Fermat Little Theorm

#1

In this vide , why we doing A%p it will not change the solution . according to fermat little theorm , a^b =a^(b%(p-1) ) not in base (that is a)

#2

hello @premkiryanastore2016
we know x^(p-1) mod p => 1 if p is prime number…fermant little theorem

now to calculate (a^b) mod p .
we can write b = (p-1) + (p-1) + (p-1) … + k

where k is remainder we are left with. i.e k=b%(p-1)

substituting value of b
(a^b)mod p = a^(p-1 + p-1 + p-1 + p-1 … + k) mod p

now using property a^(x+y) => a^x * a^y

=> ( a^(p-1) * a^(p-1) * a^(p-1) … a^k )mod p

now we know a^(p-1) mod p =1

=> 1 * 1 * 1 …a^k mod p
=> a^k mod p
substituting value of k
=> a^(b%(p-1) ) mod p