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If $gcd(n,q_1)=1$ then show that $gcd(n+tq_1,q)=1$ for some $t$

1

$begingroup$

If $q=q_1cdot q_2$ and $(q,n)=a$ and $(n,q_1)=1$ then how to select $t$ s.t. $(n+tq_1,q)=1$

bezout implies that $exists r,s$ s.t. $rn+sq_1=1$ and this means $(rn+sq_1,q)=1$ but how to get rid of $r$ ?

modular-arithmetic greatest-common-divisor

share|cite|improve this question

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

$endgroup$

add a comment | 

1

$begingroup$

If $q=q_1cdot q_2$ and $(q,n)=a$ and $(n,q_1)=1$ then how to select $t$ s.t. $(n+tq_1,q)=1$

bezout implies that $exists r,s$ s.t. $rn+sq_1=1$ and this means $(rn+sq_1,q)=1$ but how to get rid of $r$ ?

modular-arithmetic greatest-common-divisor

share|cite|improve this question

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

$endgroup$

add a comment | 

$begingroup$

If $q=q_1cdot q_2$ and $(q,n)=a$ and $(n,q_1)=1$ then how to select $t$ s.t. $(n+tq_1,q)=1$

bezout implies that $exists r,s$ s.t. $rn+sq_1=1$ and this means $(rn+sq_1,q)=1$ but how to get rid of $r$ ?

modular-arithmetic greatest-common-divisor

share|cite|improve this question

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

$endgroup$

share|cite|improve this question

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

share|cite|improve this question

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

edited Sep 21 '16 at 14:25

user133281

13.7k22752

edited Sep 21 '16 at 14:25

user133281

13.7k22752

asked Sep 21 '16 at 11:34

user1161user1161

271214

asked Sep 21 '16 at 11:34

user1161user1161

271214

2 Answers
2

active

oldest

votes

2

$begingroup$

An explicit construction: take $t$ to be the product of those primes that divide $q$ but not $n$.

We show that this works. Consider a prime $p$ dividing $q$.

• If $p$ divides $n$ then $p$ does not divide $t$ (by our choice of $t$) or $q_1$ (by $gcd(n,q_1)=1$). Hence $p$ does not divide $n+tq_1$.


• If $p$ does not divide $n$ then $p$ divides $t$, so $p$ does not divide $n+tq_1$.

In conclusion, no prime $p$ dividing $q$ can divide $n+tq_1$, so $gcd(q,n+tq_1)=1$ for this choice of $t$.

share|cite|improve this answer

edited Sep 21 '16 at 14:23

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

$endgroup$

add a comment | 

2

$begingroup$

While I do not have a direct formula for $t$, we can prove that it exists. However, it is a "bulldozer" argument that uses Dirichlet's Theorem.

It states that the sequence $ n + t q_1 _t in mathbbN$ contains infinitely many primes. Hence, all we have to do is choose a $t$ such that $n + t q_1$ is a prime larger than $q$, and we are done.

share|cite|improve this answer

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

$endgroup$

add a comment | 

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2

$begingroup$

An explicit construction: take $t$ to be the product of those primes that divide $q$ but not $n$.

We show that this works. Consider a prime $p$ dividing $q$.

• If $p$ divides $n$ then $p$ does not divide $t$ (by our choice of $t$) or $q_1$ (by $gcd(n,q_1)=1$). Hence $p$ does not divide $n+tq_1$.


• If $p$ does not divide $n$ then $p$ divides $t$, so $p$ does not divide $n+tq_1$.

In conclusion, no prime $p$ dividing $q$ can divide $n+tq_1$, so $gcd(q,n+tq_1)=1$ for this choice of $t$.

share|cite|improve this answer

edited Sep 21 '16 at 14:23

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

$endgroup$

add a comment | 

2

$begingroup$

An explicit construction: take $t$ to be the product of those primes that divide $q$ but not $n$.

We show that this works. Consider a prime $p$ dividing $q$.

• If $p$ divides $n$ then $p$ does not divide $t$ (by our choice of $t$) or $q_1$ (by $gcd(n,q_1)=1$). Hence $p$ does not divide $n+tq_1$.


• If $p$ does not divide $n$ then $p$ divides $t$, so $p$ does not divide $n+tq_1$.

In conclusion, no prime $p$ dividing $q$ can divide $n+tq_1$, so $gcd(q,n+tq_1)=1$ for this choice of $t$.

share|cite|improve this answer

edited Sep 21 '16 at 14:23

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

$endgroup$

add a comment | 

$begingroup$

An explicit construction: take $t$ to be the product of those primes that divide $q$ but not $n$.

We show that this works. Consider a prime $p$ dividing $q$.

• If $p$ divides $n$ then $p$ does not divide $t$ (by our choice of $t$) or $q_1$ (by $gcd(n,q_1)=1$). Hence $p$ does not divide $n+tq_1$.


• If $p$ does not divide $n$ then $p$ divides $t$, so $p$ does not divide $n+tq_1$.

In conclusion, no prime $p$ dividing $q$ can divide $n+tq_1$, so $gcd(q,n+tq_1)=1$ for this choice of $t$.

share|cite|improve this answer

edited Sep 21 '16 at 14:23

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

$endgroup$

share|cite|improve this answer

edited Sep 21 '16 at 14:23

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

answered Sep 21 '16 at 13:37

user133281user133281

13.7k22752

2

$begingroup$

While I do not have a direct formula for $t$, we can prove that it exists. However, it is a "bulldozer" argument that uses Dirichlet's Theorem.

It states that the sequence $ n + t q_1 _t in mathbbN$ contains infinitely many primes. Hence, all we have to do is choose a $t$ such that $n + t q_1$ is a prime larger than $q$, and we are done.

share|cite|improve this answer

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

$endgroup$

add a comment | 

2

$begingroup$

While I do not have a direct formula for $t$, we can prove that it exists. However, it is a "bulldozer" argument that uses Dirichlet's Theorem.

It states that the sequence $ n + t q_1 _t in mathbbN$ contains infinitely many primes. Hence, all we have to do is choose a $t$ such that $n + t q_1$ is a prime larger than $q$, and we are done.

share|cite|improve this answer

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

$endgroup$

add a comment | 

$begingroup$

While I do not have a direct formula for $t$, we can prove that it exists. However, it is a "bulldozer" argument that uses Dirichlet's Theorem.

It states that the sequence $ n + t q_1 _t in mathbbN$ contains infinitely many primes. Hence, all we have to do is choose a $t$ such that $n + t q_1$ is a prime larger than $q$, and we are done.

share|cite|improve this answer

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

$endgroup$

share|cite|improve this answer

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

edited Mar 26 at 6:26

Pang

14916

edited Mar 26 at 6:26

Pang

14916

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138

answered Sep 21 '16 at 12:50

ToucanNapoleonToucanNapoleon

40138