- 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

instead of any other factor.

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. - --- 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

It's a sloppy change of variables. Set m+2=n then

> 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?

(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