TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Dirichlet's theorem on primes in arithmetic progressions

Statement

If $q \geq 1$ and $(a, q) = 1$ , then there are infinitely many primes $p \equiv a (m o d q)$ .

Facts & Assumptions

Given: A modulus $q \geq 1$ and a reduced residue class $a$ modulo $q$ .

[L1]

The reciprocal-prime sum in this progression satisfies $\sum_{p \leq x, p \equiv a (q)} 1 / p = φ (q)^{- 1} \log \log x + O_{q} (1)$ (Mertens sum for primes in an arithmetic progression).

Proof

technique · direct

1.1

By [L1], the partial sums $\sum_{p \leq x, p \equiv a (q)} 1 / p$ are unbounded, because $\log \log x \to \infty$ .

L1givenalgebra

2.1

A finite set of primes would contribute a bounded reciprocal sum. Therefore the set of primes congruent to $a$ modulo $q$ cannot be finite.

step 1.1algebra∎

Depends on

Mertens sum for primes in an arithmetic progression

Used by

A noncoprime residue class has no Dirichlet conclusion Counterexample
Global approximation of finitely many square classes Lemma

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, Theorem 4.2 (standard reference, not scraped)
Andrew V. Sutherland, Number Theory I, Theorem 18.1 (standard reference, not scraped)