TITLE: Parametric solutions to $x^3+y^3+z^3 = N$?
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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
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