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I had this question at the start of my number theory class so I think it is supposed to be an easy one but I did not receive a solution. Here is the problem:

Let $n\in \mathbb{N}$ be arbitary. Prove that there exists a number $k\in\mathbb{N}$, such that $k + 1, k + 2, k+3, . . ., k+n\;$ are not primes.

I read somewhere that the gaps between the consecutive primes do not tend towards infinity so is this not a wrong statement? This is what confused me. Can someone shed some light on this?

lexhople
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    Take $k=(n+1)!+1$. –  Sep 01 '20 at 14:35
  • This smells like Wilson's Theorem: https://en.wikipedia.org/wiki/Wilson%27s_theorem – 0Interest Sep 01 '20 at 14:37
  • "Terry Tao, Ph.D. Small and Large Gaps Between the Primes" https://www.youtube.com/watch?v=pp06oGD4m00 – Marek Kryspin Sep 01 '20 at 15:04
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    Insofar as the real question here is what's asked in the final paragraph, I would say this is not a duplicate. Even if you know how to solve the assigned problem, one might still be confused by an apparent conflict with the notion that the gaps between primes do not tend to infinity. – Barry Cipra Sep 01 '20 at 15:36

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Let $g_n$ be the sequence of gaps between consecutive primes. The problem posed is to show that $\limsup_{n\to\infty}g_n=\infty$. The Twin Prime conjecture, on the other hand, says that $\liminf_{n\to\infty}g_n=2$. The 2013 breakthrough theorem of Yitang Zhang says that $\liminf_{n\to\infty}g_n\le70{,}000{,}000$ (and follow-up work by others has reduced the upper bound to $246$). Conjecturally, $g_n$ takes every even value infinitely often, but Zhang's theorem alone is enough to show that $g_n$ does not tend to infinity, even though its limsup does.

Added later: Just to be clear, I am answering what I take to be the OP's main question, which is why the assigned problem does not conflict with their having read that the sequence of gaps does not tend to infinity. As the OP surmised, the assigned problem is a relatively easy one (and as such would be a duplicate).

Barry Cipra
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  • My thanks to the downvoter who reversed their opinion of the answer. (There were three downvotes before the edit, now only two.) – Barry Cipra Sep 01 '20 at 15:41
  • I do not understand the downvotes for this excellent answer. (+1) – Peter Sep 01 '20 at 18:48
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    @Peter, thanks for your comment. My best guess is that the downvoters thought I was posting an unnecessarily advanced comment instead of a solution to an elementary problem. As indicated, one person undid their downvote after my edit (and may have changed it to an upvote); I like to think the other two simply haven't seen the edit. I don't normally comment on downvotes (and certainly not on upvotes!), but since three of them came quite quickly, I made an exception this once. – Barry Cipra Sep 01 '20 at 19:40
  • The exist an near exact method of determining the number of primes in a given interval given by \begin{equation}\pi (x) = \frac{2x}{\log_{2\times 1.05^{ \frac{1}{2x}} } 2x} \end{equation} It is a breakthrough method for more details see https://ssrn.com/abstract=4828468 – Samuel Bonaya Buya Jan 06 '25 at 06:32