There are primes $p,q$ such that $gcd(a,b)=|ap-bq|$
Clash Royale CLAN TAG#URR8PPP
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1
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Conjecture:
Given $a,binmathbb Z^+$ there are primes $p,q$ such that
$gcd(a,b)=|ap-bq|$.
I would like help with a proof or a counter-example.
Tested for millions of pseudo random numbers.
elementary-number-theory prime-numbers greatest-common-divisor computational-mathematics conjectures
asked Aug 26 at 21:10
Lehs
6,88331661
add a comment |Â
up vote
1
down vote
favorite
Conjecture:
Given $a,binmathbb Z^+$ there are primes $p,q$ such that
$gcd(a,b)=|ap-bq|$.
I would like help with a proof or a counter-example.
Tested for millions of pseudo random numbers.
elementary-number-theory prime-numbers greatest-common-divisor computational-mathematics conjectures
asked Aug 26 at 21:10
Lehs
6,88331661
2
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
2
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Conjecture:
Given $a,binmathbb Z^+$ there are primes $p,q$ such that
$gcd(a,b)=|ap-bq|$.
I would like help with a proof or a counter-example.
Tested for millions of pseudo random numbers.
elementary-number-theory prime-numbers greatest-common-divisor computational-mathematics conjectures
asked Aug 26 at 21:10
Lehs
6,88331661
Conjecture:
Given $a,binmathbb Z^+$ there are primes $p,q$ such that
$gcd(a,b)=|ap-bq|$.
I would like help with a proof or a counter-example.
Tested for millions of pseudo random numbers.
elementary-number-theory prime-numbers greatest-common-divisor computational-mathematics conjectures
asked Aug 26 at 21:10
Lehs
6,88331661
edited Aug 26 at 21:44
asked Aug 26 at 21:10
Lehs
6,88331661
asked Aug 26 at 21:10
Lehs
6,88331661
asked Aug 26 at 21:10
Lehs
6,88331661
6,88331661
2
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
2
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24
add a comment |Â
2
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
2
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24
2
2
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
2
2
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$
be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$
be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
 |Â
show 1 more comment
up vote
1
down vote
accepted
Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$
be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$
be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
Let $a, b>2$ with $b$ very larger than $a$. Suppose the twin prime conjecture true and $b-a=2$
be prime numbers, $gcd(a, b) =1=pa-qb$. If $p, q$ are prime, $ap$ and $qb$ are odd, and $ap-qb$ is even. Contradiction. You cannot have $p=2$ or $q=2$ if $a, b$ are enough big. Suppose, $a, a+2$ are prime, and $q=2$. $p$ is odd so at least $3$. $3a-2(a+2)=a-4$ if $a$ is not big it is not possible.
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
edited Aug 26 at 22:10
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
answered Aug 26 at 21:27
Tsemo Aristide
52.2k11244
52.2k11244
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
 |Â
show 1 more comment
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
1
1
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
$p$ or $q$ could be $2$.
– Lehs
Aug 26 at 21:30
1
1
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
@Lehs It's still a good starting point, because that means that either $2apm 1$ is a (prime) multiple of $b$, or $2bpm 1$ is a (prime) multiple of $a$, and it ought to be possible to find an example where that isn't the case. Say $a=11,b=13$.
– Arthur
Aug 26 at 21:42
1
1
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
a = 5, b = 29. It is impossible that $1 = |10 - 29q|$ or $1 = |5q - 58|$.
– Sasha Kozachinskiy
Aug 26 at 21:44
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
@SashaKozachinskiy: Are you sure? I've tested it for millions of prime pairs.
– Lehs
Aug 26 at 21:53
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
I guess my test program is erroneous. Have to look at it tomorrow.
– Lehs
Aug 26 at 22:32
 |Â
show 1 more comment
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2
@CharlesMadeline: $0notinmathbb Z^+$
– Lehs
Aug 26 at 21:18
2
We know that there are positive integers $u,v$ with $gcd(a,b)=|ua-vb|$, so the question is whether there exists $k$ such that both $u+kb$ and $v+ka$ are prime. Each of these arithmetic sequences contains infinitely many primes. Heuristically (i.e., with a probability-of-being-prime argument), the conjecture ought to be true
– Hagen von Eitzen
Aug 26 at 21:24