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
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
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add a comment |
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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
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
$endgroup$
1
$begingroup$
I would suggest that you go over Dirichlet's proof and see if you can modify it to work for $a=1$.
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– Mustafa Said
Dec 30 '18 at 2:32
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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.
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– John Omielan
Dec 30 '18 at 2:33
add a comment |
$begingroup$
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
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
$endgroup$
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
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
asked Dec 30 '18 at 2:27
Tejas RaoTejas Rao
34211
34211
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
add a comment |
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
add a comment |
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.
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
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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.
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
$endgroup$
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.
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
edited Dec 30 '18 at 4:05
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
answered Dec 30 '18 at 2:38
John OmielanJohn Omielan
5,1042218
5,1042218
add a comment |
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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