# Theorem of Primes in Arithmetic Progression [duplicate]

I'm am studying number theory and I came across this theorem by Dirichlet but the proof is not given and I do not know how to prove it

Can anyone help me?

$Theorem$ $of$ $Primes$ $in$ $Arithmetic$ $Progression:$ If $a$ and $b$ are integers, with $(a,b)=1$ so the arithmetic progression $an+b$, $n=1,2,3,...$ has infinite primes.

## marked as duplicate by davidlowryduda♦Oct 20 '17 at 5:24

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• many books have correct proofs. Not easy, and not something that would fit well in an answer box on this site. – Will Jagy Jul 3 '17 at 18:46
• This is not an easy result! In certain cases, such as $4n+3$, there are ways to get at it but the general theorem requires a lot of machinery. – lulu Jul 3 '17 at 18:48
• A classical reference is this book by Tom Apostol. There are very many references at MSE, e.g. here, here etc. Just look at the links on the right margin. – Dietrich Burde Jul 3 '17 at 18:48
• This is mentioned at one of Dietrich's links in a comment, several special cases that do not require any analytic approach: math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf – Will Jagy Jul 3 '17 at 19:43
• In general, the idea is that $F(s) =\displaystyle \sum_{p^k \text{ prime power }, p^k \equiv b \bmod a} p^{-sk}$ has the nice property that $F(s) =\frac{1}{\phi(a)} \sum_{\chi \bmod a} \overline{\chi(b)} \log L(s,\chi)$ where $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ are the Dirichlet L-functions. And hence to prove Dirichlet's theorem it is enough to prove $F(s)$ is not analytic at $s=1$. – reuns Jul 3 '17 at 20:16