Fermat's Last Theorem for n = 3

Date: 12/14/98 at 03:44:31
From: GUILLON
Subject: Fermat proof for n= 3

Hello,

I'm looking for the proof of Fermat's Last Theorem for the specific 
case of n = 3. Can you help?

Date: 12/14/98 at 13:09:32
From: Doctor Rob
Subject: Re: Fermat proof for n = 3

This was first proved by Leonhard Euler. He wrote about this problem
several times. At first, his proof had a gap, but eventually he filled 
the gap, and is given credit for having the first proof.

The proof starts off as follows: Assume x, y, and z are pairwise
relatively prime integers such that:

   x^3 + y^3 + z^3 = 0

with x and y odd, z even, and |z| as small as possible. Then:

   x + y = 2*a
   x - y = 2*b

where a and b are relatively prime nonzero integers, one odd and one
even. Then x = a + b, y = a - b, and:

   -z^3 = x^3 + y^3
        = (x+y)*(x^2-x*y+y^2)
        = 2*a*(a^2+3*b^2)

Then you can easily see that a^2 + 3*b^2 is odd, so 8 | 2*a, b is odd,
and GCD(2*a, a^2+3*b^2) = 1 or 3.

Case 1:  GCD(2*a, a^2+3*b^2) = 1. Then:

   2*a = r^3
   a^2 + 3*b^2 = s^3

where r is even and s is odd. Then Euler claimed that it is possible to 
write:

   s = u^2 + 3*v^2

with u and v integers. We'll come back to this claim later. Assuming it 
is valid (and it is), then u and v can be chosen so that:

   a = u*(u^2-9*v^2)
   b = 3*v*(u^2-v^2)

Then v is odd, u is nonzero, even, and not a multiple of 3, 
GCD(u,v) = 1, and:

   r^3 = 2*a = 2*u*(u-3*v)*(u+3*v)

Note that 2*u, u-3*v, and u+3*v have to be pairwise relatively prime, 
so they must be cubes of integers:

       2*u = -l^3
   u - 3*v = m^3
   u + 3*v = n^3

with none of l, m, or n equal zero (since u is not a multiple of 3).
Thus:

   l^3 + m^3 + n^3 = 0

with l even, m and n odd. Moreover, since b is nonzero and a is not a
multiple of 3:

   |z|^3 = |z^3|
         = |2*a*(a^2+3*b^2)|
         = |l^3*(u^2-9*v^2)*(a^2+3*b^2)|
         = |l^3*m^3*n^3*(a^2+3*b^2)|
         >= |l^3*(a^2+3*b^2)|
         >= |3*l^3|
         > |l|^3

which contradicts the minimality of |z|.

Case 2:  GCD(2*a, a^2+3*b^2) = 3.

Let a = 3*c. Then c is a multiple of 4, b is not a multiple of 3, and:

   -z^3 = 6*c*(9*c^2+3*b^2) = 18*c*(3*c^2+b^2)

where 18c and 3*c^2 + b^2 are relatively prime, 3*c^2 + b^2 is odd and
not a multiple of 3. This implies that 18*c and 3*c^2 + b^2 are cubes
of integers:

   18*c = r^3
   3*c^2 + b^2 = s^3

with s odd. By the same step as in Case 1, s = u^2 + 3*v^2, with
integers u and v such that:

   b = u*(u^2-9*v^2)
   c = 3*v*(u^2-v^2)

Thus u is odd, v is even and nonzero, GCD(u,v) = 1, and 2*v, u+v, and
u-v are pairwise relatively prime. From:

   (r/3)^3 = 2*v*(u+v)*(u-v)

it follows that:

   2*v = -l^3
   u + v = m^3
   u - v = -n^3

for some integers l, m, and n, with l even, m and n odd. Hence:

   l^3 + m^3 + n^3 = 0

Finally:

   |z|^3 = |z^3|
         = 27*|l^3|*|u^2-v^2|*(3*c^2+b^2)
         = 27*|l^3|*|m^3*n^3|*(3*c^2+b^2)
         >= 27*|l^3|*(3*c^2+b^2)
         >= 27*|l^3|
         > |l|^3

This contradicts the minimality of |z|.

The details of the proof of the claim are rather long. Here are the
main points.

(A) S = {a^2+3*b^2: a, b integers}.

(B) S is closed under multiplication, since:

    (a^2+3*b^2)*(c^2+3*d^2) = (a*c+3*b*d)^2 + 3*(a*d-b*c)^2
                            = (a*c-3*b*d)^2 + 3*(a*d+b*c)^2

(C) Let p be a prime and n >= 1. If p and p*n are both in S, then 
    n is in S.

(D) A prime p belongs to S if and only if p = 3 or p = 1 (mod 6).

(E) If m = a^2+3*b^2 is in S, with GCD(a,b) = 1, and if n is a divisor
    of m, then n is in S.

(F) m is in S if and only if the following condition is satisfied: if
    p is a prime, p = -1 (mod 6) or p = 2, then the exponent of the 
    exact power of p dividing m is even.

(G) If p is a prime number, p = a^2 + 3*b^2 = c^2 + 3*d^2, then
    a^2 = c^2, b^2 = d^2.

(H) Let m1, m2 be products of primes belonging to S, let m = m1*m2.
    For each expression m = u^2 + 3*v^2, there are expressions:

    m1 = a^2 + 3*b^2
    m2 = c^2 + 3*d^2

    such that:

    u = a*c - 3*b*d
    v = a*d + b*c

(I) Let s be an odd integer, such that s^3 = u^2 + 3*v^2, with nonzero
    integers u and v. Then s = t^2 + 3*w^2, with t, w integers and:

    u = t*(t^2-9*w^2)
    v = 3*w*(t^2-w^2)

Using properties (A)-(H), one can prove (I), as Euler did. If you need
details of this, they appear in Robert D. Carmichael, _Diophantine
Analysis_, Wiley, New York, 1915. That established the claim, which is
what is needed in the above proof.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/