What are the solutions in numbers of $xyz \mid x^n + y^n + z^n$, $x,y,z$ globally coprime

Question

What are globally coprime integers $x,y,z\in \mathbb Z^*$ such that $xyz$ divide $x^n + y^n + z^n$?

I have no other motivation for that problem but its inherent beauty and interest. Note that it can be assumed without loss of generality that $x\geq |y|\geq |z|$.

Here is what I've obtained so far:

• $n=1$: The solutions $(x,y,z)$ of the former relation for which $x\geq |y|\geq |z|$ is $$\{(2,1,1), (1,1,1), (3,2,1), (x,1,-1), (x,-1,1):\ x\in \mathbb N^*\} \cup \{(x,y,z): x+y+z = 0\}$$ (straightforward).

• $n = 2$: I was told in another forum that the equation $$xyz = a(x^2+y^2+z^2)$$ is possible in globally coprime numbers only if $a = 3$, in which case it is known as the Markov equation. By vieta Jumping, it is easy to obtain infinitely many solutions from a single one, but it is unclear if the set of solutions of the Markov equation stems from the trivial solution (1,1,1). I would be happy to have more information on that subject, and also how they prove that $a=3$ is the only ratio that work.

• $n >1$ odd: I've obtained the following fact:

If $xyz$ divides $x+y+z$ and $n$ is odd, $xyz$ divides $x^n + y^n + z^n$. Since the set of solution $(x,y,z)$ of the former relation for which $x\geq |y|\geq |z|$ is $$\{(2,1,1), (1,1,1), (3,2,1), (x,1,-1), (x,-1,1):\ x\in \mathbb N^*\} \cup \{(x,y,z): x+y+z = 0\},$$ these are also solutions of the proposed equation.

The proof by induction of the above proposition is based on the following formula: $$x^n+y^n+z^n = (x^{n-1}+y^{n-1}+z^{n-1})(x+y+z) - (x^{n-2}+y^{n-2}+z^{n-2})(xy+xz+yz) + (x^{n-3}+y^{n-3}+z^{n-3})xyz.$$ Despite their is no obvious reason to restrict this problem to prime exponents $n$, I would be happy to know if there are other solutions in globally coprime numbers for odd prime $n$.

Answer 1

Some solutions for $n=7$:

n=7;y=2;z=3;x_i=[1, 463, -5, -2315]
n=7;y=2;z=5;x_i=[78253, -7, -1597]
n=7;y=2;z=7;x_i=[-9, -639, -11601, -823671]
n=7;y=2;z=9;x_i=[-11, -4783097]
n=7;y=2;z=11;x_i=[559, 1499023, -13, -34861]
n=7;y=2;z=13;x_i=[62748645, -15]
n=7;y=2;z=15;x_i=[493, -17]
n=7;y=2;z=17;x_i=[-19]
n=7;y=2;z=19;x_i=[-21, -127695981]
n=7;y=3;z=2;x_i=[1, 463, -5, -2315]
n=7;y=3;z=4;x_i=[-7, -379, -18571]
n=7;y=3;z=5;x_i=[-8, -20078]

Some solutions for $n=10$:

n=10;y=5;z=13;x_i=[1, 194, 710661121, 137868257474, -1, -194, -710661121, -137868257474]
n=10;y=5;z=17;x_i=[1337954, 1506781, -1337954, -1506781]

I conjecture there are infinitely many coprime solutions for all $n$.