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I am so exited to learn math from this site. I posted the question today and I got good replies from members today itself. I will try to answer other number Theory questions in near future. With same confidence and motivation, I am sending TWO more questions to the members. These are also my observations and it may be done by simple proofs.

1) for a positive odd integer $p$, and $p_1$, $p_2$ - two different odd primes, and $p_1+p_2 - p = 1$ then $(p - p_1)!(p - p_2)! = -1 (\mod p)$ iff $p$ is prime.

2) For a prime $p > 7$, $(p,p +2)$ are said to be twin pair primes iff $4(p-3)! + (p + 2)$ divides $p$.

Please justify the above statements.

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    It seems you already know how to use $\TeX$ here. Please format the entire equations, not just the parts with subscripts; it makes for a rather bumpy reading experience when some parts are properly typeset and others aren't.2012-11-10
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    What do you mean by $p_1+p_2+p=1$? This can't hold in $\mathbb Z$, and if it's meant to be under the scope of $\pmod p$, then the summand $p$ is redundant. Also, how are you using "are said to be" in the second statement? It doesn't make sense to me as it stands. Do you mean to say simply that they *are* twin primes?2012-11-10
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    SHIT what a mistake I made. Please see my edited question 1.2012-11-10
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    @joriki!I perfectly edited my question number 1.2012-11-10
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    Why are you posting two different problems in the same thread? Also, are these problems something you are thinking about for a specific reason? It would help if you provided some context, particularly about what you have tried until now [see the guide on how to post good questions]. Have a good day!2012-11-10

2 Answers 2

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Statement 2) is strange --- $4(p-3)!+p+2$ can't possibly divide $p$, since it's bigger than $p$. Perhaps something has gone wrong during an edit.

For 1), first note that $$n!\equiv(-1)^n(p-1)(p-2)\cdots(p-n)\equiv{(-1)^n(p-1)!\over(p-n-1)!}\pmod p$$ so $$n!(p-n-1)!\equiv(p-1)!\equiv-1\pmod p$$ if $n$ is even and (using Wilson's Theorem) $p$ is prime. Now let $n=p-p_1$ (which is even) and note that $p_1-1=p-p_2$ to get one direction.

For the other direction, note that if $p$ is not prime, then it is divisible by some prime $q\le(p-1)/2$. We can't have both $q\gt p-p_1$ and $q\gt p-p_2$ (since that, together with $p_1+p_2-p=1$ would contradict $q\le(p-1)/2$), so $q$ must divide at least one of $(p-p_1)!$ and $(p-p_2)!$, whence their product can't be $-1\bmod p$.

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I do not know your statement (1), but your statement (2) is false.

Because your $$4(p-3)!+(p+2)\equiv0\pmod p$$ is actually equivalent to Wilson's theorem, i.e., $p$ is a prime iff $$(p-1)!+1\equiv0\pmod p$$

So when you choose $p=47$, then your statement (2) holds. However $p+2=49=7\times7$ is not a prime.

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    !Discuss the necessary and sufficient condition for except the case p = 7. in my second problem.2012-11-10
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    ! Kindly look my edited post.2012-11-10
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    It is an exception for $p=47$? I am not talking about $p=7$ case. If you are still not convinced try $p=103$, you will have $4(103-3)!+(103+2)\equiv0\pmod {103}$. However $p+2=105=3\times5\times7$, so another counter example because $(103, 105)$ is not a twin prime pair, but you stated that your condition is necessary and sufficient.2012-11-10