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Number thory- fermet's little theorm

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    let a and b be integers that not divisible by the prime number p
    if a^p[itex]\equiv[/itex]b^p, prove that a^p[itex]\equiv[/itex]b^p mod p^2

    2. Relevant equations

    if a^p[itex]\equiv[/itex]b^p, prove that a[itex]\equiv[/itex]b mod p

    3. The attempt at a solution
    I already get that a[itex]\equiv[/itex]b mod p , then how can I get a^p[itex]\equiv[/itex]b^p mod p^2 under the a^p[itex]\equiv[/itex]b^p mod p
     
  2. jcsd
  3. Feb 15, 2012 #2
    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.

    ap ≡ (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 ≡ rp mod p2

    similarly , bp ≡ (np + r )p mod p2 ≡ rp 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
     
    Last edited: Feb 15, 2012
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