2017/07/18 by Acquaah Peter, Peter, Acquaah
Mathematics · #11A41 (Primary) #Analytic Number Theory Research #FOS: Mathematics #History and Theory of Mathematics #Mathematics and Applications #Number Theory (math.NT)
paper · pdf · doi:10.48550/arxiv.1707.05432
openalex publication_date 2017/07/18 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
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∈ ℕ,k∈ ℕ,ak=(a,a+1,...,a+k-1) and A=\ p1,p2,...,pn\ , a finite set of primes. Then the number of components of ak that are divisible by some prime in A is less than or equal to ∑_\substack d|P(A)
dgt;1(-1)ω( d) +1\lfloor (k)/(d)\rfloor +2n where ω( d) is the number of distinct prime divisors of d and P(A)=∏p∈ Ap. We claim that the +2n in the bound can be replaced with n, the 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∈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].