Arithmetic Progression with Infinitely Many Primes

Question

What is an example of an arithmetic progression in the form bk+r, where b,k,r are integers such that b is not 2, 3, or 6, for which the arithmetic progression has infinitely many primes? The only arithmetic progression I have seen with infinitely many primes is 6k+5, but are there others with the condition stated?

Answer 1

One additional easy case is $4k+3$:

If $p_1,\dots,p_n$ are primes, then $4p_1\cdots p_n-1$ has at least one prime factor of the form $4k+3$ and is not equal to any $p_i$.

The case $4k+1$ requires a little more number theorem:

If $p_1,p_2,\dots,p_n$ are primes of the form $4k+1$ then any prime divisor $p$ of $4(p_1p_2\cdots p_n)^2+1$ is also of that form, since $-1$ is a square modulo $p$.

There is a way to generalize this to the any $bk+1$, using cyclotomic polynomials.

For example:

When $b=12$, we have the cyclotomic polynomial $\Phi_{12}(x)=x^4-x^2+1$. Then if $p_1,p_2,\dots, p_n$ are primes, then any prime factor $p$ of $\Phi_{12}(12p_1\cdots p_n)$ is of the form $12k+1$, and not equal to any of the $p_i$.

Dirichlet proved the most general result that if $b,r$ are relatively prime, there are always infinitely man primes of the form $bk+r$.

Answer 2

Since all primes except $2$ are odd, there are infinitely many primes of the form ???. Since all primes of the form $6k+5$ are of the form $3(2k)+5$ what does that tell you? In fact, there are infinitely many in any arithmetic progression of the form $bk+r$ when $b$ and $r$ are coprime, but that is a much more difficult theorem.