Are there infinitely many integer positive cubes $x^3 = a^3 + b^3 + c^3$ that are equal to the sum of three integer positive cubes? If not, how many of them are there?

4$\begingroup$ This seems to be an improved version of your previous question. Maybe you could include into the question some examples of what you would consider satisfactory answer to "how much". (For example, asymptotic density or some asymptotic estimate for the counting function?) $\endgroup$– Martin SleziakJan 3, 2022 at 9:48

10$\begingroup$ If you don't require coprimeness, then one solution leads to infinite solutions. $\endgroup$– ZeroxJan 3, 2022 at 9:49

5$\begingroup$ @MartinSleziak, the question seems to be "infinite, or finite and how many?"; and, if finite, then the format expected in the answer to "how many" seems unambiguous. (But, since @joro's answer shows that there are infinitely many, now perhaps @user473935 wants to ask one of your suggested refinements separately.) $\endgroup$– LSpiceJan 3, 2022 at 14:06

$\begingroup$ The values of x are in oeis.org/A023042. See also the reference in oeis.org/A003072 . $\endgroup$– R. J. MatharJan 5, 2022 at 15:08
6 Answers
There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010:
$$(a^42ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 bb^4)^3 = (a^4+a b^3)^3.$$
Added If you drop the positivity constraint then there is another identity for $x=v^4$:
$$v^{12}=(v^4)^3=(9u^4)^3+(3uv^39u^4)^3+(v(v^39u^3))^3.$$
I believe, but can't find it at the moment, that for all positive $x$ there exist integers $a$, $b$, $c$ such that $x^3=a^3+b^3+c^3$.

$\begingroup$ Does "can't find it at the moment" mean "can't prove it at the moment"? Without the positivity constraint, writability as a sum of three cubes is conjectured (or at least asked) for all integers, not just cubes. $\endgroup$– LSpiceOct 9, 2023 at 16:36

$\begingroup$ Doesn't the statement for positive $x$ immediately imply it for all $x$? One can just reverse the signs of $a, b, c$... $\endgroup$ Oct 9, 2023 at 18:48
The answer can also be found in Section 21.11 of HardyWright: An introduction to the theory of numbers. More precisely, it is proved there that any twelfth power $n^{12}$ can be expressed as a sum of three positive cubes in at least $[9^{1/3}n]$ ways. The underlying identity (21.11.4) is due to Gérardin (1912) as one learns from the notes to Chapter 21.
Here is a more geometric answer. Firstly it makes the most sense to view this as a surface $S \subset \mathbb{P}^3_{\mathbb{Q}}$ in projective $3$space. In which case integer solutions are the same as rational solutions as we are using homogeneous coordinates.
It is wellknown and very classical that the surface $S$ is rational over $\mathbb{Q}$, i.e. birational to $\mathbb{P}^2_{\mathbb{Q}}$. This is the geometric way to phrase some of the other answers which talk about parametrising solutions. There are purely geometric proofs of this which don't require explicit formulas, but you can find formulas e.g. here https://people.math.harvard.edu/~elkies/4cubes.html.
As $S$ is rational it satisfies weak approximation. This in particular implies that $S(\mathbb{Q})$ is dense in $S(\mathbb{R})$, where $S(\mathbb{R})$ is viewed as a real manifold.
It thus follows that any real solution to the equation can be arbitrarily well approximated by a rational solution. So for any real solution with $x,a,b,c$ positive you can find a rational solution which approximates it. Thus in a precise sense, there are incredibly many solutions to this equation with the sought after properties.
Ajai Choudhry has given a general solution for a cube represented as sum of three positive cubes.
His solution given in equation (14) of the paper is valid for all positive integers $(a,b,c)$ where $a>b$. Since there are infinitely many such triples, it is easy to see then that your equation will have infinitely many solutions.

