Well I was doing some questions and i found something. This equation $x^3+y^3+z^3=w^3$ has only one solution which is $x=3,y=4,z=5,w=6$.

And what I have have proposed is that there is not other natural number solution for all values of $x,y,z,w>1$. So my question is this proposition true and if it is can we prove that there is only one natural number solution to this.

well there is a condition for this that is these numbers add up to give the cube of a perfect number.


closed as unclear what you're asking by Karl Schwede, Chris Godsil, Ramiro de la Vega, José Figueroa-O'Farrill, Stefan Kohl Jan 22 '15 at 14:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ There certainly are other solutions, as your equation is homogeneous and symmetric. So, you should state your problem more carefully. And, please, use TeX. $\endgroup$ – Alex Degtyarev Jan 22 '15 at 10:36
  • 1
    $\begingroup$ The line $x+y=0, \, z-w=0$ is contained in the Fermat projective surface $x^3+y^3+z^3-w^3=0$, so you trivially have infinitely many integer solutions of the form $(x, \, -x, \, z, \, z)$. $\endgroup$ – Francesco Polizzi Jan 22 '15 at 10:38
  • $\begingroup$ @FrancescoPolizzi — That's right. Though OP wants natural number solutions. So $x,y,z,w \in \mathbb{Z}_{> 0}$. $\endgroup$ – jmc Jan 22 '15 at 10:42
  • 6
    $\begingroup$ We have $(a^4-2ab^3)^3 + (a^3b+b^4)^3 + (2a^3b-b^4)^3 = (a^4+ab^3)^3$. So, you can get infinitely many positive integer solutions by taking relatively large $a$ compared with $b$, e.g., setting $a=2$ and $b=1$, you get $12^3+9^3+15^3=18^3$ ($= 5832$). (I found this identity here sites.google.com/site/tpiezas/010) $\endgroup$ – Yuichiro Fujiwara Jan 22 '15 at 11:00
  • 9
    $\begingroup$ @sayanchattopadhyay: Well, 4 is not a prime... $\endgroup$ – Michael Stoll Jan 22 '15 at 11:19

Even with your additional condition that the sum is the cube of a perfect number, the following are two counterexamples:

  • $21^3 + 19^3 + 18^3 = 28^3$

  • $495^3 + 82^3 + 57^3 = 496^3$

  • 3
    $\begingroup$ On meta: Is it frowned upon to answer a question and vote to close? $\endgroup$ – joro Jan 23 '15 at 6:28
  • $\begingroup$ @joro: What I wanted to express is that on the one hand one can guess a meaning of the question and answer accordingly, but on the other it is unclear whether this guess is correct and the answer is helpful for the OP -- even more as the OP was online in the meantime and did not leave a comment. I would not have voted to close if the OP either had made a comment saying that my answer was what they were looking for, or edited the question to make clear what else they would like to know. $\endgroup$ – Stefan Kohl Jan 23 '15 at 10:09

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