More precise conjecture for linear combinations of binomial coefficients

[Roland Bacher]

Here is a more precise (and slightly more general)
version of a nice conjecture presented yesterday by Max:

Let c_1,...,c_n \in Z be n integers.
Let k_1,..k_n \in Z be a solution of the linear system given by

(*)    sum_{j=1}^n k_j {c_j x \choose x}= 0

for x=1,3,5,..,2n-3 (where {c_j x \choose x} is a binomial
coefficient).

Conjecture: We have have then
 
sum_{j=1}^n k_j {c_j p \choose p}= 0 modulo p^{2n-1}

for all except finitely many primes p.

Remarks: - Choosing c_j=0 or an equality c_j=c_i is not interesting.

- For a generic choice of c_1,...,c_n, the linear system (*)
has a unique solution (k_1,..,k_n)\in Z^n, up to a constant.

- Multiplying a given solution (k_1,..,k_n) by a suitable integer,
the identity of the conjecture (if true) can be assumed to hold for all 
primes. Working with valuations, the identity makes of course sense
for rational solutions (k_1,..k_n)\in Q^n of (*).

Roland Bacher
 


On Thu, Aug 03, 2006 at 05:18:16AM -0700, Max A. wrote:
> On 8/3/06, Roland Bacher <Roland.Bacher at ujf-grenoble.fr> wrote:
> >
> >writing (in a kind of tex notation) and using Wilsons identity
> >(p-1)!=-1 modulo p for p prime:
> >
> >C(jp,p)=(jp-1)(jp-2)....(jp-(p-1))/(p-1)!=1-jp \sum_{h=1}^{p-1}
> >h\pm (jp)^2 \sum_{1\leq h_1<h_2<p} h_1h_2 \pm (jp)^3 \sum.... modulo p for 
> >prime p
> >
> >(modulo errors,)
> >
> >one can easily prove it for a given fixed value of n.
> 
> How easily?
> I know a proof for n=2. But for larger n similar proof become too much
> sophisticated.
> Can you write down a full proof, say, for n=4 ?
> 
> Thanks,
> Max