Given an odd integer, $a$ does $gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?
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Given an odd integer, $a$ does $gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?
One can argue that there are infinitely many numbers $x$ satisfies $gcd(a,$x$)=1$. How to argue that there are infinitely many prime $p=x+1$.
Any hint?
prime-numbers greatest-common-divisor
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
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add a comment |
$begingroup$
Given an odd integer, $a$ does $gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?
One can argue that there are infinitely many numbers $x$ satisfies $gcd(a,$x$)=1$. How to argue that there are infinitely many prime $p=x+1$.
Any hint?
prime-numbers greatest-common-divisor
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
$endgroup$
2
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
3
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39
add a comment |
$begingroup$
Given an odd integer, $a$ does $gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?
One can argue that there are infinitely many numbers $x$ satisfies $gcd(a,$x$)=1$. How to argue that there are infinitely many prime $p=x+1$.
Any hint?
prime-numbers greatest-common-divisor
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
$endgroup$
Given an odd integer, $a$ does $gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?
One can argue that there are infinitely many numbers $x$ satisfies $gcd(a,$x$)=1$. How to argue that there are infinitely many prime $p=x+1$.
Any hint?
prime-numbers greatest-common-divisor
prime-numbers greatest-common-divisor
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
edited Dec 4 '18 at 22:39
kelalaka
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
asked Dec 4 '18 at 22:31
kelalakakelalaka
3301312
3301312
2
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
3
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39
add a comment |
2
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
3
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39
2
2
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
3
3
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39
add a comment |
1 Answer
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By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($nin mathbb{N}$).
If $p=2+na$ is such a prime, then $gcd(a, p-1) = gcd(a, 1+na) = 1$.
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
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1 Answer
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$begingroup$
By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($nin mathbb{N}$).
If $p=2+na$ is such a prime, then $gcd(a, p-1) = gcd(a, 1+na) = 1$.
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
$endgroup$
add a comment |
$begingroup$
By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($nin mathbb{N}$).
If $p=2+na$ is such a prime, then $gcd(a, p-1) = gcd(a, 1+na) = 1$.
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
$endgroup$
add a comment |
$begingroup$
By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($nin mathbb{N}$).
If $p=2+na$ is such a prime, then $gcd(a, p-1) = gcd(a, 1+na) = 1$.
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
$endgroup$
By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($nin mathbb{N}$).
If $p=2+na$ is such a prime, then $gcd(a, p-1) = gcd(a, 1+na) = 1$.
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
answered Dec 4 '18 at 22:45
A. PongráczA. Pongrácz
5,9631929
5,9631929
add a comment |
add a comment |
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2
$begingroup$
When $a=2$ this asks: are there infinitely many even primes?
$endgroup$
– GEdgar
Dec 4 '18 at 22:38
3
$begingroup$
If $a$ is even, no... But if $a$ is odd, my hint would be to apply the Dirichlet arithmetic progression theorem.
$endgroup$
– Daniel Schepler
Dec 4 '18 at 22:38
$begingroup$
let me correct to odd.
$endgroup$
– kelalaka
Dec 4 '18 at 22:39