Proving Prime Divisibility for {p \choose k}

[zetafunction]

Homework Statement

the idea is to prove wether a prime 'p' divides the quantities $${p \choose k}$$

for k=01,2,3,...,p-1

Homework Equations

The Attempt at a Solution

i have tried by inspection for small primes 3,5,7,11,13,17,... but can not guess a simple solution , the idea is to see if for p prime , the Cyclotomic polynomial

$$x^{p}+...+x+1$$ for p prime.

 

[slider142]

Trying more general circumstances, where p is any prime, we get the simpler question of whether the product (p-1)...(p-n) is divisible by n+1 for all n <= p - 1.
Ie., for n = 1, we question whether p-1 is divisible by 2, which is trivial. (p-1)(p-2) divisibility by 3 can be seen as well with a little more care.
There is a basic divisibility theorem about these products that you may have already proven earlier. If not, it is not difficult to prove by a little inspection.

 

[tiny-tim]

Hi zetafunction!

the idea is to prove wether a prime 'p' divides the quantities $${p \choose k}$$

for k=01,2,3,...,p-1

If p is prime, how can it not divide ^pC_k (except for k = 0) …

there's a p on the top, and nothing larger than p-1 on the bottom.

 

[slider142]

Hi zetafunction!


If p is prime, how can it not divide ^pC_k (except for k = 0) …

there's a p on the top, and nothing larger than p-1 on the bottom.

This question may only be trivial in retrospect. ;) The quotient $\frac{(r-1)!}{(r-k)!k!}$ trivially simplifies to $\frac{(r-1)\cdots(r-k-1)}{k!}$ which depends on k! successfully dividing the product on the top. This is not generally true when r is not prime. Ie., 8 does not divide 8C4.
It is then up to the OP to show why this is true for primes, but not necessarily composites.