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Hint:- Even though there are easy solutions for this problem, what Dirichlet theorem on primes in arithmetic progression has to say on this?

Is this correct??? GIVEN : n, a , d are integers and gcd(a, d) =1Since it is given that a and d are relatively prime, then by Dirichlet theorem the arithmetic sequence a, a+d, a+2d,...contain infinitely many primesLet there be some a+dq in the sequence where q is an integer Suppose a+dq=m_____(1)Since a+dq lies in the sequence it is a prime and it will be an integer. which implies there exist an integer m. From(1)we have : m=a+dqWhich implies, dq =m-a And, d divides (m-a) This implies, m congruent a (mod d) since it is given that n is an integer and we found that there exist an integer m (which is also prime) therefore gcd (n, m) = 1 Hence proved

There are some gaps in the arguemment.1. Are we taking q in with m=a+dq is prime?2. If so, how can we ensure that $gcd(n,m)=1$?

There are two gaps in the argument.1. Are we taking q with m=a+dq is prime?2. If so, how can we ensure that gcd(n,m)=1?

Replace n in the place of q so that a+nd will be in the sequence where n is an integer Then, Suppose a+nd=m_____(1)Since a+nd lies in the sequence it is a prime and it will be an integer. which implies there exist an integer m. From(1)we have : m=a+ndWhich implies , dn =m-a And, d divides (m-a) This implies, m congruent a(mod d)Since m is prime it is either divided by m or 1 This implies (a+nd) /m or (a+nd) /1Case1:(a+nd) /m then by the property of divisibiliy a/m and nd/m, but from this we cannot conclude that n/m or d/mCase 2: (a+nd) /1 then by the property ofdivisibiliy a/1 and nd/1, this implies n/1 and we can conclude that n is also a prime. As we have found that n and m are prime(provided m not equal to n) gcd (m, n) =1

The idea is there are infinitely primes of the form \(a+qd\) by Dirichlet theorem. Also the number \(n\) has a finite number of prime factors. So choose a prime of the form \(m=a+qd\) which is not a divisor of \(n\). Done!

Here is the hint for a direct solution: Can we take \(q\) as the product of all primes which divide's \(n\) but not \(a\)?

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ReplyDeleteHint:- Even though there are easy solutions for this problem, what Dirichlet theorem on primes in arithmetic progression has to say on this?

ReplyDeleteIs this correct???

ReplyDeleteGIVEN : n, a , d are integers and gcd(a, d) =1

Since it is given that a and d are relatively prime, then by Dirichlet theorem the arithmetic sequence a, a+d, a+2d,...contain infinitely many primes

Let there be some a+dq in the sequence where q is an integer

Suppose a+dq=m_____(1)

Since a+dq lies in the sequence it is a prime and it will be an integer.

which implies there exist an integer m.

From(1)we have :

m=a+dq

Which implies, dq =m-a

And, d divides (m-a)

This implies, m congruent a (mod d)

since it is given that n is an integer and we found that there exist an integer m (which is also prime) therefore gcd (n, m) = 1

Hence proved

There are some gaps in the arguemment.

Delete1. Are we taking q in with m=a+dq is prime?

2. If so, how can we ensure that $gcd(n,m)=1$?

There are two gaps in the argument.

ReplyDelete1. Are we taking q with m=a+dq is prime?

2. If so, how can we ensure that gcd(n,m)=1?

Replace n in the place of q so that a+nd will be in the sequence where n is an integer

DeleteThen,

Suppose a+nd=m_____(1)

Since a+nd lies in the sequence it is a prime and it will be an integer.

which implies there exist an integer m.

From(1)we have :

m=a+nd

Which implies , dn =m-a

And, d divides (m-a)

This implies,

m congruent a(mod d)

Since m is prime it is either divided by m or 1

This implies (a+nd) /m or (a+nd) /1

Case1:(a+nd) /m then by the property of divisibiliy a/m and nd/m, but from this we cannot conclude that n/m or d/m

Case 2: (a+nd) /1 then by the property ofdivisibiliy a/1 and nd/1, this implies n/1 and we can conclude that n is also a prime.

As we have found that n and m are prime(provided m not equal to n)

gcd (m, n) =1

The idea is there are infinitely primes of the form \(a+qd\) by Dirichlet theorem. Also the number \(n\) has a finite number of prime factors. So choose a prime of the form \(m=a+qd\) which is not a divisor of \(n\). Done!

ReplyDeleteHere is the hint for a direct solution: Can we take \(q\) as the product of all primes which divide's \(n\) but not \(a\)?

ReplyDelete