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Dirichlet's theorem on arithmetic progressions states that for positive integer $m$ and integer $a$ with $\gcd (a,m)=1$, infinitely many primes satisfy $p\equiv a(\mathrm{mod}m)$.

For example, each of the arithmetic progressions $4n+1$ and $4n+3$ contains infinitely many primes. The theorem establishes infinitude separately in every coprime residue class, rather than only in their union.

The coprimality condition is necessary. If $d=\gcd (a,m)>1$, every integer in the progression is divisible by $d$. A member can then be prime only when it equals $d$ and $d$ itself is prime. This leaves at most one prime value in that residue class. The difficult part of the theorem is sufficiency; its analytic treatment uses Dirichlet L-functions.

Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.

For every positive integer $n$, some imaginary quadratic field has an ideal class of order of a group element greater than $n$. Specifically, take a prime number $p\equiv 1(\mathrm{mod}20)$ with $p>5^n$. In $\mathbb{Z}[\sqrt{-p}]$, the Dedekind factorization theorem gives $(5)=P\overline{P}$, with distinct prime ideals $P=(5,\sqrt{-p}-2)$ and $\overline{P}=(5,\sqrt{-p}+2)$ of ideal norm five. If $P^k=(a+b\sqrt{-p})$ for $1\le k\le n$, the field norm equation $5^k=a^2+p{b}^2$ forces $b=0$. Then $k$ is even and $(a)=(5{)}^{k/2}=P^{k/2}{\overline{P}}^{k/2}$, contradicting unique factorization of ideals in a number field. Thus $[P]$ has order of a group element greater than $n$. The existence of suitable $p$ follows from Dirichlet's theorem on arithmetic progressions.