Divisible by 12: a^3 + b^3 = c^2

[James Buddenhagen]

Recently I posted here that if a+b and a^2-a*b+b^2 are 
both 3 times squares, then a parameterized solution for 
the Diophantine equation

(1)  a^3+b^3=c^2 

is: 
 
      a =  3*m^4-n^4+6*m^2*n^2
      b = -3*m^4+n^4+6*m^2*n^2
      c =  6*m*n*(3*m^4+n^4)

This gives infinitely many solutions, but not all solutions, 
as Hans Havermann has pointed out to me in private email. 
Since several sequences in OEIS are concerned with the 
Diophantine equation (1), let me say a little more about it.

First, if (a,b,c) is a solution of (1) then so is 
(k^2*a,k^2*b,k^3*c), for any k.  So let us say that two solutions 
(a,b,c) and (a',b',c') are equivalent if there is a non-zero 
rational number k such that (a',b',c')=(k^2*a,k^2*b,k^3*c).  
This is an equivalence relation and each equivalence class has a 
reduced (or primitive) representative (a,b,c) such that 
a,b,c are integers and the largest integer n such that 
n^2|a and n^2|b and n^3|c is n=1.  All this works just as 
well if a,b,c are allowed to be rational numbers and we can 
pass from a rational solution of (1) to the primitive integer 
solution in the same equivalence class by using the proper 
value for k.  Thus we may as well be looking for all rational 
points on the curve x^3+y^3=c^2.  This is actually an elliptic 
curve, and as we vary c we get a family of ellipic curves. 
So, we are really looking for all rational points on the elliptic 
surface x^3+y^3=z^2.  Is is not reasonable to expect a complete 
parametric solution, but in the case where a+b and a^2-a*b+b^2 
are both three times a square such a solution is given below.

Theorem:  (a,b,c) is a rational solution to (1) a^3+b^3=c^2 in 
which a+b and a^2-a*b+b^2 are both 3 times a rational square, 
if and only if there exist rational numbers u and v such that 

        a =  u*(3*u^2*v^4-1+6*u*v^2), 
 (2)    b = -u*(3*u^2*v^4-1-6*u*v^2), 
        c =  6*v*u^2*(3*u^2*v^4+1).

Some examples:   

     solution (a,b,c) to (1)           values for (u,v) in (2)

         (11, 37, 228)                         (16, 1/8)
         (37, 11, 228)                         (3, 2/3)
         (1, 2, 3)                             (3/4, 2/3)
         (100, 200, 3000)                      (75, 1/15)
         (71, -23, 588)                        (1, 2)
         (23/2, 1/2, -39)                      (1/2, -2)

I won't give a proof of the theorem here, but substituting (2) into 
(1) gives an identity so that direction is just a computation. Also, 
notice that replacing u by k^2*u and v by v/k changes (a,b,c) given 
by (2) to (k^2*a,k^2*b,k^3*c) so we can easily move within any equivalence 
class of solutions.  Hence it suffices to show that there exist rationals 
u,v such that the right sides of (2) give some (a',b',c') equivalent to 
the given (a,b,c).  I'll leave it at that for now.  

Some sequences related to the Diophantine equation (1) are:  
A099806, A099807, A099808, A099809, A098970, A099426.

Jim Buddenhagen