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With the recent discovery by S. Huisman (actually in 2016) of $$−284650292555885^3+ 66229832190556^3 + 283450105697727^3 = 74$$

and discussed in this post, I decided to revisit this topic. We have an infinite family of parameterizations for $N=1$,


$$(1 - 9n^3)^3 + (9n^4)^3 + (3n - 9n^4)^3 = 1$$


$$(1 + 9n^3 + 648n^6 - 3888n^9)^3 +\\ (-135n^4 + 3888n^{10})^3 + (-3n - 81n^4 + 1296n^7 - 3888n^{10})^3 = 1$$


and so on for all degrees $k=4,10,16,\dots =6m+4.$ Only one (so far) for $N=2$,

$$(-6n^2)^3 + (1 + 6n^3)^3 + (1 - 6n^3)^3 = 2$$

I went over Huisman's $15254$ solutions. The $N$ with the most number $\text{#}$ of solutions are,

$$\begin{array}{|c|c|} \hline N&\text{#}\\ \hline 792&96\\ 720&89\\ 755&83\\ 918&80\\ 638&80\\ 883&79\\ \hline \end{array}$$

So in the "small" finite range ($10^{15}$) that Huisman searched, then $$x^3+y^3+z^3 = 792$$ already has $96$ solutions. If the range is infinite, does it (and others) in fact have infinitely many solutions like $N=1$?

P.S. Using Huisman's database, I found,

$$(-2n - 3 n^4)^3 + (1 + 3 n^3 + 3 n^6)^3 + (-3 n^3 - 3 n^6)^3 = n^3 + 1$$

Of course, if we let $n\to-n$, this solves $N=n^3-1$ as well.

Q: Any other parametric solution to $x^3+y^3+z^3 = N$ where $N$ is simple in form like $N=1,\,2,\,n^3\pm1$ and is not a cube?

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  • $\begingroup$ I tried looking at $N = n^3\pm c$ for other integer $c$, but it was hard to find a pattern. $\endgroup$ – Tito Piezas III Dec 29 '17 at 15:56
  • $\begingroup$ I think your question belongs to this community, according to the fact that such field is not well explored yet and, perhaps your question is still an open problem. $\endgroup$ – user499203 Dec 29 '17 at 15:58
  • $\begingroup$ Is it possible that there is an infinite number of parametrizations for some fixed $N$? $\endgroup$ – user480281 Dec 29 '17 at 16:01
  • $\begingroup$ Or, at least, more than one? $\endgroup$ – user480281 Dec 29 '17 at 16:03
  • $\begingroup$ @AntoinePalAdeen: That is a very important question for this topic. It is known only for $N=1$. But other $N$ have many, like $N=720$ has 89 solutions, and is just begging to be parameterized, if indeed possible. $\endgroup$ – Tito Piezas III Dec 29 '17 at 16:04
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Seiji Tomita has written an article in which he has given multiple parametrization for which the sums of three cubes is $N=(square)$ The link is given below.

     http://www.maroon.dti.ne.jp/fermat/dioph87e.html
| cite | improve this answer | |
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  • $\begingroup$ Welcome to MSE. This should be a comment, not a answer. $\endgroup$ – José Carlos Santos Jan 7 '18 at 17:43

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