Proof of No Integer Solutions for a^3 + b^3 = c^3Date: 02/08/2004 at 02:15:41 From: Phil Subject: Algebraic Proof In the equation a^3 + b^3 = c^3, how is it possible to prove that there are no integers that satisfy the equation? I have tried everything, from simple algebra to calculus to applied geometry, but I can't find a way to prove it. Date: 02/08/2004 at 08:09:30 From: Doctor Rob Subject: Re: Algebraic Proof Thanks for writing to Ask Dr. Math, Phil! 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. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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