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Everyone's familiar with the identity $a^3+b^3+c^3=d^3$ where $(a,b,c,d)=(3,4,5,6)$. Pretty as it is it suffers the defect that two of these four integers are divisible by $3$ while two are even. The first defect is relieved by using $(370,518,763,859)$. And $(361,3317,5053,5491)$ takes care of both. But can one find a quadruple of pairwise prime positive integers satisfying the identity?

Disclaimer--I found the identities by looking through Table 6 of Selmer's famous Acta Mathematica paper searching for elliptic curves $x^3+y^3= N$ whose Mordell-Weil rank is $> 1$, and using the generators of the group that Selmer provides. But his tables only go up to $N=499$. This, I hope, explains the title of my question. Maybe the tables have been pushed further, providing an answer. To make this something beyond a recreational exercise, one might ask if there are infinitely many such quadruples or even for how the number of such quadruples with $d< N$ grows with $N$ as $N\longrightarrow$ infinity.

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    $\begingroup$ You can get new solutions from two old points by intersecting the cubic surface with lines through these two points. Alas, the three that you give are already on one line. But if you have more you could use that to get even more quadruples. $\endgroup$ Commented Apr 25 at 8:23
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    $\begingroup$ This is one with pairwise coprime (13903, 113345, -98687, 79249) but not all positive (obtained using the above method and the taxicab point (1,12,-10,9) ). $\endgroup$ Commented Apr 25 at 8:39

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The complete integer solution to equation $x^3+y^3+z^3+t^3=0$ is given by the union of families $$ (x,y,z,t) = (v,w,-w,-v), \quad\quad v,w \in {\mathbb Z} $$ and \begin{equation} \begin{split} x &= \frac{u}{d}(c(a+3b)(a^2+3b^2)-c^4) \\ y &= -\frac{u}{d}(c(a-3b)(a^2+3b^2)-c^4) \\ z &= \frac{u}{d}((a^2+3b^2)^2-(a+3b)c^3) \\ t &= - \frac{u}{d}((a^2+3b^2)^2+(3b-a)c^3) \end{split} \end{equation} where $a,b,c,$ and $u$ are arbitrary integers and $d$ is any common divisor of the expressions in the parentheses, see, e.g. Proposition 4.67 in my recent book

Polynomial Diophantine Equations: A Systematic Approach https://link.springer.com/book/10.1007/978-3-031-62949-5

We need solutions with $xyzt<0$ pairwise coprime. This obviously implies that $|u|=1$ and $|d|$ is the greatest common divisor of the expressions in the parentheses. We then just try all small $a,b,c$ and check the conditions. It returns a lot of the requested solutions. The one with the smallest sum of variables is $$ 107^3 + 209^3 + 337^3 = 365^3. $$ I was even able to find a solution $$ 599^3 + 691^3 + 823^3 = 1033^3 $$ in which all variables are not only coprime, but in fact prime numbers.

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  • $\begingroup$ Many thanks--I like the parametrization, though it might be better adapted to practice than to theory And I gather that 365 is the smallest d providing my desired example. Can one work out a reasonable lower bound for how the number of d < N that provide examples grows with N? $\endgroup$ Commented Apr 25 at 15:40
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    $\begingroup$ (a) Yes, $365$ is the smallest $d$ (in your notation) that I obtained by trying all possible small $a,b,c$ (in my notation). To prove this rigorously, one should either prove that no large parameters can lead to small solutions, or eliminate cases from $1$ to $364$ by direct search. (b) Sorry, I do not see how to easily obtain the requested lower bound from my parametrization. $\endgroup$ Commented Apr 25 at 16:03
  • $\begingroup$ Another (and smaller) all-prime solution is $193^3+461^3+631^3=709^3$. $\endgroup$ Commented May 7 at 14:38

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