Are there any minor extensions of Dirichlet's theorem?












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For example, can we say that for $k$ $odd$, there are infinitely many primes of the form $a+bk$, for a fixed $a,b$ with $gcd(a,b)=1$?



How about for $k$ $odd$, there are infinitely many primes of the form $1+bk$, for a fixed $a,b$?






elementary-number-theory dirichlet-series







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    I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 2:32










  • $begingroup$
    Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:33


















1












$begingroup$


For example, can we say that for $k$ $odd$, there are infinitely many primes of the form $a+bk$, for a fixed $a,b$ with $gcd(a,b)=1$?



How about for $k$ $odd$, there are infinitely many primes of the form $1+bk$, for a fixed $a,b$?






elementary-number-theory dirichlet-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 2:32










  • $begingroup$
    Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:33
















1












1








1





$begingroup$


For example, can we say that for $k$ $odd$, there are infinitely many primes of the form $a+bk$, for a fixed $a,b$ with $gcd(a,b)=1$?



How about for $k$ $odd$, there are infinitely many primes of the form $1+bk$, for a fixed $a,b$?






elementary-number-theory dirichlet-series







share|cite|improve this question









$endgroup$




For example, can we say that for $k$ $odd$, there are infinitely many primes of the form $a+bk$, for a fixed $a,b$ with $gcd(a,b)=1$?



How about for $k$ $odd$, there are infinitely many primes of the form $1+bk$, for a fixed $a,b$?






elementary-number-theory dirichlet-series




elementary-number-theory dirichlet-series






share|cite|improve this question













share|cite|improve this question











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asked Dec 30 '18 at 2:27









Tejas RaoTejas Rao

33711




33711








  • 1




    $begingroup$
    I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 2:32










  • $begingroup$
    Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:33
















  • 1




    $begingroup$
    I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
    $endgroup$
    – Mustafa Said
    Dec 30 '18 at 2:32










  • $begingroup$
    Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:33










1




1




$begingroup$
I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
$endgroup$
– Mustafa Said
Dec 30 '18 at 2:32




$begingroup$
I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
$endgroup$
– Mustafa Said
Dec 30 '18 at 2:32












$begingroup$
Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
$endgroup$
– John Omielan
Dec 30 '18 at 2:33






$begingroup$
Note that for either case, for $k$ being odd, there won't be infinitely many primes if $a$ and $b$ are both odd as $a + bk$ will then always be even.
$endgroup$
– John Omielan
Dec 30 '18 at 2:33












1 Answer
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$begingroup$

For certain cases, you can say that $k$ must be odd, but it doesn't really add any restriction. For example, if $a$ is even and $b$ is odd, with $gcd left(a, bright) = 1$, then $k$ must be odd for $a + bk$ to be a prime, except from the one possible case of $2$. Apart from such relatively basic cases, I don't believe there is much more that restricting $k$ to be odd, or just $1$, will do. Nonetheless, I agree with Mustafa Said that you may wish to check Dirichlet's proof (e.g., do an online search about any extensions to it) to see about this yourself.






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    $begingroup$

    For certain cases, you can say that $k$ must be odd, but it doesn't really add any restriction. For example, if $a$ is even and $b$ is odd, with $gcd left(a, bright) = 1$, then $k$ must be odd for $a + bk$ to be a prime, except from the one possible case of $2$. Apart from such relatively basic cases, I don't believe there is much more that restricting $k$ to be odd, or just $1$, will do. Nonetheless, I agree with Mustafa Said that you may wish to check Dirichlet's proof (e.g., do an online search about any extensions to it) to see about this yourself.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For certain cases, you can say that $k$ must be odd, but it doesn't really add any restriction. For example, if $a$ is even and $b$ is odd, with $gcd left(a, bright) = 1$, then $k$ must be odd for $a + bk$ to be a prime, except from the one possible case of $2$. Apart from such relatively basic cases, I don't believe there is much more that restricting $k$ to be odd, or just $1$, will do. Nonetheless, I agree with Mustafa Said that you may wish to check Dirichlet's proof (e.g., do an online search about any extensions to it) to see about this yourself.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For certain cases, you can say that $k$ must be odd, but it doesn't really add any restriction. For example, if $a$ is even and $b$ is odd, with $gcd left(a, bright) = 1$, then $k$ must be odd for $a + bk$ to be a prime, except from the one possible case of $2$. Apart from such relatively basic cases, I don't believe there is much more that restricting $k$ to be odd, or just $1$, will do. Nonetheless, I agree with Mustafa Said that you may wish to check Dirichlet's proof (e.g., do an online search about any extensions to it) to see about this yourself.






        share|cite|improve this answer











        $endgroup$



        For certain cases, you can say that $k$ must be odd, but it doesn't really add any restriction. For example, if $a$ is even and $b$ is odd, with $gcd left(a, bright) = 1$, then $k$ must be odd for $a + bk$ to be a prime, except from the one possible case of $2$. Apart from such relatively basic cases, I don't believe there is much more that restricting $k$ to be odd, or just $1$, will do. Nonetheless, I agree with Mustafa Said that you may wish to check Dirichlet's proof (e.g., do an online search about any extensions to it) to see about this yourself.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 4:05

























        answered Dec 30 '18 at 2:38









        John OmielanJohn Omielan

        4,1691215




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