March 27, 2015

Homework Help: A number thoery question

Posted by kate on Wednesday, November 15, 2006 at 6:04pm.

Please help me! Thank you very much.

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.

* 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.

The 200 page proof overloads the buffer, sorry!

Can you show me the web link to this proof please?

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