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
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
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 won't divide p.
So within the factors of , knowing that , is there any divisor of p ?
Within the factors of , knowing that , 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...