The 450 Pound Problem (x^3 + y^3 = 6z^3)

On New Year's Eve, 1994, an announcement appeared on the Internet
stating that The Sunday Telegraph of London had offered a cash prize 
(450 pounds sterling) to the first person to send them a solution 
in co-prime positive integers greater than 100 of the equation 

                 (A^3/B^3)  +  (C^3/D^3)  =  6

Just for fun I emailed them the first such solution that came to
mind:

A = 792220572662549608190252926112121617686087939438245\
              \665806051608621113641830336450448115419524772568639
C = 677959805103821424723263992665061838773573375138707\
              \37934706199386093375292356829747318557796585767361
B = D = 436066841882071117095002459324085167366543342937477\
              \344818646196279385305441506861017701946929489111120

The offer turned out to be legitimate, and the Telegraph actually did 
send me L450 ($706.50).

Problems of this type were "solved" in principle by Viete, Fermat,  
Euler, etc.  In fact, Diophantus dealt with similar equations.  However,
it seems Legendre once stated that this particular equation had 
no solutions, although it's not clear why he thought so.  In any case,
Lame' pointed out that  (17/21)^3 + (37/21)^3 = 6.  (The Telegraph's
stipulation that the integers be greater than 100 was clearly intended 
to exclude this "easy" solution.)

Of course, given one solution, the tangent-chord method, or some 
variation, yields infinitely many other solutions.  As Poincare 
conjectured and Mordell proved, all the rational points on a curve 
such as x^3 + y^3 = 6 can be generated by tangent and chord 
constructions applied to a finite set of points.  Do we know the set 
of points that generate all solutions in this particular case?  We
know that ALL the rational solutions are based on Lame's point 
(17/21,37/21), because the group of rational points on this curve
has rank 1.  (Chris Thompson points out that "one can grind this 
out following the proof of Mordell's theorem.")

Another question is: What led Legendre to believe there were NO 
solutions?  Interestingly, it can _almost_ be proved by descent that 
the equation can have no solutions.  Starting from scratch, we observe 
that (AD)^3 + (BC)^3 = 6(BD)^3 implies that D^3 divides B^3 and that 
B^3 divides D^3, so we have B=+-D and the equation can be written in 
the form A^3 + B^3 = 6C^3 where A,B,C are integers.

Since A^3 + B^3 is even, whereas A and B cannot both be even, 
they must both be odd, so we have co-prime integers u,v such that 
A=u+v and B=u-v.  Making these substitutions gives u(u^2 + 3v^2) 
= 3C^3.  Clearly u is divisible by 3.  Also, since u and (u^2+3v^2)/3 
are co-prime, they must both be cubes, so we have integers a,b such 
that  u = 27a^3  and  u^2 + 3v^2 = 3b^3.  From the theory of binary 
quadratic forms it's known that all the solutions of the latter 
equation are of the form

   u = 9y(x^2 - y^2)       v = x(x^2 - 9y^2)       b = x^2 + 3y^2

where gcd(x,y)=1.  Noting that u must be of the form 27a^3 we have

                  3a^3 = y(x+y)(x-y)

This shows that 3 must divide exactly one of the quantities y, (x+y), 
or (x-y).  In the first case it follows that (y/3), (x+y), and (x-y) 
must all be cubes.  Setting 

       (y/3) = q^3      (x+y) = r^3      (x-y) = s^3

we have (r)^3 + (-s)^3 = 6(q)^3, which is a solution of the original 
equation in absolutely smaller integers.  Therefore, if this were the 
only possible case, we would be forced to conclude that no solution 
is possible (by the principle of "infinite descent").  Legendre may 
have been thinking along these lines when he stated the equation 
was insoluable.

However, if y is not divisible by 3, then 3 divides either (x+y) or 
(x-y), in which case we arrive at

  (s)^3 + 2(q)^3 = 3(r)^3      or      (r)^3 + 2(-q)^3 = 3(s)^3

respectively.  At this point the "descent" argument no longer applies, 
i.e., a solution to this equation does not necessarily imply an 
absolutely smaller solution.

It follows that every solution of A^3 + B^3 = 6C^3 can be reduced 
down through a sequence of successively smaller triples (q,r,s) that 
satisfy the equation r^3 +s^3 = 6q^3 until reaching a triple that 
satisfies instead the equation r^3 + 2s^3 = 3q^3.  The smallest 
solution (r,s,q) of the latter equation is (1,1,1), and the tangent 
line through that point gives (-5,4,1).  Every solution of the 
Telegraph's problem that I've seen so far can be reduced to one of 
these two irreducible cases.  However, there are other solutions 
that reduce to, for example, (655,-488,253), which is on the tangent 
line through (-5,4,1).

Conversely, given a solution of either  r^3 + s^3 = 6q^3   or   r^3 + 
2s^3 = 3q^3  we can generate the solution of A^3 + B^3 = 6C^3 
"above it" by working backwards.  For example, in the latter case 
we would compute y=s^3 and x=3q^3 - s^3, and then the "higher" 
solution of A^3+B^3=6C^3 is given by

        A = 9y(x^2-y^2) + x(x^2-9y^2)
        B = 9y(x^2-y^2) - x(x^2-9y^2)

Has anyone tried constructing a "family tree" of all the solutions 
of A^3 + B^3 = 6C^3 generated from the irreducible solutions of the
related equation r^3 + 2s^3 = 3q^3 ?