# Proof of a cubic equation problem [closed]

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 KohlJan 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.

• There certainly are other solutions, as your equation is homogeneous and symmetric. So, you should state your problem more carefully. And, please, use TeX. – Alex Degtyarev Jan 22 '15 at 10:36
• 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 $(x, \, -x, \, z, \, z)$. – Francesco Polizzi Jan 22 '15 at 10:38
• @FrancescoPolizzi — That's right. Though OP wants natural number solutions. So $x,y,z,w \in \mathbb{Z}_{> 0}$. – jmc Jan 22 '15 at 10:42
• 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) – Yuichiro Fujiwara Jan 22 '15 at 11:00
• @sayanchattopadhyay: Well, 4 is not a prime... – Michael Stoll Jan 22 '15 at 11:19

• $21^3 + 19^3 + 18^3 = 28^3$
• $495^3 + 82^3 + 57^3 = 496^3$