Dirichlet's Theorem on arithmetic progressions is often stated as something like:
Every arithmetic progression where the first term and the difference are coprime contains an infinite amount of primes.
But can be rewritten as:
If $(a,m) = 1$ then there are infinite primes $p$ such that $p\equiv a\pmod m$.
I'm trying (fruitlessly) to come up with an elementary proof of its weak version.
If $(a,m) = 1$ then there is at least one prime $p$ such that $p\equiv a\pmod m$.
Note that if this is stated in terms of arithmetic progressions then it would not be a weak version (once you find one prime, consider the sequence with difference $m$ and first term $p + m$).