# Show that $4((n-1)!+1)+n$ is divisible by $n(n+2)$ [duplicate]

I have to prove the following statement:

If $$n$$ and $$n+2$$ are prime numbers, then $$4((n-1)!+1)+n$$ is divisible by $$n(n+2)$$.

My approach is to show that $$4((n-1)!+1)+n$$ is divisible by $$n$$ and by $$n+2$$ seperately and using the chinese remainder theorem I can conclude that it is divisible by the product $$n(n+2)$$. I have already managed to show that $$4((n-1)!+1)+n\equiv 0$$ mod $$n$$, hence divisible by $$n$$, but I can't figure out how to show that $$4((n-1)!+1)+n\equiv 0$$ mod $$n+2$$.

I tried expanding the term to $$4\cdot (n-1)!+4+n$$ and working with that but I'm not getting anywhere.

Any help is appreciated.

• – Martin R Oct 3 at 11:16