n^2 + k^2 = c^3

[Joshua Zucker]

On 5/24/06, T. D. Noe <noe at sspectra.com> wrote:
> 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

Yeah, I think the k>n restriction is artificial too, but the original
def. of the seq. includes that restriction.  So n = 0 shows up once,
and then in general cubes don't show up paired with 0.  (The k > n
restriction also explains why things like 11 and 52 are missing.)  If
it's not already in there -- and it doesn't appear to be -- then I
agree with you that at least the sequence of n that work with any k
should be submitted as well.  But first I want to make sure that I
know I'm not missing any terms.  I still feel uncomfortable being so
unsure about whether other numbers should be in the list!

Thanks for the Dickson reference.  That might be a good start!

--Joshua