# Using the Dirichlet's Theorem to find the number of primes in an arithemetic progression.

Dirichlet's Theorem says that the sequence of integers {Ak+B}, where A, B have no common divisor other than +-1, contains infinitely many primes. It does not say that all such numbers Ak+B are prime, nor does it say how many integers in such a sequence are themselves primes in an arithmetic progression.

(a) Find four primes in arithmetic progression. Find five primes in arithmetic progression.

(b) If you wish to discover six or more primes in arithmetic progression {Ak+B}, where k= 0,1,2,3..., why must A be divisible by 10?

(c)Find six primes in arithmetic progression.

My Answer may not be correct. Please give suggestion on my following answer and correct me if I am wrong.

(a) gcd(4,3) =1, so By Dirichlet's Theorem, there are infinitely many prime of the form 4k+3 where k=0,1,2... and the numbers of the form 4k+3 form an arithmetic progression with a=3 and d=4. AP of 4k+3={3,7,11,15,19,23,...}. Therefore, the first four primes in AP of the form 4k+3 are 3,7,11,19.

gcd(2,1)=1, By Dirichlet's Theorem, there are infinitely many prime of the form 2k+1 where k=0,1,2... and the numbers of the form 2k+1 form an arithmetic progression with a=1 and d=2. AP of 2k+1={1,3,5,7,9,11,13,...}. Therefore, the first five primes in AP of the form 2k+1 are 3,5,7,11,13.

(b)Primes are odd and since even+even is even, odd+odd is even, and odd+even is odd, B must be odd and A must be even. Proof by contrapositive: Suppose 5 does not divides A, so A is not a multiple of 5 and A is also not a multiple of 10. Thus 10 does not divides A for any A >= 10. For A to be an even A must also be multiple of 10 since even integers contain multiple of 10 and 2|10k, k=1,2,3,... but A can never be a multiple of 10 since 10 does not divides A for any A >= 10. Hence A cannot be even.

Therefore, A must be divisible by 10 for some A >= 10 for A to be an even integer.

(c) gcd(10,1)=1, By Dirichlet's Theorem, there are infinitely many prime of the form 10k+1 where k=0,1,2... and the numbers of the form 10k+1 form an arithmetic progression with a=1 and d=10. AP of 10k+1={1,11,21,31,41,51,61,71,81,91,101,...}. Therefore, the first six primes in AP of the form 2k+1 are 11,31,41,61,71,101.

• Please write your question using latex. – Marco Cantarini Mar 11 '15 at 8:39
• When we say primes in arithmetic progression, we mean primes with a common difference, like $3,5,7$ which have a difference of $2$. This is not directly related to dirichlets theorem – Mathmo123 Mar 11 '15 at 11:20
• I'm not sure I understand your solution to (b). Why can't $A$ be $6$ for example? – Mathmo123 Mar 11 '15 at 11:24

If $$A,B\in \Bbb N$$ and if $$S=\{A+nB: 0\le n\le 5\}$$ is a set of primes then (i) $$2|B,$$ else $$S$$ has 3 even members, (ii) $$3|B,$$ else $$S$$ has 2 members divisible by $$3,$$ (iii) $$5|B,$$ else $$S$$ has 2 members divisible by $$5.$$ So $$30|B.$$ Hence $$\gcd (A,30)=1,$$ else $$A+B$$ is greater than $$30$$ and is divisible by a member of $$\{2,3,5\}.$$
If we try the smallest allowable prime $$A=7$$ and the smallest allowable $$B=30,$$ we are rewarded with $$S=\{7,37,67,97,127,157\}.$$