Number thory- fermet's little theorm

[yeland404]

Homework Statement

let a and b be integers that not divisible by the prime number p
if a^p$\equiv$b^p, prove that a^p$\equiv$b^p mod p^2

Homework Equations

if a^p$\equiv$b^p, prove that a$\equiv$b mod p

The Attempt at a Solution

I already get that a$\equiv$b mod p , then how can I get a^p$\equiv$b^p mod p^2 under the a^p$\equiv$b^p mod p

 

[blazeatron]

Hi ... i dono how to get the result a^p≡b^p mod p^2 using Fermat's theorem . But , the result can be proven using Binomial series.
Let a = mp + r , b = np + r where m and n are arbitrary integers.r is also an integer that denotes the remainder when a or b is divided by p.It is clear that our definition for a and b satisfy the intermediate result a≡b mod p = r.

a^p ≡ (mp + r )^p mod p²

Expanding RHS by binomial series and removing the terms which is exactly divisible by p^2 , we get ,

(mp + r )^p mod p² ≡ r^p mod p²

similarly , b^p ≡ (np + r )^p mod p² ≡ r^p mod p²

Thus , it is proved that a^p and b^p will produce the same remainder when divided by p^2

Hope that u can understand this explanation... if not , post me the steps that you didnt understand ... i ll try to explain in detail