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Theorem: For every positive integer $n$, there is a sequence of $n$ consecutive positive integers containing no primes. (Another MSE post about this Theorem)

Proof: Since we desire "a sequence of $\color{#1560BD}{n}$ consecutive positive integers containing no primes,"
thus denominate these $\color{#1560BD}{n}$ numbers as: $x + \color{#1560BD}{0}, x + \color{#1560BD}{1}, ..., x+(\color{#1560BD}{n - 2}), x+(\color{#1560BD}{n - 1})$.
Thus the objective is to prove: None of these are prime. $\equiv$ All of these are composite.

Define $x := (n + 1)! \color{ #FF4F00}{+ 2}.$ Then for all $\color{green}{0 \leq i \leq n - 1}$:
$$\begin{align}x \quad \color{green}{+ i} & = 1\cdot2\cdot3\cdot...\cdot(i + 1)(i+2)(i+3)...n(n+1)\color{ #FF4F00}{+ 2} \quad \color{green}{+ i} \\ & = (\color{green}{i} \color{ #FF4F00}{+ 2})\left[1\cdot2\cdot3\cdot...\cdot(i + 1)\quad(i+3)...n(n+1) \qquad + 1\right] \qquad \qquad\blacksquare \end{align} $$

How would you divine/previse to define $x := (n + 1)! \color{ #FF4F00}{+ 2}$ ?

Supplementary dated Jan 25 2014: $1.$ Yury's answer uncloaks the easier choice of $x := (n + 1)!$. Thus, why did Velleman add/be concerned with $\color{ #FF4F00}{+ 2}$ in his $x$ ?

$2$. Which variable in my question is $i > 1$ in Yury's answer? It differs from my $\color{green}{0 \le i \le n - 1}$?

$3$. Would someone please elucidate Yury's answer starting from "The problem now is ..."?

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    It works! And is the standard example used. Another explicit version that you can use is $\prod p_k + i$, where $p_k$ are all the primes less than $n$. – Calvin Lin Sep 21 '13 at 16:11
  • I guesss the fact that $;n!;$ is divisible by all the naturals $,1\le k\le n;$ makes it a quite natural candidate... – DonAntonio Sep 21 '13 at 16:19
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    We have to find $x$ s.t. $x+i$ is not prime. So $x+i$ must have a non-trivial factor. What could it be? We don't know much about $x+i$. The factor will depend on $i$. The most natural choice for the factor is $i$ (when $i>1$). Now $i$ divides $x+i$ if and only if $i$ divides $x$. The problem now is to find $x$ that is divisible by $2,\dots, n+1$. One option is to let $x=(n+1)!$. – Yury Sep 21 '13 at 16:20

1 Answers1

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(moved my answer from comments)

We have to find $x$ s.t. $x+i$ is not a prime number. So $x+i$ must have a non-trivial factor.
What could it be? We don't know much about $x+i$. The factor will depend on $i$.
The most natural choice for the factor is $i$ (when $i>1$).
Now $\color{#009900}{i}$ divides $x+\color{#009900}{i}$ if and only if $\color{#009900}{i}$ divides $x$.
The problem now is to find $x$ that is divisible by $2,\dots,n+1$.
One option is to let $x=(n+1)!$ and consider consecutive numbers $x+\color{#009900}2, \dots, x + \color{#009900}{(n+1)}$.

Caution: The $x = (n + 1)!$ in the last line differs from the OP's $x := (n + 1)! \color{ #FF4F00}{+ 2}$.

Yury
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  • +1. Thank you very much. I made an ancillary edit to distinguish your $x$ from mine. Is it right? Is there any reason why Velleman added $\color{ #FF4F00}{+ 2}$ to your $x$? Yours seems easier and more natural? –  Jan 07 '14 at 08:20
  • @LePressentiment $(n+1)!+1$ may be prime. For example, $1!+1$, $2!+1$, $11!+1$... So the 2 really is necessary here, and if you look closely, Yury is using the same set of numbers as Velleman. – Andrew Dudzik Jan 07 '14 at 10:06
  • @User-33433: Thank you. Would you like to rewrite your comment as an answer for which I can upvote? I'll ruminate over this some more. Also, how did you recognise that $11! + 1$ is prime? I wouldn't think that you had computed it? –  Jan 12 '14 at 12:03
  • @LePressentiment I think this answer is already fine; I'm mostly just clarifying it. As for $11!+1$ (and $150209!+1$, for that matter), I found it here: http://en.wikipedia.org/wiki/Factorial_prime – Andrew Dudzik Jan 12 '14 at 18:53
  • @User-33433: Thank you for your comment. I'm still bewildered so would be grateful if you would please answer my supplementary in my OP? –  Jan 25 '14 at 14:16
  • @LePressentiment Your $i$ and Yury's $i$ differ by $2$. But you are saying the same thing. Try it out with small values of the variables, and all will be clear. – Andrew Dudzik Jan 25 '14 at 21:15