Compute all integer solutions to the equation


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

(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 $a,b,c$,

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


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

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,


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


$$(d-18) = 0$$

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!

  • $\begingroup$ 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? $\endgroup$ – Kieren MacMillan Oct 14 '14 at 15:38
  • 1
    $\begingroup$ @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. $\endgroup$ – Tito Piezas III Jan 27 '15 at 5:48
  • $\begingroup$ 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 $\endgroup$ – 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:





Solutions have the form:





Solutions have the form:





Solutions have the form:








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

  • $\begingroup$ Thank you very much! But, why? $\endgroup$ – ziang chen May 3 '14 at 11:19
  • $\begingroup$ I do not understand. Why what? $\endgroup$ – individ May 3 '14 at 11:48
  • $\begingroup$ How to prove that these Solutions have the form? $\endgroup$ – ziang chen May 3 '14 at 11:50
  • $\begingroup$ 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. $\endgroup$ – individ May 3 '14 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.