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Proof of Fermat's Theorm

  1. Jun 21, 2014 #1
    1. The problem statement, all variables and given/known data
    In proving Fermats Theorm, I got confused trying to follow one of the steps.

    Theorm: If p is prime, then:
    a^p~a (mod p).


    part of proof:
    multiply all the nonzero elements in Z_p by a to obtain
    [a][1],[a][2],.....,[a][p-1].
    These are all distinct elements and none equal [0], so they must be the set of all nonzero elements [1],[2],....,[p-1] in some order. In particular, the products are the same and we obtain
    ([a]^(p-1))([1],[2],....,[p-1]) = ([1],[2],....,[p-1])

    That fact that those last parts equal eachother is confusing me, help?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 21, 2014 #2
    Don't you mean ([a]^(p-1))([1][2]....[p-1]) = ([1][2]....[p-1])?

    And if that is what you mean, does your question still stand?
     
  4. Jun 21, 2014 #3
    Firstly, you can consider Euler's theorem too.
    So we know that the set Q = { a*a_1 , a*a_2, .., a*a_p-1 } includes all elements in the set B = {a_1, a_2, .., a_p-1 } in mod p.
    We know that from the property that
    " if (a,p) = 1, we can get rid of the a in the equation :
    a*a_i = a*a_j (mod p) ⇔ a_i = a_j (mod p) "

    Which means that all elements in Q are distinct. And they are equal to one and only one of the elements in B.

    So we get

    (a*a_1) * (a*a_2) * .. * (a*a_p-1 ) = a_1 * a_2 *..* a_p-1 (mod p)
    =>
    a^(p-1) * A = A (mod p)
    =>
    a^(p-1) = 1 (mod p).

    end of the proof.
     
  5. Jun 21, 2014 #4
    In addition to what is said above, you will need to state this restriction on the part you are doing: ##p## does not divide ##a##, or ##p \nmid a##.

    The case of p divides a, ##p \mid a##, will need to be discussed separately.
     
  6. Jun 21, 2014 #5
    As I told above;
    " if (a,p) = 1, we can get rid of the a in the equation :
    a*a_i = a*a_j (mod p) ⇔ a_i = a_j (mod p) "
    you can't get rid of a if you don't have (a,p) = 1 property. And that prop. is given in the theorem.
     
  7. Jun 21, 2014 #6
    This is true, it may be given in the assignment since this case is trivial.

    OP, the key here is that if p is prime, and a and b are both not multiples of p, then ab is not a multiple of p. Do you see why this makes the left product one of p-1 distinct classes none of which are [0]?
     
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