Provided the equation
![xyz = p^n(x + y + z)](/media/m/b/e/5/be5413442813e0592cca460ea71a771d.png)
where
![p \geq 3](/media/m/9/f/0/9f0c935d7738faea35703b7f65c29563.png)
is a prime and
![n \in \mathbb{N}](/media/m/2/b/a/2ba27c6141ca415bb86bae1d237f1fac.png)
. Prove that the equation has at least
![3n + 3](/media/m/7/0/f/70f1845af83cd1154bd14805cd47c027.png)
different solutions
![(x,y,z)](/media/m/5/2/e/52e5ec762ff5263770c6e6c12cb9838e.png)
with natural numbers
![x,y,z](/media/m/b/7/2/b72c022e9d438802d328d34eb61bb4ba.png)
and
![x < y < z](/media/m/f/8/2/f8230ecd7f3fe97f52c4f0ed42914597.png)
. Prove the same for
![p > 3](/media/m/d/1/4/d14b51fb9611c16eac843d287f8c9fdf.png)
being an odd integer.
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Provided the equation $xyz = p^n(x + y + z)$ where $p \geq 3$ is a prime and $n \in \mathbb{N}$. Prove that the equation has at least $3n + 3$ different solutions $(x,y,z)$ with natural numbers $x,y,z$ and $x < y < z$. Prove the same for $p > 3$ being an odd integer.