devotion & mathematics

   Wilson’s Theorem :   Statement :   If p is a prime then (p-1)! + 1 = 0 ( mod p ) Proof :          Suppose p is a prime number..     For p=2, we have (p-1)! + 1 = 2 and 2 ≡ 0 ( mod 2 )    i.e. the statement is true for p=2.   Let p > 2 and a 𝟄 Z such that 1 ≤ a ≤ p-1.   Since p is prime , we have (a,p) = 1 and hence the linear congruence ax ≡ 1 ( mod p) has    unique solution, say x 0 .   Let a’ 𝟄 x 0 — and   1 ≤ a’ ≤ p-1 where x 0 — is the congruence class of x 0 . Clearly aa’ ≡ 1 ( mod p ) a’ = a ð   a 2 ≡ 1 ( mod p ) ð   p / a 2 -1 ð   p/(a-1) or p/ (a+1)   for p/(a-1) and a > 0 =>   a+1 = p ð   a = p-1   ∴ a’ = a ð   either a = 1 or a= p-1 similarly if a’ ≠ a then a 𝟄 { 2,3,…,p-2}. ∴ the distinct a , a’ belong to the set { 2,3,…,p-2} containing (p-3) elements. Th...

Fermat's Theorem : Statement :                       If  p is a prime and (a,p) = 1 then a p-1 ≡ 1 (mod p). Proof :                Suppose p is a prime number and (a,p) = 1.         Since (a,p) = 1 then the numbers a,2a,3a,…,(p-1) are divided by p and the remainders are 1,2,3 …,                 (p-1) ; not necessarily in this order.          Let a ≡ r 1 ( mod p), 2a ≡ r 2 ( mod p) , … , (p-1)a ≡ r p-1 ( mod p)         Since r 1 , r 2 , … r p-1 are remainders obtained when a,2a,…,(p-1)a are divided by p.         ∴ r 1. r 2 . … .r p-1 = 1 . 2 .3 …. . (p-1).         Multiplying the above congruent relations : a . 2a. … . (p-1) ≡ r 1 r 2 …r p-1 ( mod p ) ð   {1.2…....