$\begingroup$ Signature at the end of the paper: Ambassador, Embassy of India, Kantari Street, Sahmarani Building, P.O. Box 1135240, Beirut, Lebanon $\endgroup$ Jan 12, 2022 at 13:22

$\begingroup$ @GH from MO, Author of paper, Ajai Choudhry's email info is, [email protected]. His address is, 13/4 Clay square, Lucknow, UP, 226001,India. He has also posted on arxiv.org. $\endgroup$– RemyJan 13, 2022 at 11:25

1$\begingroup$ I just found it nice that an ambassador published mathematics. $\endgroup$ Jan 13, 2022 at 12:24

2$\begingroup$ @GH from MO. Mathematician Ajai Choudhry going from being a diplomat to being a mathematician is an interesting story told by George Szpiro in his book Mathematical Medley the link is here books.google.com.au/… and the list of his math papers are given here, zh.booksc.eu/g/Ajai%20Choudhry $\endgroup$– RemyJan 14, 2022 at 1:23

2$\begingroup$ @GH from MO. The link for the book by Sziro was pasted wrong but is shown again. books.google.com.au/… $\endgroup$– RemyJan 14, 2022 at 9:39
There is a parametric solution of $w^3=x^3+y^3+z^3$ with second degree where $s^3=p^3+q^3+r^3$ as follows.
Substitute $x=pt+a, y=qta, z=rt+b, w=st+b$ to $w^3=x^3+y^3+z^3$ where $(a,b)$ are arbitrary integers.
Let $t=\frac{(rb^2+qa^2+pa^2sb^2)}{q^2a+p^2as^2b+r^2b},$ then we obtain a parametric solution
$x= (q^2+pq)a^2+(r^2+s^2)ba+(prps)b^2$
$y= (p^2+pq)a^2+(s^2+r^2)ba+(qrqs)b^2$
$z= (qr+pr)a^2+(p^2+q^2)ba+(s^2rs)b^2$
$w= (ps+qs)a^2+(p^2+q^2)ba+(rsr^2)b^2.$
We can get the positive solutions by choosing $(a,b)$ so that $(x,y,z,w)$ are positive.
For instance, taking $(p,q,r,s) = (1,6,8,9)$ and we obtain
$(x,y,z,w) =(42a^2+17bab^2, 7a^217ba6b^2, 56a^2+35ba+9b^2, 63a^2+35ba+8b^2).$
We know if $\frac{a}{b} > \frac{17+\sqrt{457}}{14}$ then $x,y$ are positive.
a, b w x y z
[ 3, 1,],[ 340, 214, 3, 309]
[ 4, 1,],[1156, 739, 38,1045]
[ 5, 1,],[ 293, 189, 14, 264]
[ 6, 1,],[2486,1613, 144,2235]
[ 7, 1,],[1670,1088, 109,1499]
[ 7, 2,],[1203, 764, 27,1090]
[ 8, 1,],[1440, 941, 102,1291]
[ 9, 1,],[2713,1777, 204,2430]
[ 9, 2,],[5765,3704, 237,5202]
Similarly, taking $(p,q,r,s) = (3,4,5,6)$ and we obtain
$(x,y,z,w) =(28a^2+11ba3b^2, 21a^211ba4b^2, 35a^2+7ba+6b^2, 42a^2+7ba+5b^2).$
If $\frac{a}{b} > \frac{11+\sqrt{457}}{42}$ then $x,y$ are positive.
a, b w x y z
[ 1, 1,],[ 9, 6, 1, 8]
[ 2, 1,],[ 187, 131, 58, 160]
[ 3, 1,],[ 202, 141, 76, 171]
[ 3, 2,],[ 440, 306, 107, 381]
[ 4, 1,],[ 235, 163, 96, 198]
[ 4, 3,],[ 801, 553, 168, 698]
[ 4, 5,],[ 937, 593, 16, 850]
[ 5, 1,],[ 545, 376, 233, 458]
[ 5, 2,],[ 20, 14, 7, 17]
[ 5, 3,],[ 600, 419, 162, 517]
[ 5, 4,],[1270, 872, 241,1111]
[ 5, 6,],[1440, 922, 51,1301]
[ 6, 1,],[1559,1071, 686,1308]
[ 6, 5,],[1847,1263, 326,1620]
[ 6, 7,],[ 293, 189, 14, 264]
[ 7, 1,],[ 352, 241, 158, 295]
[ 7, 2,],[2176,1514, 859,1837]
[ 7, 3,],[1125, 788, 381, 958]
[ 7, 4,],[ 778, 544, 219, 669]
[ 7, 5,],[1214, 841, 272,1055]
[ 7, 6,],[2532,1726, 423,2225]
[ 7, 8,],[2770,1796, 157,2491]
[ 7, 9,],[1452, 911, 6,1321]
[ 8, 1,],[2749,1877,1252,2302]
[ 8, 3,],[2901,2029,1044,2462]
