Results 1 to 4 of 4

Math Help - Can anyone show me the proof for Fermat Last Theorem for n=3

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    83

    Can anyone show me the proof for Fermat Last Theorem for n=3

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by beta12 View Post
    Thank you very much.
    See here

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Ha ha. I actually have that page stored as my favorite.

    I cannot follow what they say, so poorly written. I do however know that proof for n=4 (I even have it memorized ).
    ----
    The problem is I think he wants it to be solved in terms of quadratic fields.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2006
    Posts
    83
    Hi Perfecthacker and Captainblack,

    I went to the above given link. I can't fully figure out the whole approach for the proof of Fermat Theorem n=3 after reading the information provided.

    I think my question wants me to solve it in different approach.
    Let me put down the whole question in here:

    Prove Fermat's Last theorem for n=3 : X^3 + Y^3 = Z^3
    where X, Y, Z are rational integers, then X, Y, or Z is 0.

    Hint:
    * Show that if X^3 + Y^3 = Epsilon* Z^3, where X, Y, Z are quadratic integers in Q[sqrt(-3)], and epsilon is a unit in Q[sqrt(-3)], then X, Y, or Z is 0. (recall how to find all solutions to X^2 + Y^2 =Z^2 since it similar.)

    *Begin with if X^3 + Y^3 = Epsilon* Z^3, where X, Y, Z are quadratic integers in Q[sqrt(-3)], where epsilon is a unit, then lambda divides X, Y or Z, where lambda is (3+sqrt(-3))/2. Also show that (lambda)^2 is an associate of 3.

    *It will be useful to show that if X is congruent to 1 mod lambda, then X^3 is congruent to 1 (mod lambda)^4). Work out a similar desciption for when X is congruent to -1 mod lambda.
    Use these fact to show that if X^3 + Y^3 = Epsilon* Z^3, if X and Y are not multiples of lambda, but Z is , then Z is a multiple of (lambda)^2. Do this by reducing (mod lamda)^4.

    note that X^3 + Y^3 = Epsilon* Z^3 can be factored:
    (x+y)(x+wy)(x+w^2*y)= Epsilon* Z^3 ,
    where w is an appropriately chosen quadratic integer.

    Consider each of these factors as quadratic integers p, q, r. Express x and y in terms of p, q, and r. The fact that we have three equations and two unknowns indicates there will be some extra constraint on p, q, and r.

    Consider how many time lambda occurs in the prime factorization of p, q, and r. Use unique factorization of Q[sqrt(-3)] to show that except for the factors of lambda that you computed , p, q, and r are cubes times units.

    Use the extra constraint on p, q, and r to find another solution
    X^3 + Y^3 = Epsilon* Z^3 ,
    where Z has one less factor of lambda. Note that it may be necessary to exchange the roles of x, y, z, -x, -y, or -z.

    Derive a contradiction along the lines of Fermat's method of descent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 10th 2011, 09:51 AM
  2. Replies: 2
    Last Post: October 30th 2010, 03:04 PM
  3. How do you show that a Fermat number is not prime?
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 21st 2010, 09:11 AM
  4. Replies: 1
    Last Post: March 9th 2009, 01:51 PM
  5. Fermat's Last Theorem: an amateur proof
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: December 19th 2008, 07:43 AM

Search Tags


/mathhelpforum @mathhelpforum