Difference between revisions of "Dirichlet's Theorem"
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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. | ||
==Stronger Result== | ==Stronger Result== | ||
| − | For any positive integers <math>a</math> and <math>m</math> such that <math>(a,m)=1</math>, | + | For any positive integers <math>a</math> and <math>m</math> such that <math>\gcd(a,m)=1</math>, |
<cmath> | <cmath> | ||
\sum_{\substack{p\leq x\\ p\equiv a\mod m}}\frac{1}{p}=\frac{1}{\phi(m)}\log\log x+O(1) | \sum_{\substack{p\leq x\\ p\equiv a\mod m}}\frac{1}{p}=\frac{1}{\phi(m)}\log\log x+O(1) | ||
Latest revision as of 11:22, 3 February 2025
Theorem
For any positive integers 
and 
such that 
, there exists infinitely many prime 
such that 
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 and such that ,
![\[\sum_{\substack{p\leq x\\ p\equiv a\mod m}}\frac{1}{p}=\frac{1}{\phi(m)}\log\log x+O(1)\]](http://latex.artofproblemsolving.com/8/0/e/80ef31b95ca7e5263e507c1dc1def8c05ca66183.png)
where the sum is over all primes less than 
that are congruent to mod , and 
is the totient function.
