Dirichlet's theorem states that arithmetic sequence with first term and common difference relatively prime, contains infinitely many prime numbers.
Assume that we only want infinitely many numbers in such an arithmetic sequence such that every two of them are relatively prime - is it possible to prove it in an elementary way?
Yes. If $a_k = ak + d$ where $\gcd(a, d) = 1$, pick an increasing subsequence $b_k$ as follows. First, $b_1 = 1$. Next, $b_{i+1}$ is the smallest positive solution to the system of congruences
$$b_{i+1} \equiv d \bmod a$$ $$b_{i+1} \equiv 1 \bmod b_1 b_2 \dots b_i.$$
By hypothesis, $b_1 b_2 \dots b_i \equiv d^i \bmod a$, so $a$ and $b_1 b_2 \dots b_i$ are relatively prime and hence the above system has solutions by CRT. $b_{i+1}$, by construction, is relatively prime to $b_1, b_2, \dots b_i$.
