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Dirichlet's theorem on arithmetic progressions

Theorem on the number of primes in arithmetic sequences

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Johann Peter Gustav Lejeune Dirichlet
Unknown author Unknown author · Public domain

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Through Coprime integers, Integer and Prime number

a,a+d,a+2d,a+3d,…,

and Dirichlet's theorem states that this sequence contains infinitely many prime numbers. The theorem extends Euclid's theorem that there are infinitely many prime numbers (of the form 1 + 2n). Stronger forms of Dirichlet's theorem state that for any such arithmetic progression, the sum of the reciprocals of the prime numbers in the progression diverges and that different such arithmetic progressions with the same modulus have approximately the same proportions of primes. Equivalently, the primes are evenly distributed (asymptotically) among the congruence classes modulo d containing a coprime to d.

The theorem is named after the German mathematician Peter Gustav Lejeune Dirichlet, who proved it in 1837.

Examples

The primes of the form 4n + 3 are (sequence A002145 in the OEIS)

3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, 199, 211, 223, 227, 239, 251, 263, 271, 283, ...

They correspond to the following values of n: (sequence A095278 in the OEIS)

0, 1, 2, 4, 5, 7, 10, 11, 14, 16, 17, 19, 20, 25, 26, 31, 32, 34, 37, 40, 41, 44, 47, 49, 52, 55, 56, 59, 62, 65, 67, 70, 76, 77, 82, 86, 89, 91, 94, 95, ...

The strong form of Dirichlet's theorem implies that

13+17+111+119+123+131+143+147+159+167+⋯

is a divergent series.

Sequences dn + a with odd d are often ignored because half the numbers are even and the other half is the same numbers as a sequence with 2d, if we start with n = 0. For example, 6n + 1 produces the same primes as 3n + 1, while 6n + 5 produces the same as 3n + 2 except for the only even prime 2. The following table lists several arithmetic progressions with infinitely many primes and the first few ones in each of them.

We can generate some forms of primes by using an iterative method. For example, we can generate primes of the form 4n + 3 by using the following method:

Let a0 = 4(1) + 3 = 7. Then we let a1 = 4a0 + 3 = 4(7) + 3 = 31, which is prime. We continue by computing 4(7)(31) + 3 = 871 = 13(67). Because 4(7)(31) + 3 is of the form 4n + 3, either 13 or 67 is of the form 4n + 3. We have that 67 = 4(16) + 3 and is prime, so a3 = 67. We then continue this process to find successive primes of the form 4n + 3 (Silverman 2013).

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