Proof of No Integer Solutions for a^3 + b^3 = c^3

Date: 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/