# Thread: Prime number and bonmial coefficient

1. ## Prime number and bonmial coefficient

Let p>0 be a prime number. prove that p divides the binomial coefficient

(p i)= p!/i!(p-i)! for each integer 0<i<p.

( p over i ) = P factorial over i factorial times (p minus i) factorial.

and use that to show that (a + b)^p= a^p+ b^p mod p

2. For the first part note that $(p!)=p[(p-1)!]$ and $\left( {p - 1} \right)! = \prod\limits_{i = 1}^{p - 1} i$.
The result follows from those two observations.

Then note that $\left( {a + b} \right)^p = \sum\limits_{k = 0}^p {\binom{p}{k}} a^k b^{p - k} = a^p + b^p + \sum\limits_{k = 1}^{p - 1} {\binom{p}{k}} a^k b^{p - k}$

3. ## prime numer and binomail coefficient

I am really thankful for the hints but i still cant go further with
shown observations. I would appreciate a little more elaboration. Thanks.

4. Hello,

To continue what Plato said, you should also know that a prime number has only 2 divisors : 1 and himself. Hence every integer in $\{2, 3, \dots, p-1 \}$ won't divide p.

So within the factors of $i!=2 \times 3 \times 4 \times \dots \times i$, knowing that $i, is there any divisor of p ?
Within the factors of $(p-i)!$, knowing that $p-i, is there any divisor of p ?

You can imagine that "p remains as a unique element and won't be divided/separated by anything in the denominator".

And I think that a bit of Gauss's lemma can help for the formal demonstration...

5. Originally Posted by JCIR
I am really thankful for the hints but i still cant go further with
shown observations. I would appreciate a little more elaboration. Thanks.
You have to prove that $p|{p \choose k}$(like Moo did), then you can continue in Plato's argument:
$\left(\sum\limits_{k = 1}^{p - 1} {\binom{p}{k}} a^k b^{p - k}\right) \,\, mod \, p = 0$---------------------since $p|{\binom{p}{k}}$