Question

There exist an (infinite) non-constant arithmetical progression whose terms are all prime numbers
If true then enter $1$ and if false then enter $0$

Answer 1

The answer is negative. Indeed, if such progression exists, denote its common difference by r, and consider the consecutive r integers $(r+1)!+2,(r+1)!+3\cdots ,(r+1)!+(r+1).$ Each of them is a composite number, but since the progression has the common difference r, one out of any r consecutive integers must be a term of the progression.This is a contradiction.