Show that a number is not prime? Show that for any integer $n>1$, all the numbers $n!+2, n!+3, \ldots, n!+n$ are composite (i.e. not prime).
 A: Hint:
Try to show, that if for two numbers $a$ and $b$, $a$ is divisible by $d$ and $b$ is divisible by $d$, then so is their sum. Then go looking for such a common divisor in your sums.
A: HINT $\rm\quad  k\: $ divides $\rm\: k\:m + k\:,\ $ and, of course, $\rm\ k\:$ divides $\rm\: n!\ $ for $\rm\:k =2,3,\:\ldots\:,n\:$.
A: $8! = 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8$ so combine $k|8!$ and $k|k$ to get $k|8!+k$

Definition: $a|b :\iff \exists k, a k = b.$
Theorem: $a|m$ and $a|n$ imply that $a|m+n$.
Proof: Our hypothesis says that $a k = m$ and $a h = n$, add this and use distributivity to get $a (k + h) = n + m$ which proves that $a|n+m$.
Note that if $k|b$ with ($k$ not $1$ or $b$) then $b$ has a divisor, so it is not prime.
A: Take an arbitrary number $n! + k$, where $n > 1, 1 < k \leq n$.
The following is the definition of the factorial function:
$$n! = \prod_{i=1}^n i$$
Hence every i where $1 \leq i \leq n$ is a factor of $n!$. Since the range k covers is within the range of i, this is clearly true for all k.
We know that k is a factor of $n!$, so let us say $n! = qk, k \in \mathbb{Z}$. Hence $n! + k = qk + k = (q + 1)k$, which implies q + 1 and k are integral factors of $n! + k$. QED
This is a reasonably rigorous proof. When you become more familiar with the concepts of divisibility and factorials the result should be fairly apparent from first sight.
