Given an odd integer, $a$ does $\gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?

Question

Given an odd integer, $a$ does $\gcd(a,p-1)=1$ has infinitely many prime $p$ solutions?

One can argue that there are infinitely many numbers $x$ satisfies $\gcd(a,$x$)=1$. How to argue that there are infinitely many prime $p=x+1$.

Any hint?

Answer 1

By Dirichlet's theorem, there are infinitely many primes in the arithmetic series $2+na$, ($n\in \mathbb{N}$). If $p=2+na$ is such a prime, then $\gcd(a, p-1) = \gcd(a, 1+na) = 1$.