Fermat's Last Theorem for n = 3Date: 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/ |
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