[ 8, 5,],[1031, 719, 268, 890]
[ 8, 7,],[ 25, 17, 4, 22]
[ 8, 9,],[3597,2341, 228,3230]
[ 9, 1,],[1735,1182, 799,1452]
[ 9, 2,],[3548,2454,1487,2985]
[ 9, 4,],[3734,2616,1241,3183]
[ 9, 5,],[1921,1344, 553,1650]
[ 9, 7,],[ 292, 201, 58, 255]
[ 9, 8,],[4226,2868, 653,3723]
There is another method to express a given cube as a sum of three cubes.
First we tranform the original equation $a^3 + b^3 + c^3 =d^3$ into $a^3 + b^3 = d^3  c^3$. Then the following well known identity is used:
$$d^3c^3= (dc)(d^2 +dc +c^2).$$
At this point we pick a number $N$ we want to express as a sum of three cubes. Let's pick the first number given above by Tomita $N=9$. At this point we don't know the value of the largest $c$. We try the value $c=d1=91=8$ to reduce the identity above to a quadratic equation:
$$d^3c^3= (c+1)^2 + c(c+1) + c^2 = 3c^2 +3c +1 = 9^3  8^3 = 217 = 7\cdot31.$$
After factoring $217$, we take the small factor (not necessarily a prime) and decompose it into the following way: $7=1+6=2+5=3+4$. If $217$ is the sum of two cubes, then it can be written as $1^3 + 6^3$ or $2^3 + 5^3$ or $3^3 + 4^3$. The correct form is $217=1^3 + 6^3$. So now we can write the decomposition of $9^3$ as:
$$9^3 = 8^3 + 6^3 + 1^3.$$
At this point we can also get the decomposition of $9^16^3=513$ as a sum of two cubes. $513=27\cdot19$ simply by writing:
$9^16^3=513= 1^3 + 8^3$. We know that $513$ is divisible by $8+1=9$.
It's not always the case that the largest element $c$ is $c=d1$. In that case the quadratic equation will not have integer solutions. So we can try $c=d2$ or $c=d3$…until we hit on a quadratic solution with integers. If the difference $d^3c^3$ is a prime, we know there is no solution since primes cannot be decomposed into a sum of two cubes.
I have no idea how this method will perform compared to other methods using complex mathematics. I don't know if it can be improved by zooming on $x$ in $c=dx$ that will provide a number that can be written as a sum of two cubes.

1$\begingroup$ This seems not to address "how many cubes are a sum of three positive cubes?", replacing it instead by "given a specific cube that we suspect is a sum of three positive cubes, what is a (more or less heuristic) method for finding them?" \\ Locationbased references like 'above' do not work well on SO, when answers are in different positions on the page for different people, and at different times. I have edited in a reference to the answer that you seemed to mean. $\endgroup$– LSpiceOct 9, 2023 at 16:33

$\begingroup$ Thanks for the edit. I suspect there are an infinite number of numbers that are the sum of three positive cubes though I can't prove. Each equation which is based on $d=c+1$ produces two more equations for two different numbers besides the original. We have also $96$ and $91$ corresponding to $N=513$ and $N=728$. Beside, we can almost always ( of course no proof of that ) find a number by finding a solution involving some $c$ with $c=d1,d2,d3....$ $\endgroup$ Oct 9, 2023 at 17:40

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