Find $n$ for which $\frac{(n!)^2-(n+2)!}{(n+1)!}$ is an integer

Question

I think there is no integer $n$ for which the above expression is an integer, but I am not sure.

It is an integer for $n=5,7,8,9,11,13,14,15,17,19,20,\dots$. Indeed it is an integer for all $n\ge5$ such that $n+1$ is not prime (which are the integers for which $n+1$ divides $n!$. — Greg Martin, Apr 26 '16 at 21:09

GaussTheBauss · Answer 1 · 2016-04-26T22:30:57.117

6

With a little bit of work, you can show that

$$\frac{(n!)^2-(n+2)!}{(n+1)!} = \frac{n!}{(n+1)} - (n+2)$$

so any $n$ for which $(n+1)$ divides $n!$ is a solution.

See Greg Martin's comment above.

edited Apr 26 '16 at 22:30

answered Apr 26 '16 at 21:13

GaussTheBauss

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You mean $n+1$ divides $n! $. I think it is a typo! – Mhenni Benghorbal Apr 26 '16 at 22:30
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@MhenniBenghorbal Yes, Thank you! – GaussTheBauss Apr 26 '16 at 22:30
You are welcome! – Mhenni Benghorbal Apr 26 '16 at 22:31
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Note that $n+1\mid n!$ iff $n+1$ is not prime (see e.g. here). Cheers! – Bart Michels Apr 26 '16 at 22:44
@barto Yes! which has been pointed out by Greg Martin above. – GaussTheBauss Apr 26 '16 at 23:20

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