does / does this prove my question, though?
            $endgroup$
            – Tejas Rao
            Dec 29 '18 at 23:18










          • $begingroup$
            The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:21










          • $begingroup$
            @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
            $endgroup$
            – Peter
            Dec 30 '18 at 10:02














          3












          3








          3





          $begingroup$

          That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.






          share|cite|improve this answer











          $endgroup$



          That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 23:10

























          answered Dec 29 '18 at 23:09









          user3482749user3482749

          4,276919




          4,276919








          • 3




            $begingroup$
            $gcd(2^n+1,2^{n+1}) = 1$ is important
            $endgroup$
            – mathworker21
            Dec 29 '18 at 23:09










          • $begingroup$
            Indeed it is. Added something about that.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:11










          • $begingroup$
            How does / does this prove my question, though?
            $endgroup$
            – Tejas Rao
            Dec 29 '18 at 23:18










          • $begingroup$
            The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:21










          • $begingroup$
            @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
            $endgroup$
            – Peter
            Dec 30 '18 at 10:02














          • 3




            $begingroup$
            $gcd(2^n+1,2^{n+1}) = 1$ is important
            $endgroup$
            – mathworker21
            Dec 29 '18 at 23:09










          • $begingroup$
            Indeed it is. Added something about that.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:11










          • $begingroup$
            How does / does this prove my question, though?
            $endgroup$
            – Tejas Rao
            Dec 29 '18 at 23:18










          • $begingroup$
            The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
            $endgroup$
            – user3482749
            Dec 29 '18 at 23:21










          • $begingroup$
            @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
            $endgroup$
            – Peter
            Dec 30 '18 at 10:02








          3




          3




          $begingroup$
          $gcd(2^n+1,2^{n+1}) = 1$ is important
          $endgroup$
          – mathworker21
          Dec 29 '18 at 23:09




          $begingroup$
          $gcd(2^n+1,2^{n+1}) = 1$ is important
          $endgroup$
          – mathworker21
          Dec 29 '18 at 23:09












          $begingroup$
          Indeed it is. Added something about that.
          $endgroup$
          – user3482749
          Dec 29 '18 at 23:11




          $begingroup$
          Indeed it is. Added something about that.
          $endgroup$
          – user3482749
          Dec 29 '18 at 23:11












          $begingroup$
          How does / does this prove my question, though?
          $endgroup$
          – Tejas Rao
          Dec 29 '18 at 23:18




          $begingroup$
          How does / does this prove my question, though?
          $endgroup$
          – Tejas Rao
          Dec 29 '18 at 23:18












          $begingroup$
          The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
          $endgroup$
          – user3482749
          Dec 29 '18 at 23:21




          $begingroup$
          The $a$ and $b$ in your question are precisely the constant term and coefficient of $m$ in my answer.
          $endgroup$
          – user3482749
          Dec 29 '18 at 23:21












          $begingroup$
          @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
          $endgroup$
          – Peter
          Dec 30 '18 at 10:02




          $begingroup$
          @TejasRao For every $m$ for which $2^{n+1}cdot m+2^n+1$ is prime, you can choose $k=2m+1$ and get $2^ncdot k+1$ with an odd $k$. Since infinite many $m$ do the job, infinite many primes of the desired form must exist.
          $endgroup$
          – Peter
          Dec 30 '18 at 10:02


















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