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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.

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