Show that if $n$, $n+2$ and $n+6$ are a prime triplet then
$4320(4((n-1)!+1)+n)+361n(n+2)\equiv0\ \pmod{ (n(n+2)(n+6)}.$
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1Use Chinese Remainder theorem and Wilson's theorem. – Math.StackExchange May 28 '14 at 10:47
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Maybe this helps $(m, m+2)$ is twin prime, iff $4((m-1)! + 1) \equiv -m \pmod {m(m+2)}$... – draks ... May 28 '14 at 11:02
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the equation is valid mod n(n+2) for the theorem in the previous comment, to see it stands mod n+6 and then combine them, use the fact that 6!(n-1)!=(n+5)! mod n+6, and wilson theorem i.e (n+5)!+1=0 the initial equation becomes 6!(24)(n-1)!+(24)(6!)+6(6!)n+361(4)=0 next important step, you notice 24(6!)= -4(6!)n since 4(6!)(n+6)=0 so we end up with 24(n+5)!+1440n -1444n =0, i.e 24(n+5)!-4n =0 but n=-6 so 24(n+5)!+24=0 which is true by multiplying 24 to wilsons theorem in mod n+6
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