I have found the claim that given that $n\geq2$, we have that the sequence of $n-1$ numbers $n!+2,n!+3,...,n!$ is made up of only composite numbers. Is there a proof of this? I found this pretty fascinating but I am not sure how to go around it. It seems to hold for the first few examples
$n=2$: $S=\{4\}$
$n=3$: $S=\{8,9\}$
$n=4$: $S=\{26,27,28\}$
On
For a given $n$, the $i$-th term is:
$$ i+n!=i(1+1\times 2 \times \cdots \times (i-1) \cdots (i+1)\times \cdots n) $$
Thus the $i$-th term is the product of $i$ and $(1+n!/i)$, thus it is a composite number.
On
I feel like I've answered this question before.
So $n!$ is divisible by $n$, right? Then $n! + n$ must also be divisible by $n$. This much should be obvious.
Since $$n! = \prod_{i = 1}^n i,$$ $n!$ must also be divisible by $n - 1$. In fact, $$\frac{n!}{n - 1} = (n - 2)! n.$$ Then $$\frac{n! + n - 1}{n - 1} = \ldots$$ Sorry, I forgot where I was going with that, I'm rushing to answer before this question gets closed on account of being a duplicate.
The point is that $n! + n - 1$ must be divisible by $n - 1$. And so we keep going down until to we conclude that both $n!$ and $n! + 2$ are even numbers. $n! + 1$ is divisible by $1$, but primes are divisible by $1$, so taht doesn't tell us much
$n!$ is a rather special integer. It is composed of 1 factor of each from the list $\{1,2,3, \cdots, n-1, n\}.$
So,
$$n \geq 2 \Rightarrow 2|n! \Rightarrow n!+2 \equiv 0 \mod(2)$$
$$ n \geq 3 \Rightarrow 3|n! \Rightarrow n!+3 \equiv 0 \mod(3)$$
$$\vdots$$
$$ n \geq k \Rightarrow k|n! \Rightarrow n! + k \equiv 0 \mod(k)$$
$$\vdots$$
$$ n \geq n-1 \Rightarrow n-1|n! \Rightarrow n! + (n-1) \equiv 0 \mod(n-1)$$
$$n \geq n \Rightarrow n|n! \Rightarrow n! + n \equiv 0 \mod(n).$$
The last two steps shown here are rather obvious but I want to make it reasonably clear why every term in the list is composite.
Note that one of the nice things about this is that we have shown that there exists arbitrarily long lists of consecutive composite integers!