For a prime , show .

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- Aug 10th 2010, 07:20 PMchiph588@Binomial Coefficient
For a prime , show .

- Aug 11th 2010, 02:13 AMsimplependulum
I try not to consider the fact , i just regard it as the coefficient of a term in

We know is the coefficient of in the expansion of while is the coefficient of in the expansion of . Therefore , is the coefficient of of .

I would like to show that all the terms that the degrees are the multiple of prime are congruent to zero modulo .

I consider

Note that all the terms in are divisible by .

Then i consider the second polynomial , the annoying one .

I use the fact to simplify :

Then we are done , haha , let's get back the first polynomial whose degree ranges from to so they are prime to . Consider , we find that the degrees are all mutiple of so they become useless in the expansion if we are evaluating the coefficient of since no terms in could make the 'connection' to them . Therefore , the coefficient we are considering ( of ) is necessarily the mutiple of , one from and one from .

Remarks :

I don't like this result ... i am going to show that

For a prime we always have

:

Let's check my solution , write or

Then the coefficient of in is congruent to ( ) that in . It suffices to show that the coefficient of in is congruent to zero modulo .

ie :

.

Please see this post , http://www.mathhelpforum.com/math-he...nt-149645.html

It is not a help begging but a challenge , let's finish the proof by solving my challenge . - Aug 11th 2010, 10:15 AMNonCommAlg
let by the Fermat's little theorem, in we have

as a result which is Wilson's theorem, and for all in particular for any integer

thus and the result now follows from the trivial identity

**simplependulum**'s question is also solved by this method and using an old result due to Wolstenholmes which says that for

thus for all integers therefore for all primes and all integers we have - Aug 11th 2010, 07:02 PMsimplependulumQuote:

**simplependulum**'s question is also solved by this method and using an old result due to Wolstenholmes which says that for

thus for all integers therefore for all primes and all integers we have

For

Proof :

Assume

so we are going to show we have

Therefore , - Aug 12th 2010, 07:34 PMmeleseAn attempt proof.
For any integer let , then (*) ; this will be justified later.

With this we can write for some positive integer the following: , and so

(^) (I applied the congruence (*)).

Set for brevity, so and . Consider the following , the last equality is because . I write the main equality: (#)

Now computing (#) modulo and using result (^) we have ; . Since is relatively prime to , we can arrive at , or .

Everything depends now on proving (*). From the expansion of it can be noticed that taking from more than one parentheses gives terms that are divisble by , therefore , where .

Consider the polynomial . Then, , where . and because we have ; or . Divide by , so , and is divisible by .

We now can conclude that .

I hope I'm not missing something... - Aug 15th 2010, 09:50 AMPaulRS
We proceed by induction on , for it's trivial that for all .

Now assume the assertion holds for , let's show it the holds for . - in what follows we consider a > b since it's trivial otherwise.

We have

Note that if divides we are done, because by the inductive hypothesis and Pascal's identity.

Now . If we denote by the greatest integer r such that then:

;

since ;

where is the greatest r such that since - you can check this by using Polignac's formula -

Thus and so divides for all k = 1,...,p-1 .

But also is a multiple of p for and so we are done