Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ..."?

share|improve this question
1  
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
1  
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 Answer 1

(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 $x := (n + 1)! \color{ #FF4F00}{+ 2}$.

share|improve this answer
    
+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? –  LePressentiment Jan 7 at 8:20
    
@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. –  you-sir-33433 Jan 7 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? –  LePressentiment Jan 12 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: en.wikipedia.org/wiki/Factorial_prime –  you-sir-33433 Jan 12 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? –  LePressentiment Jan 25 at 14:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.