Lecture 22

Lecture 22

Proof of Fermat's (Little) Theorem

The form of the theorem that we will use is

a^p=a (mod p)

for every prime number p and every integer a < p. The following proof is based on induction, and the binomial theorem.

Proof of Fermat's Little Theorem:
First, note that the theorem is obvious if a=1, since 1^p=1, and 1 = 1 (mod p)

Next, assume the theorem is true for a>1.

Note: the binomial expansion of (a+1)^p is given by

(a+1)^p = a^p + pa^p-1 +p(p-1)/2! a^p-2+...+pa+1

Consequently, after subtracting the first and last terms of the right hand side from the left, we have (a+1)^p-a^p-1 = 0 (mod p), since each of the remaining factors is divisible by p. (Note: the leading term p in p(p-1)(p-2)...(p-i+1)/i! cannot be cancelled by the denominator if p is prime).

By the induction hypothesis, a^p-a=0 (mod p), therefore the sum

(a+1)^p-(a+1) =
(a+1)^p-a^p-1+a^p-a = 0 (mod p)

The conclusion that we reach is that if a^p-a != 0 (mod p) then p is composite (not prime). This provides a quick test.

Choose a large number p.
Choose another random integer a < p.
Calculate a^p mod p.
If you do not get a, then p is prime.
If you get a^p = a(mod p), then try another random integer a < p.
Keep trying.
If it fails, for any number a < p, then p is composite.

On the other hand, if a^p = a(mod p) for all a < p, that does not mean that p is prime. p=561=3*11*17 is a counter example. It is the smallest such number.

Other proofs of Fermat's little theorem.