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Let $n$ and ($n+2$) be two prime numbers. If any real value of $n$ satisfies that condition, then prove that $$\frac{4{[(n-1)!+1]}+n}{n(n+2)} = k$$ where $k$ is a positive integer.

SOURCE: BANGLADESH MATH OLYMPIAD (Preparatory Question)

My Attempt:

I know a basic formula of modular arithmetic only and that is:

$(n-1)! \equiv -1\pmod n$ ....(but I don't know why it's true)

However, out of curiosity and after gaining nothing but few knowledge about modulus, I noticed that $(n-1)! +1$ would be divisible by n. Moreover, we will get same result from $4[(n-1)!+1]$. Similarly, $4[(n-1)!+1]+n \equiv 0 \pmod n$.

But how can I show the divisibilty of the above term by $(n+2)$ and in the same case what will be its remainder? Further, how can I prove the above condition and what is the essence that $n$ and $(n+2)$ must be supposed to be prime number and nothing else? And if they are not so, can't the desired proof be solved?

Any reference according to post or small hint will be very helpful for me. Thanks in advance.

Anirban Niloy
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    No need to apologize, this is a standard process here - if you don't find it, there is chance others will. And still that does not guarantee that found question has information you need, in which case you can post new question with specific details you want to be addressed (ideally with referring to that question and why it does not address it). – Sil Feb 21 '19 at 08:13
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    @Sil I've got all the information whatever I need in that question and I've no specific detailes even. So, I would better let the question remaining duplicate. – Anirban Niloy Feb 21 '19 at 08:25

2 Answers2

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The hint: $$(n+1)!+1\equiv0\pmod{n+2}.$$

Jyrki Lahtonen
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Michael Rozenberg
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If $P$ is a prime then, $P|(P-1)!+1$,or $P|(P-3)!(P²-3P+2)+1$.So these means $P|2(P-3)!+1$.

In your question $N+2$ is prime. If you arrange the numerator then you get $2[2(N-1)!+1]+N+2$.

Now as $N+2$ is a prime then $(N+2)|2(N-1)!+1$. So, $N+2$ the divides the numerator.

Alapan Das
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