Compute all integer solutions to the equation

$$x^3+y^3+z^3=w^3$$

  • 4
    Have you tried something, or at least taken a look at Tito Pieza's collection? – ccorn Aug 16 '13 at 17:37
  • 1
    It's a bit like Fermat's last theorem :D – gen Aug 16 '13 at 17:40
  • 2
    @gen, unlike Fermat's equation this one surely has infinite number of solutions: $(x,y,-y,x)$ ($x,y \in \mathbb{Z}$) etc. – njguliyev Aug 16 '13 at 17:48
  • 1
    There are many known families of parametric solutions. But as far as I know there is no known parametric representation of all the solutions. – André Nicolas Aug 16 '13 at 18:15
up vote 11 down vote accepted

(I revisited and revised this old question because of this post.)

The complete rational solution to, $$x_1^3+x_2^3+x_3^3+x_4^3=0\tag1$$

is known and there are various formulas (due to Euler, Binet, Steggall, Elkies, etc).

I. Integer case:

However, if we wish to solve the form,

$$x_1^3+x_2^3+x_3^3 = x_4^3\tag2$$

where $0<x_1<x_2<x_3$, then there is also a complete positive integer solution given by Choudhry's On Equal Sums of Cubes (1998). For those who want the summarized version, for positive integers $a,b,c$,

$$\begin{aligned} d\,x_1 &= (-a^3 - b^3 + c^3)c\\ d\,x_2 &= -(a^2 - a b + b^2)^2 + (a + b)c^3\\ d\,x_3 &= (a^2 - a b + b^2)^2 + (2a - b)c^3\\ d\,x_4 &= (a^3 + (a - b)^3 + c^3)c\end{aligned}$$

and,

$$a>b,\quad c >(a^3+b^3)^{1/3}$$

where $d=1$, or $d$ is chosen such that $\text{GCD}(x_1,x_2,x_3,x_4)=1$.

II. Example:

Given any solution $x_1,x_2,x_3,x_4$, then one can recover $a,b$ as,

$$(x_2 + x_3) (a x_2 - b x_3)^3 + (a^2 - a b + b^2)\, x_1^3\, \big((2 a - b) x_2 - (a + b) x_3\big)=0\tag3$$

then $c,d$,

$$3 a c^2 x_1 + (a^3 + b^3 - c^3) (x_2 + x_3)=0\tag4$$

$$3 a (a^2 - a b + b^2)^2 + d \, \big((2 a - b) x_2 - (a + b) x_3\big)=0\tag5$$

For example, the smallest is the well-known,

$$3^3+4^3+5^3 = 6^3$$

Substituting $x_1,\,x_2,\,x_3 = 3,4,5$ into $(3)$, one finds the linear factor,

$$(a-2b)=0$$

so $a=2,\;b=1$. Then into $(4)$ and $(5)$,

$$(c-3)=0$$

$$(d-18) = 0$$

III. Statistics:

A lot of positive integers $N^3$ can be expressed as a sum of three positive cubes. The sequence A023042 begins as,

$$N= 6, 9, 12, 18, 19, 20, 24, 25, 27, 28, 29, 30, 36, 38, 40,\dots$$

and for a higher range,

$$N=\dots,9990, 9991, 9992, 9993, 9994, 9995, 9996, 9997, 9998, 9999, 10000,\dots$$

which shows no skip. In fact, based on the table here, it is $94.2\text{%}$ of all $N<10000$. Extrapolating from the trends in the table, it is easily $99\text{%}$ of all $N<1000000$. Thus, choosing a random $N$ in the high end of the range, chances are very good there will be a positive integer solution to,

$$x_1^3+x_2^3+x_3^3 = N^3$$

P.S. Thanks, ccorn!

  • Do you know if the 3.3.3 equation $$x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$$ has — or, indeed, can have — a complete solution/parameterization? – Kieren MacMillan Oct 14 '14 at 15:38
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    @KierenMacMillan: The substitution, $$(a+b)^3+(c+d)^3+(e+f)^3 =(a-b)^3+(c-d)^3+(e-f)^3$$ reduces the problem to solving a quadratic. So there might be a complete solution, but I haven't really explored this thoroughly yet. – Tito Piezas III Jan 27 '15 at 5:48
  • Choudhry has a complete parameterization in the special case where $x_1+x_2+x_3=y_1+y_2+y_3$. See hal.archives-ouvertes.fr/hal-01112553/document – Kieren MacMillan Apr 13 '16 at 19:40

For Diophantine equations: $X^3+Y^3+Z^3=R^3$

Symmetric solutions can be written as: $\left|\frac{X+Y}{R-Z}\right|=\frac{b^2}{a^2}$

Solutions have the form:

$X=b(a^6-b^6)p^3+3ba^6p^2s+3ba^6ps^2+b(a^6-b^6)s^3$

$Y=(a^7-ab^6)p^3+3(a^7-a^4b^3-ab^6)p^2s+3(a^7-2a^4b^3)ps^2+(a^7-3a^4b^3+2ab^6)s^3$

$Z=(2ba^6+3a^3b^4+b^7)p^3+3(2ba^6+a^3b^4)p^2s+3(2ba^6-a^3b^4)ps^2+(2ba^6-3a^3b^4+b^7)s^3$

$R=(a^7+3a^4b^3+2ab^6)p^3+3(a^7+2a^4b^3)p^2s+3(a^7+a^4b^3-ab^6)ps^2+(a^7-ab^6)s^3$

Solutions have the form:

$X=3(b^6-7a^6)as^2-9a^4ps-ap^2$

$Y=(6ab^6+21a^7-27b^3a^4)s^2-3(2ab^3-3a^4)ps+ap^2$

$Z=(3b^7+33ba^6-18a^3b^4)s^2+3(4ba^3-b^4)ps+bp^2$

$R=(3b^7+6ba^6-9a^3b^4)s^2+3(2ba^3-b^4)ps+bp^2$

Solutions have the form:

$X=(b^6-7a^6)ap^2+9a^4ps-3as^2$

$Y=(2ab^6+9b^3a^4+7a^7)p^2-3(2ab^3+3a^4)ps+3as^2$

$Z=(3b^4a^3+2ba^6+b^7)p^2-3(2ba^3+b^4)ps+3bs^2$

$R=(6b^4a^3+11ba^6+b^7)p^2-3(4ba^3+b^4)ps+3bs^2$

Solutions have the form:

$X=(aj-bt)^2sp^6-3t^2j^2p^2s^5+3(aj+bt)jtps^6-(a^2j^2+abtj+b^2t^2)s^7$

$Y=2(aj-bt)^2sp^6-6(aj-bt)jtp^4s^3+3(a^2j^2-b^2t^2)p^3s^4+3t^2j^2p^2s^5-3(aj+bt)tjps^6+$

$+(a^2j^2+abtj+b^2t^2)s^7$

$Z=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)btp^4s^3+3t^2j^2p^3s^4-6bjt^2p^2s^5-$

$-(a^2j^2-2abtj-2b^2t^2)ps^6$

$R=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)ajp^4s^3+3t^2j^2p^3s^4-6atj^2p^2s^5-$

$-(b^2t^2-2abtj-2a^2j^2)ps^6$

$a,b,j,t,p,s$ - what some integers.

  • Thank you very much! But, why? – ziang chen May 3 '14 at 11:19
  • I do not understand. Why what? – individ May 3 '14 at 11:48
  • How to prove that these Solutions have the form? – ziang chen May 3 '14 at 11:50
  • There are no such solutions only. I brought these, but there are others when relationships are not squares. I just like to write such a formula. – individ May 3 '14 at 12:03

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