A new proof of Dirichlet's theorem concerning prime numbers in arithmetic progressions

TL;DR

This paper presents a new elementary proof of Dirichlet's theorem on primes in arithmetic progressions using sieving techniques and bounds on the prime counting function, avoiding complex analytic methods.

Contribution

It introduces an elementary sieving argument and a novel bound on divisibility counts to prove Dirichlet's theorem, improving understanding of prime distribution in arithmetic progressions.

Findings

Provides a new elementary proof of Dirichlet's theorem

Establishes bounds on the number of divisible components in integer intervals

Suggests potential for tighter bounds in prime divisibility estimates

Abstract

It is known that there are infinitely-many prime numbers which take the form of a polynomial of degree one with integer coefficients, this is Dirichlet's theorem. We use an elementary sieving argument together with bounds on the prime number counting function to provide a new proof of Dirichlet's theorem. We show that if $a\in \mathbb{N},k\in \mathbb{N},a_{k}=(a,a+1,...,a+k-1) $ and $A=\left\{ p_{1},p_{2},...,p_{n}\right\} $, a finite set of primes. Then the number of components of $a_{k}$ that are divisible by some prime in $A$ is less than or equal to $$ \sum\limits_{\substack{ d|P(A)\\ d>1}}(-1)^{\omega \left( d\right) +1}\left\lfloor \frac{k}{d}\right\rfloor +2n $$ where $\omega \left( d\right)$ is the number of distinct prime divisors of $d$ and $P(A)=\prod_{p\in A}p$. We claim that the $+2n$ in the bound can be replaced with $n$, the \texttt{best possible bound}. However, we did not demonstrate our claim in this paper since the $+2n$(bound) is enough for the new proof of Dirichlet's theorem. This result effectively means that given $[1,x], x\in\mathbb{R}$; if the primes in $A$ divide $h$ integers in $[1,x]$ then for every $g>0$, they will divide at most $h+2|A|$ integers in $[1+g,x+g].$