Difference between revisions of "Dirichlet's Theorem"

(Stronger Result)
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==Theorem==
 
==Theorem==
For any positive integers <math>a</math> and <math>m</math> such that <math>(a,m)=1</math>, there exists infinitely many prime <math>p</math> such that <math>p\equiv a\mod m</math>
+
For any positive integers <math>a</math> and <math>m</math> such that <math>\gcd(a,m)=1</math>, there exists infinitely many prime <math>p</math> such that <math>p\equiv a\mod m</math>
  
 
Hence, for any arithmetic progression, unless it obviously contains finitely many primes (first term and common difference not coprime), it contains infinitely many primes.
 
Hence, for any arithmetic progression, unless it obviously contains finitely many primes (first term and common difference not coprime), it contains infinitely many primes.

Revision as of 11:22, 3 February 2025

Theorem

For any positive integers $a$ and $m$ such that $\gcd(a,m)=1$, there exists infinitely many prime $p$ such that $p\equiv a\mod m$

Hence, for any arithmetic progression, unless it obviously contains finitely many primes (first term and common difference not coprime), it contains infinitely many primes.

Stronger Result

For any positive integers $a$ and $m$ such that $(a,m)=1$, \[\sum_{\substack{p\leq x\\ p\equiv a\mod m}}\frac{1}{p}=\frac{1}{\phi(m)}\log\log x+O(1)\] where the sum is over all primes $p$ less than $x$ that are congruent to $a$ mod $m$, and $\phi(x)$ is the totient function.

See Also