## Wilson's sexy twin cousin? and new function?

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• I am reading The Little Book of Big Primes by Paulo Ribenboim. I m on page 145 trying to figure out the steps for proving: (== means congruent) 4((n-1)!+1)+n
Message 1 of 2 , Jan 6, 2004
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I am reading The Little Book of Big Primes by Paulo Ribenboim. I'm on
page 145 trying to figure out the steps for proving:
(== means congruent)

4((n-1)!+1)+n == 0 (mod n(n+2))

the first step is stating Wilson's theorem as the following:

(n-1)!+1 == 0 (mod n) which is has been proved earlier in the book.

the next thing is one of the places where I am getting lost:

4(n-1)!+2 == 0 (mod (n+2))

It seems like this statement was pulled out of the blue. Where did it
come from?

I see that the '+2' inside the mod function and the (n-1)! comes from
dealing with the prime (p+2) with Wilson's, but why where did the term
4 (n-1)! and the +2 come in? It would seem like 'something'(n-1)! +3
would be added but for what ever reason the 4(n-1)! and +2 is used

Things get better for a little bit, then he starts talking about the
converse.

There he states Wilson's for n and (n+2) then states:
"But n(n+1)=(n+2)(n-1)+2, so 2(n-1)! +1 = k(n+2), where k is an
integer." On the first eq. I see that the n(n+1) is needed for the
(n-1)! to be changed into the term(n+1)! and the right hand side on
both eq's are needed for dividing by (n+2). The left hand of the
second eq. is where I am lost the second time. Where did this come
from? It looks like the first one times 2, which begs the question why
not 2 instead of 4 for this term?

Now for what I would like to do.

My thought is that there is some way to gereralize this so it works
for cousins (n+4), sexies (n+6), and even better sexy twin cousins.
hee hee.

I see that I can use n(n+1)(n+2)(n+3) = (n-1)(n+1)(n+2)(n+4)+4 for
cousins. But what would is the form of second eq. from above?

{my guess for the sexy is n(n+1)(n+2)(n+3)(n+4)(n+5) =
(n-1)(n+1)(n+2)(n+3)(n+4)(n+6)+6 or (n+5)!/ n! =
(n+6)!/((n-2)!*n*(n+5))+6 }??

Is there a function simply reduces n! to factors which with work can
use Fermat's Little Theorum as to use (mod P#)?

Something like
P = 11, and 6! = 2^4*3^2*5
so 2^k(1) (mod 2310) + 3^k(2) (mod 2310) + ... 11^k(8) (mod 2310) == x
(mod 2310) which just finds x? Note that P and n! are independent
varibles.
• ... It s a sloppy change of variables. Set m+2=n then (m+2-1)! +1 == 0 (mod m+2) (m+1)(m)(m-1)!+1 == 0 (mod m+2) (-1)(-2)(m-1)!+1 == 0 (mod m+2) 2(m-1)!+1
Message 2 of 2 , Jan 7, 2004
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--- In primenumbers@yahoogroups.com, "John W. Nicholson"
<johnw.nicholson@s...> wrote:
> I am reading The Little Book of Big Primes by Paulo Ribenboim. I'm
> on page 145 trying to figure out the steps for proving:
> (== means congruent)
>
> 4((n-1)!+1)+n == 0 (mod n(n+2))
>
> the first step is stating Wilson's theorem as the following:
>
> (n-1)!+1 == 0 (mod n) which is has been proved earlier in the book.
>
> the next thing is one of the places where I am getting lost:
>
> 4(n-1)!+2 == 0 (mod (n+2))
>
> It seems like this statement was pulled out of the blue. Where did
> it come from?

It's a sloppy change of variables. Set m+2=n then
(m+2-1)! +1 == 0 (mod m+2)
(m+1)(m)(m-1)!+1 == 0 (mod m+2)
(-1)(-2)(m-1)!+1 == 0 (mod m+2)
2(m-1)!+1 == 0 (mod m+2)
QED

Rick
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