Hi,

I need help for proving :

let p be a prime, and a and b be any integers such

that a ≥ b ≥ 0. Prove that the binomial pa over pb is congruent modulu p to the binomial a over b.

Thank's in advance

Printable View

- March 9th 2014, 04:32 PMhedicongruence of binomial coefficients
Hi,

I need help for proving :

let p be a prime, and a and b be any integers such

that a ≥ b ≥ 0. Prove that the binomial pa over pb is congruent modulu p to the binomial a over b.

Thank's in advance - March 9th 2014, 04:52 PMromsekRe: congruence of binomial coefficients
not touching a zip file sorry, post the image if you like

- March 10th 2014, 08:40 AMjohngRe: congruence of binomial coefficients
Hi,

Here's the result you need: Legendre's Theorem - The Prime Factorization of Factorials. Just apply this to (pa)!, (pb)! and (p(a-b))! and it should be clear. - March 10th 2014, 10:15 AMjohngRe: congruence of binomial coefficients
Oops. Sorry, as far as I can tell, the previous post merely shows:

$$\nu_p{pa\choose pb}=\nu_p{a\choose b}$$

So one binomial is 0 mod p iff the other is also 0, but this does not solve the problem generally. - March 10th 2014, 12:24 PMhediRe: congruence of binomial coefficients
Yes,this was my difficulty.It can be solved by the binomial expansions of to p-congruent binomials.