How many cubes are the sum of three positive cubes?

Question

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?

Answer 1

There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010:

$$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3.$$

Added If you drop the positivity constraint then there is another identity for $x=v^4$:

$$v^{12}=(v^4)^3=(9u^4)^3+(3uv^3-9u^4)^3+(v(v^3-9u^3))^3.$$

I believe, but can't find it at the moment, that for all positive $x$ there exist integers $a$, $b$, $c$ such that $x^3=a^3+b^3+c^3$.

Answer 2

The answer can also be found in Section 21.11 of Hardy-Wright: 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.

Answer 3

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 well-known 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.

Answer 4

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.

Answer 5

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 $x=pt+a, y=qt-a, z=rt+b, w=st+b$ to $w^3=x^3+y^3+z^3$ where $(a,b)$ are arbitrary integers.

Let $t=\frac{-(rb^2+qa^2+pa^2-sb^2)}{-q^2a+p^2a-s^2b+r^2b},$ then we obtain a parametric solution

$x= (q^2+pq)a^2+(-r^2+s^2)ba+(pr-ps)b^2$
$y= (p^2+pq)a^2+(-s^2+r^2)ba+(qr-qs)b^2$
$z= (qr+pr)a^2+(-p^2+q^2)ba+(s^2-rs)b^2$
$w= (ps+qs)a^2+(-p^2+q^2)ba+(rs-r^2)b^2.$

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
$(x,y,z,w) =(42a^2+17ba-b^2, 7a^2-17ba-6b^2, 56a^2+35ba+9b^2, 63a^2+35ba+8b^2).$
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 $(x,y,z,w) =(28a^2+11ba-3b^2, 21a^2-11ba-4b^2, 35a^2+7ba+6b^2, 42a^2+7ba+5b^2).$
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]

Answer 6

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:

$$d^3-c^3= (d-c)(d^2 +dc +c^2).$$

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 $c=d-1=9-1=8$ to reduce the identity above to a quadratic equation:

$$d^3-c^3= (c+1)^2 + c(c+1) + c^2 = 3c^2 +3c +1 = 9^3 - 8^3 = 217 = 7\cdot31.$$

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 $9^1-6^3=513$ as a sum of two cubes. $513=27\cdot19$ simply by writing:
$9^1-6^3=513= 1^3 + 8^3$. We know that $513$ is divisible by $8+1=9$.

It's not always the case that the largest element $c$ is $c=d-1$. In that case the quadratic equation will not have integer solutions. So we can try $c=d-2$ or $c=d-3$…until we hit on a quadratic solution with integers. If the difference $d^3-c^3$ 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 $c=d-x$ that will provide a number that can be written as a sum of two cubes.