There are primes $p,q$ such that $gcd(a,b)=|ap-bq|$

The name of the pictureThe name of the pictureThe name of the pictureClash 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




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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














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




share|cite|improve this question


















  • 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












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




share|cite|improve this question














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





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 21:44

























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












  • 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










1 Answer
1






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oldest

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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.






share|cite|improve this answer


















  • 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










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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.






share|cite|improve this answer


















  • 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














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.






share|cite|improve this answer


















  • 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












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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 26 at 22:10

























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












  • 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

















 

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