Twin primes and modulo 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.
 A: 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.
A: 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 $-1\bmod p$.  
