Twin primes satisfy the congruence? I need a justification for my observation.
In general, we can list twin prime pairs in $(6n-1, 6n+1)$, where $n$ is some positive number. Of course, this is valid except $(3, 5)$. Now, I construct, for any such twin primes pair will satisfy the following my observation. 
$$4(6n-2)! = -3(1+2n)\pmod{ 36n^2 - 1}$$
Is my observation is true or not? I have checked for many pairs. Fortunately, the equation holds for any such twin pair. Could you explain the generalization of my statement or observation.
 A: By Wilson's theorem, $(p-1)!\equiv-1\bmod p$ if and only if $p$ is a prime. So assume that $6n-1$ and $6n+1$ are primes. Then $(6n-2)!\equiv-1\bmod(6n-1)$, and $(6n)!\equiv-1\bmod(6n+1)$. The second congruence can be rewritten as $(6n-2)!\equiv-1(-1)^{-1}(-2)^{-1}=-2^{-1}\bmod(6n+1)$.
Multiplying both congruences by $4$ yields
$$4(6n-2)!\equiv-4\bmod(6n-1)\;,\\4(6n-2)!\equiv-2\bmod(6n+1)\;.$$
Now we just have to check that these are indeed the residues of $-3(1+2n)$.
A: Writing your twin primes as $(p, p+2)$, your observation can be rewritten as:
$4(p-1)! = -4-p$ (mod $p^2 + 2p$).
This is known as Clement's Theorem.
Note, by the way, that this is an if and only if statement: the above relation is satisfied if and only if $(p, p+2)$ are twin primes.
(And yes, Wilson's Theorem is typically used to prove Clement's Theorem. Cf. Theorem 2 here.)
A: I will give few details about Clement's Theorem. The rest you can complete.
For n > 1, consider n, n + 2 are twin primes iff 4(n-1)!+1 +n = 0 (mod n(n+2)). From the Wilson’s theorem we have (n-1)!+ 1= 0 (mod n) . 4(n-1)! + 1 + n = 4(0)+n = 0 (mod n). Since n + 2 also prime, (n+1)!+ 1 = 0 (mod n+2) and 4(n-1)! + 1 +n = 2[2(n-1)!+2] + n = 2[(-1)(-2)(n-1)!+2]+n = 0 (mod (n+2)).
