My teacher only wrote this on the board.
\begin{align*} k \mid (n + 1)! &\implies k \mid (n + 1)! + k \\ &\implies (n + 1)! + k\text{ is not prime} \end{align*}
Not exactly sure what he means by this and how it relates to the problem.
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$$(n+1)!+2,(n+1)!+3,\ldots (n+1)!+(n+1)$$ are all composite. This is because
$$ a|b \text{ and }a|c \implies a|b+c$$
$ \\ $
For example, for $n > 2,\ 3|(n+1)!\ $ and $\ 3|3 \ $ so $ \ 3|(n+1)!+3$
David Reed
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2$n!+1$ doesn't need to be composite. – Andrés E. Caicedo Dec 07 '17 at 00:54
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yes typo. will fix right now – David Reed Dec 07 '17 at 00:54
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1@AndrésE.Caicedo good catch. Thank you. Been very sedated this week because of post surgical pain medication. – David Reed Dec 07 '17 at 00:59
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Setting $x_k:=(n+1)!+k$ for $k=2,\ldots,n+1$, we have that $x_2,\ldots,x_{n+1}$ are $n$ consecutive composite integers. Indeed, for each $k=2,\ldots,n+1$, we have that $$x_k=k\left(\frac{(n+1)!}{k}+1\right).$$
John Griffin
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