Dirichlet's theorem states that for any coprime $k$ and $m$ there exists infinitely many primes $p$ such that $p \equiv k \pmod m$.

Some special cases of this theorem are easy to prove without any analytic methods. Those cases include, for example, $m=4, k=1$ and $m=4, k=3$.

Both cases could be proved by considering first $t$ prime numbers $p_i \equiv k \pmod m$ and constructing a new number which is proved to have prime divisor $p \equiv k \pmod m$ that is not equal to any $p_i$.

For case $m=4, k=1$ we can consider number $(p_1 p_2 \cdots p_t)^2 + 1$. And for case $m=4, k=3$ number $4p_1 p_2 \cdots p_k + 3$.

Those constructions could also be applied to some other special cases as well.

Are there any other special cases for which there exists a simple non-analytic proof which don't use any of those two constructions?

Dirichlet's theorem states that for any coprime $k$ and $m$ there exists infinitely many primes $p$ such that $p \equiv k \pmod m$.

Some special cases of this theorem are easy to prove without any analytic methods. Those cases include, for example, $m=4, k=1$ and $m=4, k=3$.

Both cases could be proved by considering first $t$ prime numbers $p_i \equiv k \pmod m$ and constructing a new number which is proved to have prime divisor $p \equiv k \pmod m$ that is not equal to any $p_i$.

For case $m=4, k=1$ we can consider number $(p_1 p_2 \cdots p_t)^2 + 1$. And for case $m=4, k=3$ number $4p_1 p_2 \cdots p_t + 3$.

Those constructions could also be applied to some other special cases as well.

Are there any other special cases for which there exists a simple non-analytic proof which don't use any of those two constructions?