mod:a mod p is the remainder of dividing a by p
Fermat's little theorem: a ^(p- 1 )≡ 1 (mod p)
Premise: p is a prime number, and a, p are relatively prime
Coprime: a and p have the same factor of 1.
Let's take a look at what ≡ is:
a≡b(mod p) <=> a mod p=b mod p
Note: <=> Both sides are equal
Before the proof, the lemma is given:
( 1 ) If p, c are relatively prime, and a*c≡b* c(mod p)
proving process:
∵a*c mod p = b*c mod p
∴(a*c - b*c) mod p = 0
∴(a-b)*c mod p=0;
∴(a -b)* c is a multiple of p
∵p,c coprime
∴k*p*c mod p = 0
∴(a -b)=k*p // It is recommended that you push with a pen
∴(a -b)%p= 0
( 2 ) If a1, a2, a3, a4, am are a complete residual system of mod m, and m, b are coprime, then
b *a1,b*a2,b*a3,b*a4...b* am is also a complete residual system of mod m.
Complete Residual System: Take a number from each residual class modulo n to obtain a set of n numbers, called a complete residual system modulo n.
proving process:
Using proof by contradiction:
Assuming that there exists a b *ai≡b*aj(mod p), it can be proved by lemma ( 1 ) that ai≡aj(mod p)
So this assumption does not hold. So Lemma ( 2 ) holds.
Begin the proof of Fermat's little theorem:
0 , 1 , 2 , 3 , 4 ... p- 1 is the complete residual system of p
∵a,p coprime
∴a, 2 *a, 3 *a, 4 *a.......(p- 1 )* a is also a complete residue system mod p
∴1*2*3.........*(p-1)*a≡a*2*a*3*a......(p-1)*a (mod p)
∴ (p-1)! ≡ (p-1)!*a^(p-1) (mod p)
Make an appointment on both sides at the same time (p - 1 )!
a^(p-1)≡1(mod p)