From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: rec.puzzles,sci.math
Subject: A^3+B^3+C^3=D^3 [Re: 3(cubed) + 4(cubed) + 5(cubed) = 6(cubed)]
Date: 19 Jun 1997 14:34:37 GMT

[x-posted to sci.math]

In article <sacerdos-1906972155230001@port229-tnt-ak-2.ihug.co.nz>,
Sarah <sacerdos@ihug.co.nz> wrote:

>Help!  I need to find examples of three cubes adding to make a fourth cube
>(integer values), and then make some sort of general summary for the
>problem.  Can you find anymore examples like the one in the subject, or do
>you know of some general solution?

This is a famous Diophantine problem, to which Dickson's
_History of the Theory of Numbers, Vol. II_ devotes many pages.
Yes, there are plenty of examples, also of A^3+B^3=C^3+D^3
(which is mathematically the same thing if you change a sign) --
most famously 1729 = 12^3 + 1^3 = 10^3 + 9^3 [note the hostname
I'm posting from!] -- and even known formulas for finding all
solutions.  Here is one which I worked out last year:

Write the equation symmetrically as

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

so for instance your 3^3+4^3+5^3=6^3 is (x,y,z,t)=(3,4,5,-6).  Assume
that we do not have x+z = y+t = 0 (such solutions are trivial anyhow).
Then for some u1, u2, u3 we have

  x = -(u2+u1) u3^2 + (u2^2 + 2 u1^2) u3 - u2^3 + u1 u2^2 - 2 u1^2 u2 - u1^3,
   
  y =  u3^3 - (u2+u1) u3^2 + (u2^2 + 2 u1^2) u3 + u1 u2^2 - 2 u1^2 u2 + u1^3,
    
  z = -u3^3 + (u2+u1) u3^2 - (u2^2 + 2 u1^2) u3 + 2 u1 u2^2 - u1^2 u2 + 2 u1^3,
     
  t = (u2-2*u1) u3^2 + (u1^2 - u2^2) u3 + u2^3 - u1 u2^2 + 2 u1^2 u2 - 2 u1^3,

possibly after removing a common factor.  Specifically we can take

  u1 = tz - xy,
  u2 = xz - xy + ty + x^2 - tx + t^2,
  u3 = z^2 - yz + xz + y^2 - xy + ty.

For instance (3,4,5,-6) yields u1 = -42, u2 = +42, u3 = 0; since the equations
are homogeneous we may as well take (u1,u2,u3)=(1,-1,0), and then recover
x=3, y=4, z=5, t=-6.
    
ObPuzzle: find all solutions of

  (u2-2*u1) u3^2 + (u1^2 - u2^2) u3 + u2^3 - u1 u2^2 + 2 u1^2 u2 = 2 u1^3

in integers u1, u2, u3.

ObMath: The above map (x:y:z:t) -> (u1:u2:u3) blows down the lines:

  x+wt = y+wz = 0  to  (-w:1:1);
  x+w*t = y+w*z = 0  to  (-w*:1:1);
 
  y+wt = z+wx = 0  to  (0:1:-w);
  y+w*t = z+w*x = 0  to  (0:1:-w*);

  z+wt = x+wy = 0  to (1:-w*:-w);
  z+w*t = x+w*y = 0  to (1:-w:-w*);

w being a cube root of unity.  We have thus explicitly represented the
Fermat cubic surface as P^2 blown up at six points; I do not know if
such formulas have been previously exhibited.  The usual approach to
rationally parametrizing that surface, using a Q(w)-conjugate pair of
skew lines, amounts to getting that surface as P1^2 blown up at 5 points.

--Noam D. Elkies (elkies@math.harvard:edu)
  Dept. of Mathematics, Harvard University