A103255 x: x^3+y^3=z^2

[Richard Mathar]

Some clue of improving the status of A103255 is given in
a paper by Darmon et al in the "Links" entry of the OEIS:

"On the Equations z^m=F(x,y) and Ax^p+By^q=Cz^r",
Bull Lond Math Soc 27 (6) (1995) 513-543.
http://www.dms.umontreal.ca/~andrew/PDF/superelliptic.pdf

which says in section 7.2

"The general solution of x^3+y^3=z^2 separates into two
 parametrizations: x=a(a^3-8b^3)/t^2, y=4b(a^3+b^3)/t^2, z=(a^6+20a^3b^3-8b^6)/t^3
 where (a,b)=1, a is odd and t=(3,a+b), and x=(a^4+6a^2b^2-3b^4)/t^2,
 y=(3b^4+6a^2b^2-a^4)/t^2, z=6ab(a^4+3b^4)/t^3,
 where (a,b)=1, 3 does not divide a, t=(2,a+1,b+1).
"

Examples of these general solutions are in the following table
(Solutions to the parametrization II obviously come in groups of
4 by toggling the signs of a and b, and solutions to the parametrization
I in groups of 2 by toggling the signs of a and b at the same time):

x     y     z  a  b  parametrization
------------------------------------
  1  -1     0 -1  0 II
  1  -1     0  1  0 II
  1   2     3 -1 -1 II
  1   2     3  1  1 II
  1   2    -3 -1  1 II
  1   2    -3  1 -1 II
 37  11   228 -2 -1 II
 37  11  -228 -2  1 II
 37  11  -228  2 -1 II
 37  11   228  2  1 II
-11 443 -9324 -2  3 II
-11 443  9324  2  3 II
-11 443 -9324  2 -3 II
 -7   8    13 -1 -1 I
 -7   8    13  1  1 I
 -7   8   -13  1  2 I
 -7   8   -13 -1 -2 I
-23  71  -588 -1  2 II
-23  71   588  1  2 II
-23  71   588 -1 -2 II
... and continuing only with x>0,y>0 solutions...
57 112 -1261 -1 4 I
57 112 -1261 1 -4 I
57 112 1261 -3 -1 I
57 112 1261 3 1 I
65 56 -671 -1 2 I
65 56 -671 1 -2 I
65 56 671 -5 -1 I
65 56 671 5 1 I
193 3482 -205485 -5 7 II
193 3482 -205485 5 -7 II
193 3482 205485 -5 -7 II
193 3482 205485 5 7 II
217 312 -6371 -1 3 I
217 312 -6371 1 -3 I
217 312 6371 -7 -2 I
217 312 6371 7 2 I
305 1064 -35113 -1 7 I
305 1064 -35113 1 -7 I
305 1064 35113 -5 -2 I
305 1064 35113 5 2 I
433 242 -9765 -5 3 II
433 242 -9765 5 -3 II
433 242 9765 -5 -3 II
433 242 9765 5 3 II
781 4019 -255720 -4 5 II
781 4019 -255720 4 -5 II
781 4019 255720 -4 -5 II
781 4019 255720 4 5 II
877 851 -35928 -4 3 II
877 851 -35928 4 -3 II
877 851 35928 -4 -3 II
877 851 35928 4 3 II
889 4440 -297037 -1 10 I
889 4440 -297037 1 -10 I
889 4440 297037 -7 -3 I
889 4440 297037 7 3 I
913 182114 -77716821 -13 19 II
913 182114 -77716821 13 -19 II
913 182114 77716821 -13 -19 II
913 182114 77716821 13 19 II

[It seems that A103255 and A103254 include only solutions with x>0, y>0,
such that (x,y,z)=(-7,8,13) is not found in A103255 as x=-7 or x=8.
A121980 does not impose this restriction, so z=13 is included there.]

To check out the most general cases of x and y it might be helpful to 
factorize the first parametrization as quoted above in the form
"The general solution of x^3+y^3=z^2 separates into two
 parametrizations: x=a(a-2b)(a^2+2ab+4b^2)/t^2, y=4b(a+b)(a^2-ab+b^2)/t^2,
 z=(a^2+2ab-2b^2)(a^4-2ba^3+6a^2b^2+4ab^3+4b^4)/t^3
 where (a,b)=1, a is odd and t=(3,a+b), and x=(a^4+6a^2b^2-3b^4)/t^2,
 y=(3b^4+6a^2b^2-a^4)/t^2, z=6ab(a^4+3b^4)/t^3,
 where (a,b)=1, 3 does not divide a, t=(2,a+1,b+1).
"

Although it seems (to me) that this clue on the format of the general
term x in A103255 does not provide an instant solution to the problem
of proving absence of some x, someone might take a look at it.

PS: The workshop on "Solvability of Diophantine Equations"
is held in Leiden (Netherlands) in May 2007:
http://lorentzcenter.nl/lc/web/2007/230/description.php3?wsid=230