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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\ (\text{where}\ k\le x\le n)\ $ is not prime.

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

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Let $q=p(k)$. Using Dirichlet's theorem on primes in arithmetic progressions, there is $n\geq m$ large enough so that for any $a$ not divisible by $q$, there is some $k\leq x\leq n$ such that $p(x)\equiv a\pmod q$. Also taking $a=0$ and $x=k$, we see this is true for all residue classes mod $q$.

Take any natural $s$ (which presumably also requires $s>0$ for you). Pick some $k\leq x\leq n$ such that $p(x)\equiv -s\pmod q$ (which exists by construction). Then $p(x)+s$ is divisible by $q$ and greater than $q$.

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  • $\begingroup$ Simple and nice. ### (Frankly speaking, somehow, I made a huge "typo". I am leaving this question as it is but you may try my new "typo"-free version -- see a new post). $\endgroup$ – Wlod AA Sep 20 at 8:57
  • $\begingroup$ @WlodAA Do you mean including $k=0$? I don't see how that makes any difference really. My solution should still work. $\endgroup$ – Wojowu Sep 20 at 9:04
  • $\begingroup$ there was ASLO a huge typo, see my new post. That other typo, 0 < or = k, was trivial. $\endgroup$ – Wlod AA Sep 20 at 9:11
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    $\begingroup$ Ah sorry, I see you made a new post. I thought you just edited that one. I will take a look at the new one. $\endgroup$ – Wojowu Sep 20 at 9:19

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