Diophantine Equation $a^3+b^3+c^3=d^3$

Question

There are lots of claims that general parametric solution of Diophantine equation $a^3+b^3+c^3=d^3$ was found but I did not find any identity or set of identities that would give all solutions.

So was the general solution really found?

$\begin{array}{cccc} \{3,4,5,6\} & \{1,6,8,9\} & \color{gray}{{6,8,10,12}} & \color{gray}{{2,12,16,18}} \\ \color{gray}{{9,12,15,18}} & \{3,10,18,19\} & \{7,14,17,20\} & \color{gray}{{12,16,20,24}} \\ \{4,17,22,25\} & \color{gray}{{3,18,24,27}} & \{18,19,21,28\} & \{11,15,27,29\} \\ \color{gray}{{15,20,25,30}} & \color{gray}{{4,24,32,36}} & \color{gray}{{18,24,30,36}} & \color{gray}{{6,20,36,38}} \\ \color{gray}{{14,28,34,40}} & \{2,17,40,41\} & \{6,32,33,41\} & \color{gray}{{21,28,35,42}} \\ \{16,23,41,44\} & \color{gray}{{5,30,40,45}} & \{3,36,37,46\} & \{27,30,37,46\} \\ \color{gray}{{24,32,40,48}} & \color{gray}{{8,34,44,50}} & \{29,34,44,53\} & \color{gray}{{6,36,48,54}} \\ \{12,19,53,54\} & \color{gray}{{27,36,45,54}} & \color{gray}{{36,38,42,56}} & \color{gray}{{9,30,54,57}} \\ \{15,42,49,58\} & \color{gray}{{22,30,54,58}} & \color{gray}{{21,42,51,60}} & \color{gray}{{30,40,50,60}} \\ \color{gray}{{7,42,56,63}} & \color{gray}{{33,44,55,66}} & \{22,51,54,67\} & \{36,38,61,69\} \\ \{7,54,57,70\} & \{14,23,70,71\} & \color{gray}{{8,48,64,72}} & \{34,39,65,72\} \\ \color{gray}{{36,48,60,72}} & \color{gray}{{12,51,66,75}} & \{38,43,66,75\} & \color{gray}{{12,40,72,76}} \\ \{31,33,72,76\} & \color{gray}{{39,52,65,78}} & \color{gray}{{28,56,68,80}} & \color{gray}{{9,54,72,81}} \\ \{25,48,74,81\} & \color{gray}{{4,34,80,82}} & \color{gray}{{12,64,66,82}} & \{19,60,69,82\} \\ \{28,53,75,84\} & \color{gray}{{42,56,70,84}} & \color{gray}{{54,57,63,84}} & \{50,61,64,85\} \\ \{20,54,79,87\} & \{26,55,78,87\} & \color{gray}{{33,45,81,87}} & \{38,48,79,87\} \\ \{21,43,84,88\} & \{25,31,86,88\} & \color{gray}{{32,46,82,88}} & \{17,40,86,89\} \\ \color{gray}{{10,60,80,90}} & \{25,38,87,90\} & \color{gray}{{45,60,75,90}} & \{58,59,69,90\} \\ \color{gray}{{6,72,74,92}} & \color{gray}{{54,60,74,92}} & \{32,54,85,93\} & \color{gray}{{15,50,90,95}} \\ \{19,53,90,96\} & \color{gray}{{48,64,80,96}} & \{45,69,79,97\} & \color{gray}{{11,66,88,99}} \\ \color{gray}{{16,68,88,100}} & \color{gray}{{35,70,85,100}} & \color{gray}{{51,68,85,102}} & \{12,31,102,103\} \\ \{33,70,92,105\} & \color{gray}{{58,68,88,106}} & \color{gray}{{12,72,96,108}} & \{13,51,104,108\} \\ \{15,82,89,108\} & \color{gray}{{24,38,106,108}} & \color{gray}{{54,72,90,108}} & \{29,75,96,110\} \\ \end{array}$

None of these (and similar ones) give them all: $$a^3 \left(a^3+b^3\right)^3=b^3 \left(2 a^3-b^3\right)^3+b^3 \left(a^3+b^3\right)^3+a^3 \left(a^3-2 b^3\right)^3\\ a^3 \left(a^3+2 b^3\right)^3=a^3 \left(a^3-b^3\right)^3+b^3 \left(a^3-b^3\right)^3+b^3 \left(2 a^3+b^3\right)^3\\ \left(4 x^2-4 x y+6 y^2\right)^3+\left(5 x^2-5 x y-3 y^2\right)^3+\left(3 x^2+5 x y-5 y^2\right)^3=\left(6 x^2-4 x y+4 y^2\right)^3$$

https://mathworld.wolfram.com/DiophantineEquation3rdPowers.html

We can read there: "The general rational solution to the 3.1.3 equation was found by Euler and Vieta". Can you provide me that solution? I do not have the books in references.

