How would one show that any given prime p_i must be a factor of some (p_j - 1)? Is that a true property of primes even?

Question

In short, what I'm asking is, if you were to go through the whole set of positive primes term by term and find for each prime p the prime factorization of (p - 1), whether all prime numbers would eventually show up as a factor of some (p_i - 1) term for some i. So the question I suppose can be phrased instead as whether for any prime p_i, there must exist some p_j such that p_i is a factor of (p_j - 1).

Answer 1

As stated in the comments, Dirichlet's theorem on arithmetic progressions implies that, for any prime $p,$ then since $p$ and $(p+1)$ are coprime, there is a prime of the form $p_j= (p+1) + np;\ $ in other words, $ p_j - 1\ $ is a multiple of $p.$ In fact, Dirichlet's theorem implies that there are infinitely many primes in the set $\{ (p+1) + np: n\in\mathbb{N}. \}.$