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
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
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add a comment |
$begingroup$
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
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
$endgroup$
add a comment |
$begingroup$
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
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
$endgroup$
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
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
edited Dec 30 '18 at 0:32
Tejas Rao
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
asked Dec 29 '18 at 23:06
Tejas RaoTejas Rao
31811
31811
add a comment |
add a comment |
1 Answer
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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.
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
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3
$begingroup$
$gcd(2^n+1,2^{n+1}) = 1$ is important
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– mathworker21
Dec 29 '18 at 23:09
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Indeed it is. Added something about that.
$endgroup$
– user3482749
Dec 29 '18 at 23:11
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How does / does this prove my question, though?
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– Tejas Rao
Dec 29 '18 at 23:18
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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
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@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
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
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.
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
$endgroup$
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.
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
edited Dec 29 '18 at 23:10
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
answered Dec 29 '18 at 23:09
user3482749user3482749
4,276919
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
add a comment |
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
add a comment |
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