Finding rational solutions to $x^3+5y^3+25z^3-15xyz=0$, or proving that none exist

Question

I'm trying to find values $x,y,z \in \mathbb{Q}$ such that $$x^3+5y^3+25z^3-15xyz=0$$ or prove that no such numbers exist.

I'm not sure what direction to go in because I can't isolate any one of the variables. Any help would be greatly appreciated.

Answer 1

Except for the trivial solution there are no other ones. First suppose there is a nonzero integer solution $x,y,z\in\mathbb Z$ of the equation. If each of $x,y,z$ has a factor $5$, then $x/5,y/5,z/5$ is also a nonzero integer solution. Repeat this procedure until at least one of them is not a multiple of $5$.

Now mod $5$ on both sides of the equation you get $x^3\equiv 0\pmod 5$, which means $x$ is a multiple of $5$. Let $x=5k,k\in\mathbb Z$. Then the equation becomes $$125k^3+5y^3+25z^3-75kyz=0\\\iff y^3+5z^3+25k^3-15yzk=0$$ Mod 5 again gives you $y^3\equiv0\pmod 5\implies y$ is a multiple of $5$. Let $y=5j,j\in\mathbb Z$. Then the equation becomes $$125j^3+5z^3+25k^3-75kjz=0\\\iff z^3+5k^3+25j^3-15jzk=0$$ which implies $z$ is a multiple of $5$. Hence $x,y,z$ are all multiples of $5$, contradicting our assumption.

Finally, the equation has no nonzero rational solution because it has no nonzero integer solution (think why).

Answer 2

$x^3 = 15xyz - 25z^3 - 5y^3$

The RHS is divisible by $5.$ The LHS (therefore $x$) must also be divisible by $5.$

$x = 5u$

$125u^3 = 75uyz - 25z^3 - 5y^3\\ y^3 = 15uyz - 5z^3 - 25u^3$

And we have cycled back an equation that is equivalent to where we started.

This kicks of an argument of infinite descent. $y$ must be divisible by 5, then we will get that $z$ must be divisible by $5.$ And then it will cycle around to x divisible by 25, etc.

Other than the trivial $(x,y,z) = (0,0,0)$, there are no other solutions.