Let $x,y$ be odd integers and let $z$ be an integer. The question is to find all solutions to the equation, $$x^3+y^3+4z^3=0$$ Of course we have the trivial solution $(x,-x,0)$. Are there any others?

By considering the equation modulo $4$ we see that wlog $x=4k+1$ and $y=4m+3$. Numerical experimentation shows that there are no solutions with $|k|, |m| < 10000$. In fact since the cubic residues modulo $9$ are $0,\pm 1$ then we see that $z$ is a multiple of $9$ and moreover, dividing by $9^3$ if necessary, we see that $x^3$ and $y^3$ must have opposite residue modulo $9$. The same holds by considering the equation modulo $7$. I'm also aware of Broughan's theorem, but it's not clear to me whether it helps.


1 Answer 1


Let $X=x/z, Y=y/z$ then we get $X^3+Y^3=-4$.
In general, $X^3+Y^3=n$ can be transformed to elliptic curve $v^2=u^3-432n^2$.
Hence we get $v^2=u^3-6912$.
According to LMFDB, this elliptic curve has rank $0$ and has no integer solution.
Hence $x^3+y^3+4z^3=0$ has no nontrivial solution.

  • $\begingroup$ Thanks a lot for your answer! Perhaps you can help me understand the last point. After transforming from $(x,y,z)$ to $(X,Y)$ we're now looking for rational solutions. The fact that an elliptic curve has rank $0$ means that it only has finitely many rational points. How can I conclude that actually there are no rational points at all? $\endgroup$
    – John Donne
    Nov 21, 2022 at 8:33
  • $\begingroup$ @John ,Elliptic curve of rank $0$ means that all rational points are torsion (has integer coordinates). However according to LMFDB, elliptic curve has no integer solution. So, elliptic curve has no nontrivial solution. $\endgroup$
    – Tomita
    Nov 21, 2022 at 11:48
  • $\begingroup$ Ah I see, of course! Thanks a lot $\endgroup$
    – John Donne
    Nov 21, 2022 at 14:37
  • $\begingroup$ @Tomita Hi Tomita. Do you have the computing power to find three primitive integers $(b,c,d)$ such that $(-b^6-c^6+d^6)^{1/3}=a$ for integer $a$ with $b<c<d<B$ for some bound $B$? Because of an elliptic curve, this in fact has an infinite family of solutions. However, there is at least one that does not belong to this family, so we don't know if our present solutions are the smallest. Can you kindly check this post? $\endgroup$ Aug 21 at 12:40

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