Let us look at the whole quote "he general rational solution to the 3.1.3 equation
A^3=B^3+C^3+D^3
(5) was found by Euler and Vieta (Hardy 1999, pp. 20-21; Dickson 2005, pp. 550-554)" — user64494, Oct 22 '23 at 17:07
And what??? I wrote that I do not have these books to read them. — azerbajdzan, Oct 22 '23 at 17:08
Have you read in the book that: "...but neither of these is the general solution."??? — azerbajdzan, Oct 22 '23 at 17:35
As written in the linked by you Math World article, " No general solution giving all positive integral solutions is known (Dickson 2005, pp. 550-561).". Deep regard again. — user64494, Oct 22 '23 at 17:40
But at the same time you can read on the same link: "The general rational solution to the 3.1.3 equation was found by Euler and Vieta" - and this is exactly what is written in my original question. So if you have nothing valuable to comment then do not do so... — azerbajdzan, Oct 22 '23 at 17:45
you can look at the solutions in the following post: https://math.stackexchange.com/questions/4776322/sums-of-three-cubes-of-form-a3b3c3-c13/4778242#4778242 — user25406, Oct 22 '23 at 20:15
@user25406: Thanks. The post is really interesting and a bit related to the topic of my question, but I do not think it provides any formula that could be considered as a general solution similar to Euler one in my answer. — azerbajdzan, Oct 22 '23 at 20:42
Although you ask about a general rational solution, you may also be interested in the general integer solution in Choudhry A (1998) On Equal Sums of Cubes (Theorem 3) which may be dowloaded here. — Adam Bailey, Oct 24 '23 at 19:04
@Adam Bailey: Actually I am interested in both general solutions and any of the two answers would be accepted. Thanks for the reference of the book. — azerbajdzan, Oct 24 '23 at 19:10
@user64494 Although the Dickson quotation is dated 2005 it is I think just a reprint of his 1919 book - see here. That no general integer solution is known was true in 1919 but no longer true in 2005. — Adam Bailey, Oct 24 '23 at 19:12

azerbajdzan · Accepted Answer · 2023-10-22T18:42:00.463

2

This is the general rational solution found by Euler and simplified by Binet as stated in G. H. Hardy's and E. M. Wright's book An Introduction to the Theory of Numbers.

$$\left(\lambda \left((a+3 b) \left(a^2+3 b^2\right)-1\right)\right)^3+\left(\lambda \left(1-(a-3 b) \left(a^2+3 b^2\right)\right)\right)^3=\left(\lambda \left(-\left(a^2+3 b^2\right)^2+a+3 b\right)\right)^3+\left(\lambda \left(\left(a^2+3 b^2\right)^2-a+3 b\right)\right)^3$$

It is general solution of both versions of equations (as for odd power the sign does not play any role):

$$a^3+b^3=c^3+d^3\\ a^3+b^3+(-c)^3=d^3$$

edited Oct 22 '23 at 18:42

answered Oct 22 '23 at 18:36

azerbajdzan

1,166

I think that there is more than one general solution. I can't say anything about the parametric solutions but a general solution can be found for when $d=c+1$. But this general solution doesn't apply to the case of $d=c+2$ or $d=c+3$...For example $25^3 + 38^3 +87^3=90^3$ corresponds to the case when $d=c+3$. If you try $d=c+1$ or $d=c+2$, the coresponding quadratic equation will not have integer solutions ( the determinant not being a square). But with $d=c+3$, you get the equation ( after some simplification ) $c^2 + 3c -7830=0$ with solutions $c=87, -90$ or $d=c+3=87+3=90$. (cont). – user25406 Oct 23 '23 at 10:15
Surely, if you have one general rational parametric solution then there are infinite number of them, but they are all equivalent, so no need for more than one formula. – azerbajdzan Oct 23 '23 at 10:25
(cont) but also $d=c+3=-90+3=-87$. so we can write $25^3+38^3=90^3-87^3$ or $25^3 + 38^3 + 87^3 = 90^3$. But this is going backward, that is given $25^3 + 38^3$, can we find a solution as a difference of two cubes. The difference $90^3- 87^3$ is divisible by $25+38=63=7*9$. So I like to start with choosing a number, in this case $63$ and try to find is one of its decomposition $63=1+62=2+61=...=25+38=...$ can be expressed as say $25^3 + 38^3= d^3 - c^3$. The answer is yes but we needed to do a lot of work to get that decomposition (cont). – user25406 Oct 23 '23 at 10:26
It's of course easier to write a program to get the solution, or one solution. I still don't know if we can find more than one solution with some other decomposition of $63$. But when we find one solution, we can turn it into a $3$ solutions. We can simply write the two other solutions : $25^3 + 87^3 = 90^3 - 38^3$ corresponding to $d=c+52$ and $38^3 + 87^3= 90^3 - 25^3$ corresponding to $d=c+65$. – user25406 Oct 23 '23 at 10:34
The other interesting fact is to combine two equations $a^3+b^3+c^3=d^3$ and $e^3 +f^3+g^3 =h^3$ into $d^3 + h^3= a^3 +b^3 +c^3+e^3 +f^3 +g^3$ to get a sum of two cubes expressed as a sum of six cubes but also a sum of four cubes expressed as a sum of 4 different cubes $a^3+b^3+c^3+h^3=e^3+f^3+g^3+d^3$. – user25406 Oct 23 '23 at 10:44

Diophantine Equation $a^3+b^3+c^3=d^3$

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