Let $m$ is prime in form $13k+1$ and $ m \neq 8191$ Then $\dfrac{2^{m-1}-1}{8191m}$ is integer

Are there infinitely primes of the form 13k+1?

Yes, it is Dirichlet prime number theorem

RagvaloD wrote: Yes, it is Dirichlet prime number theorem What does that theorem say? Can't find anything about it.

@above Let $a,b$ be positive integers such that $\gcd(a,b)=1$. Then there are infinitely many primes in form of $an+b$. The proof is elementary but requires analytic arguments and some complex numbers.

For primes of the form $pk+1$, where $p$ is an odd prime, there is an elementary proof. Look at the set of prime factors of $\frac{(np)^p-1}{np-1}$ where $n$ runs over all positive integers, and show that this set is infinite.

RagvaloD wrote: Let $m$ is prime in form $13k+1$ and $ m \neq 8191$ Then $\dfrac{2^{m-1}-1}{8191m}$ is integer Hi I get it $8191=2^{13}-1$ and hence $8191|2^{13k}-1$ but i don't get how $m| 2^{m-1}-1$, do you mind elaborate a bit? Thanks.

oompaloompa1012 wrote: RagvaloD wrote: Let $m$ is prime in form $13k+1$ and $ m \neq 8191$ Then $\dfrac{2^{m-1}-1}{8191m}$ is integer Hi I get it $8191=2^{13}-1$ and hence $8191|2^{13k}-1$ but i don't get how $m| 2^{m-1}-1$, do you mind elaborate a bit? Thanks. That is Fermatt's little theorem for $m$ is odd prime .

there exists infinite number of primes in the form of 8190k+1 and 8191 is a prime therefore, by fermats little thm and set m to be a prime, we have 2^(m-1)=1 mod m and 8191 therefore, there exists infinite number of m st the number is an integer