
> 
> I've grossly overstated things in my two posts before the last one. Thank you for providing references that have returned me to reality.
> 
> 
> 


**Conjecture** For arbitrary integers $\ 0 \le k \le m\ $ there exists
integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at
least one of the numbers


$$\ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$


where $\ k\le x < n$.



> 
> Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.
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> 
> 


An assembler-like equivalent formulation:


**Conjecture'**


$$ \forall\_{m\in\mathbb Z\_{\ge 0}}\,\forall\_{k\in 0..m}\,
\exists\_{n\in\mathbb Z\_{\ge m}}\,\forall\_{s\in\mathbb N}\,
\exists\_{x\in k..n\!-\!1} $$
$$ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$



> 
> Here (Perl notation),
> 
> 
> 


$$\ u..v\ :=\ \{x\in\mathbb Z:\ u\le x\le v\} $$



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.


