# Proving A Theorem Concerned With Prime Numbers

I am in the process of reading this brilliant little book Prove It: A Structured Approach--very brilliant, have I mention that already?

Anyways, here is the theorem: For every positive integer $n$,there is a sequence of $n$ consecutive positive integers containing no primes.

I understand the prove, for the most part; but there is a little bit giving me difficulty.

Here is the entire proof:

Suppose $n$ is a positive integer. Let $x=(n+1)!1+2$.We will show that none of the numbers $x,x+1,x+2,\ldots,x+(n - 1)$ is prime. Since this is a sequence of n consecutive positive integers, this will prove the theorem. To see that $x$ is not prime, note that $x=1\cdot2\cdot3\cdot4\cdots(n+1)+2\rightarrow x = 2\cdot(1\cdot3\cdot4\cdots(n+1)+1)$ Thus, $x$ can be written as a product of two smaller positive integers, so $x$ is not prime. Similarly, we have $x+1=1\cdot2\cdot3\cdot4\cdots(n+1)+3\rightarrow x=3\cdot(1\cdot2\cdot4\cdots(n+l) +1)$, so $x+1$ is also not prime. In general, consider any number $x+i$, where $0\leq i \leq n-1$. Then we have

$$x+i = 1\cdot2\cdot3\cdot4\cdots(n+1)+(i +2)$$ $$x+i= (i+2)\cdot(1\cdot2\cdot3\cdots(i+1)\cdot(i+3)\cdots(n+1)+1)$$

The last step is the one I can't quite apprehend. I see that $(i+2)$ is factored out, but I why are there remaining $i$'s in the second factor?

-
For example, if $i=5$, $i+2=7$, $$x + 5 = 7 \cdot (1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 8 \cdot \ldots \cdot (n+1)) + 1)$$ The $6$ is $i+1$ and the $8$ is $i+3$. –  Robert Israel Nov 21 '12 at 21:30
Robert provided a nice response. If you are interested in other proof like books, read my response here: math.stackexchange.com/questions/187340/… –  Amzoti Nov 21 '12 at 21:45
They're just pointing out that $(i+2)$ is skipped in the product. For instance you can write $n! = 1 \cdot 2 \cdots n$, but if you want to remove say $k < n$ from that product you can do $1 \cdot 2 \cdots (k-1) \cdot (k+1) \cdots (n-1) \cdot n$. –  roliu Nov 21 '12 at 22:50
The book may be very good. But it has obscured a nice idea. –  André Nicolas Nov 22 '12 at 2:09
@AndréNicolas Really? Would you mind pointing out this obscured idea, please? –  Mack Nov 22 '12 at 13:34

$i+2$ is factored out and that's the reason why there is written the rpoduct of all numbes from $1$ to $i+1$, inclusive, followed by the product of all numbers from $i+3$ to $n+1$ inclusive. Actually, I find it more natuurally to obsere that $x+i$ is $i+2$ more than $(n+1)!$ and the factorial is of course a nontrivial multiple of $i+2$.

-

Hint $\$ More clearly, let $\rm\:m = (n\!+\!1)! > 0.\$ Then

$\qquad\quad\rm 2\mid m\:\Rightarrow\:2\mid m\!+\!2\ne 2\:\Rightarrow\:m\!+\!2\:$ not prime
$\qquad\quad\rm 3\mid m\:\Rightarrow\:3\mid m\!+\!3\ne 3\:\Rightarrow\:m\!+\!3\:$ not prime
$\qquad\quad \ \ldots\qquad\qquad\ \ldots\qquad\qquad\qquad\ldots$
$\quad\ \ \rm n\!+\!1\mid m\:\Rightarrow\:\quad\quad\,\ldots\quad\quad\,\Rightarrow\:m\!+\!n\!+\!1\:$ not prime

So $\rm\: m\!+\!2,\,m\!+\!3,\ldots,m\!+\!n\!+\!1\:$ are $\rm\:n\:$ consecutive nonprimes.

-