- I don't suppose anyone can sketch a proof of it, could they?

A quick search on the interweb failed to yield anything, and I don't

remember seeing it in any of the books on my limited bookshelf at home.

I suspect it's in H&W. (DB - can you issue a challenge that's worth a

book to the winner where only I can solve it, please ;-) )

Phil

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http://mail.yahoo.com - Monday, September 04, 2006 6:21 PM [GMT+1=CET],

Phil Carmody <thefatphil@...> escribió:

> I don't suppose anyone can sketch a proof of it, could they?

I think there are two or more versions of the theorem, related I suppose. In

>

> A quick search on the interweb failed to yield anything, and I don't

> remember seeing it in any of the books on my limited bookshelf at

> home. I suspect it's in H&W. (DB - can you issue a challenge that's

> worth a book to the winner where only I can solve it, please ;-) )

>

"Introduction to Analytic Number Theory" of T. M. Apostol, there is this:

If p >= 5 is prime, then

S_{p-2} = Sum((p - 1)!/k, k, 1, p-1) = 0 (mod p^2)

Dem.:

That sum is the coefficient of -x in the polynome

g(x) = (x - 1)(x - 2)...(x - (p-1))

= x^(p-1) - S_1*x^(p-2) + S_2*x^(p-3) - ...

+ S_{p-3}*x^2 - S_{p-2}*x + (p-1)!

The roots of g(x) are 1, 2, ..., p-1. They also are roots of the congruence

h(x) = x^(p-1) - 1.

But then, the difference f(x) = g(x) - h(x) has degree p-2 and p-1 roots

module p. Then, for Lagrange's th., all of coefficients of f(x) are zero

(mod p) (Wilson's th. is a corolary).

Therefore, S_k, k = 1, .. ,p-2 are zero (mod p) and

g(p-1) = (p-1)! = p^(p-1) - S_1*p^(p-2) + ... S_{p-3}*p^2 - S_{p-2}*p +

(p-1)!

(here p > 3 is important)

===> S_{p-2}*p = 0 (mod p^3)

===> S_{p-2} = 0 (mod p^2) (q.e.d.)

Best regards,

Ignacio Larrosa Cañestro

A Coruña (España)

ilarrosa@... - --- Ignacio Larrosa Cañestro <ilarrosa@...> wrote:
> Monday, September 04, 2006 6:21 PM [GMT+1=CET],

...

> Phil Carmody <thefatphil@...> escribió:

> > I don't suppose anyone can sketch a proof of it, could they?

>

> I think there are two or more versions of the theorem, related I suppose. In

> "Introduction to Analytic Number Theory" of T. M. Apostol, there is this:

>

> If p >= 5 is prime, then

>

> S_{p-2} = Sum((p - 1)!/k, k, 1, p-1) = 0 (mod p^2)

> ===> S_{p-2} = 0 (mod p^2) (q.e.d.)

Many thanks Ignacio! The formulation of the theorem I have seen was different,

but closely related (namely the C(2p-1,p-1) one), and hopefully not too much of

a leap from the above.

In particular, I think I noticed that on Dave Rusin's archive of useful

sci.math.* posts, there's a discussion which seems to relate the two

formulations.

I thought I spotted a generalisation (of the other formulation), and wished to

prove it before relying on its truth!

Thanks again,

Phil

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