n^2 + k^2 = c^3

[T. D. Noe]

>Solving n^2 + k^2 = c^3 -- for which n is it possible?
>
>For the sequencess A077059 etc., I think there are several mistakes --
>namely, terms that are not in the seq but for which k and c exist.
>
>However, my fear is that my list still has mistakes!
>
>How can I prove that a particular value of n cannot be written as n^2
>+ k^2 = c^3 for any k > n?
>
>For example, the original A077059 is missing 38, but 38^2 + 10582^2 =
>482^3 is the smallest solution.  The prime factorizations certainly
>aren't revealing (10582 = 2*11*13*17).
>
>I can find ways of determining, say, that any solution must equal
>certain numbers mod this or that, which eventually forces any possible
>solutions to be pretty big, but I can't yet find a way to prove that
>there really isn't any solution.
>
>Can anyone help prove that I'm not missing any numbers in this list?
>Or find values of k and c that show that I *am* missing some numbers?
>
>I get, for A077059,
>0 2 5 7 9 10 16 17 18 26 30 35 36 37 38 40 44 47 50 51 54 56 58 60 65
>72 73 74 75 80 82 88 95 97 101 102 106 107 115 122 128 130 135 136 140
>142 143 144 145 146 150 154 164 170 174 182 187 189 190 191 197 198
>208 210 215 219 226 234 238 240 242 243 250 257 259 260 265 270 280
>285 286 288 290 296 297 298 299
>which differs from the current seq. in my inclusion of 38, 73, 95, 97,


IMHO, I think the restriction k>1 is artificial.  You should allow k=0,
which would allow all cubes to be members of this sequence.  Also, I think
11 should be included because 11^2+4^2=5^3.  Same with 52 because
52^2+47^2=17^3.

Dickson's History of the Theory of Numbers, Vol 2, page 229 gives a family
of solutions that includes your 38 and 97.  Quote from Dickson:
"It is stated that a set of solutions of x^2+y^2=z^3 is given by
  x=(m^2-3n^2)m, y=(3m^2-n^2)n, z=x^2+y^2."

Perhaps the other numbers belong to another family.

Tony