Diophantine Equations – Solution for $z_1^3 +z_2^3+z_3^3=\alpha^3$ for Every Integer $\alpha>0$

Question

I should have asked this question before the other one since Moore's tables for $5$th powers have radius $<17700$, while Wrobleski's tables for $3$rd powers go up to one million, hence more data to work with. And since this is Diophantine, then it is not required that the $z_k$ be positive integers.

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I. Form $x_1^3+x_2^3 = x_3^3+x_4^3$

We focus on primitive solutions and consider $(x_1,x_2) = (x_3,x_4)$ as trivial. The first ones are,

\begin{align}
1^3 & + 12^3 \,=\, 10^3 + 9^3\\
2^3 & + 16^3 \,=\, 15^3 + 9^3\\
3^3 & + 60^3 \,=\, 59^3 + 22^3\\
4^3 & + 110^3 = 101^3 + 67^3\\
5^3 & + 76^3 \,=\, 69^3 + 48^3\\
6^3 & + 552^3 = 551^3 + 97^3
\end{align}

and so on, for all $x_1<3000$ (which is quite a long stretch). Note that for some $x_1 = 13m$, it seems terms are a bit larger,

$$13^3 + 5288^3 = 5148^3 + 2253^3$$

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II. Form $x_1^3+x_2^3+x_3^3=x_4^3$

Likewise, we disregard trivial solutions $x_i = x_4$. Hence,

\begin{align}
1^3 & + \,6^3\, + \,8^3 \,=\, 9^3\\
2^3 & + 17^3 + 40^3 = 41^3\\
3^3 & + 10^3 + 18^3 = 19^3\\
4^3 & + 17^3 + 22^3 = 25^3\\
5^3 & + \,4^3\, + \,3^3\, =\, 6^3\\
6^3 & + 32^3 + 33^3 = 41^3
\end{align}

and so on, also for all $x_1<3000$. Again, for some $x_1 = 13m$, terms are a bit larger,

$$(4\times13)^3 + 321^3 + 3327^3 = 3328^3$$

P.S. Note that for squares, we have the identity,

$$n^2 + (n + 1)^2 + (n^2 + n)^2 = (n^2 + n + 1)^2$$

which proves every positive integer $n$ appears at least once. It may be true for cubes as well.

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III. Questions

1. For any positive integer $x_1$, is it true there are primitive positive solutions to both forms?
2. If we can't prove it in general, how high can Wroblewski's tables extend the confirmed range to a bound $x_1>3000$? (Less than $10^6$ for sure.)

Answer 1

For form $x_1^3+x_2^3+x_3^3=x_4^3$, the following parametric form (also used in this answer) generates solutions for any $x_1$:

$\qquad x_1=m$

$\qquad x_2=3m^2+2m+1$

$\qquad x_3=3m^3+3m^2+2m$

$\qquad x_4=3m^3+3m^2+2m+1$