Mixing
Are there infinite many primes p such that 2p-1 is also prime?
$begingroup$
I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
number-theory prime-numbers
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
$endgroup$
add a comment |
$begingroup$
I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
number-theory prime-numbers
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
$endgroup$
5
$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06
add a comment |
$begingroup$
I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
number-theory prime-numbers
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
$endgroup$
I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
number-theory prime-numbers
number-theory prime-numbers
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
edited Aug 28 '14 at 5:11
Martin Sleziak
44.8k10119273
44.8k10119273
asked Aug 28 '14 at 5:09
Li LiLi Li
241
asked Aug 28 '14 at 5:09
Li LiLi Li
241
asked Aug 28 '14 at 5:09
Li LiLi Li
241
241
5
$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06
add a comment |
5
$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06
5
5
$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
$endgroup$
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
add a comment |
$begingroup$
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
add a comment |
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2 Answers
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$begingroup$
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
$endgroup$
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
add a comment |
$begingroup$
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
$endgroup$
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
add a comment |
$begingroup$
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
$endgroup$
I have managed to find a proof such that all Sophie Germain primes are of the prime form $6n-1$, but have failed to get further. My proof is:
If the primes created by the formula $2x + 1$ equals the prime type $4y + 1$, the equation of $2x + 1 = 4y + 1$ is created, which simplifies to $x = 2y$. But as $x$ is prime, $x$ cannot equal anything but $2$. Therefore, the Germain safe primes are of the prime type $4y – 1$. This creates the equation $2x + 1 = 4y – 1$, which simplifies to $x = 2y - 1$. That means both $4y – 1$ and $2y – 1$ needs to be prime.
This will never be true of the values of $y$ that are of the formula $3n + 2$, excluding $2$. This is because the first value of $y$ in the equation $2y – 1 (5)$ makes $9$, and each new term in the sequence is the last term $+ 2$. After $3$ places, you add $6$, making a new multiple of $3$.
This leads the terms in the formula $3n$. Let us substitute $y$ for $3n$. $2(3n) – 1$ simplifies to $6n – 1$. For $3n – 1$, the other formula, you produce $6n + 1$.
For hypothetical $6n + 1$ Germain primes, the safe prime would be $12n + 3$. As terms from the formula $12n + 3$ will always divide by three, the only possible Germain prime is where $n = 0$. As the Germain "prime" would be $1$, and as $1$ is not prime, there are no safe primes of the formula $12n + 3$. Therefore, all Germain primes must be from the formula $6n - 1$.
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
edited Aug 31 '18 at 10:59
Björn Friedrich
2,68461831
2,68461831
answered Aug 31 '18 at 10:05
mngimngi
94
answered Aug 31 '18 at 10:05
mngimngi
94
answered Aug 31 '18 at 10:05
mngimngi
94
94
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
add a comment |
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
1
1
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
$begingroup$
Every odd prime is of the form $6npm 1$ except for the prime $3.$ If the prime $p$ is greater than $3$ and $p=6n+1$ then $2p+1=12n+3=3(4n+1),$ which is $>3$ and divisible by $3,$ so is not prime..... On the other hand, $3 $ $is$ a Germain prime.
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:14
add a comment |
$begingroup$
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
add a comment |
$begingroup$
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
add a comment |
$begingroup$
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
edited Sep 2 '18 at 7:44
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
answered Aug 31 '18 at 10:15
Gerry MyersonGerry Myerson
147k8149302
147k8149302
add a comment |
add a comment |
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$begingroup$
Yes, I'm afraid that many people have thought of this, and it is just as unsolved as the $2p+1$ case (it is widely believed that there are infinitely many). Sophie Germain didn't get primes named after her just for asking the question; those primes are named in honour of their role in her work on Fermat's Last Theorem.
$endgroup$
– Erick Wong
Aug 28 '14 at 5:49
$begingroup$
More generally if you take any natural nymber $K>1$ and any non-zero integer $A$ such that $gcd(K,A)=1,$ and ask whether the set of primes $p,$ such that $Kp+A$ is also prime, is an infinite set, the correct answer is "Nobody knows".
$endgroup$
– DanielWainfleet
Sep 2 '18 at 15:06