NT3

Prove that there are $n\,$ consecutive composite numbers, for any $n > 0\,$ .

Let $m = n + 1\,$ . Consider the sequence of $n\,$ numbers

$m! + 2, m! + 3, \ldots, m! + m\,$

Then for any $k\,$ such that $2 \leq k \leq m\,$

$m! + k\,$	$= \left(1\cdot 2\cdot\ldots\cdot (k - 1)\cdot k\cdot (k + 1)\cdot\ldots\cdot m\right) + k\,$
	$= k\left[\left(1\cdot 2\cdot\ldots\cdot (k - 1)\cdot (k + 1)\cdot\ldots\cdot m\right) + 1\right]\,$

Therefore, $m! + k\,$ is composite, and since $k\,$ was arbitrary, every number in the sequence is composite.

As an interesting side note, now consider taking $n\to\infty\,$ .