Well, I'll take a guess... perhaps (n-1)!+1 is not

congruent to 0 mod (n+2)?

Joseph.

--- "John W. Nicholson" <

reddwarf2956@...>

wrote:

>

>

> == means congruent

>

> Wilson's theorem for twins is

>

> Let n>=2. The integers n and n+2 form a pair of twin

> primes iff

>

> 4[(n-1)!+1] + n == (mod n(n+2))

>

> But,

>

> (n-1)!+1 == 0 (mod n)

>

> and by a Corollary of Fermat treorem with (a,n)=1,

>

> a^(n) - a == 0 (mod n)

>

> so,

>

> a^(n)-a == (n-1)!+1 == 0 (mod n).

>

> Therefore,

>

> 4(a^(n)-a) + n == 0 (mod n(n+2))

> or

> 4*a^(n) - 4*a + n == 0 (mod n(n+2))

>

>

> If it has not been stated before, I am surprised.

>

>

>

>

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