A typo-free detail oriented prime conjecture

Like in the previous post, let $q=p(k)$ and let $n\geq m$ be such that primes $p(x),k\leq x<n$ cover all residue classes mod $q$.

Suppose this $n$ doesn't work. This means that $p(x+s+1)-p(x+s)=p(x+1)-p(x)$ for all $k\leq x<n$. Adding up a bunch of such equalities we get $p(x+s)-p(k+s)=p(x)-p(k)$ for all $k\leq x\leq n$. As $s>0$, we have $p(k+s)>q$ so it is indivisible by $q$. There is some $k<x<n$ such that $p(x)\equiv -p(k+s)\pmod{q}$, and we get $p(x+s)=p(k+s)+p(x)-p(k)\equiv 0\pmod q$, which is impossible as $p(x+s)>q$, hence a contradiction.