Show that for all integers k, there is a solution with
$x=3k^2-1$ and
$z=k^2+1$
You will need to calculate $y$ to show that there is such a solution, and show that the solution $(x,y,z)$ is relatively prime.
So I calculated that $y=k^3-3k$, but I have to show that $x,y,z$ are all relatively prime. Halp.
Let $k$ be even. We show that no prime $p$ can divide both $x$ and $z$. Suppose to the contrary that $p$ divides $3k^2-1$ and $p$ divides $k^2+1$. Then $p$ divides $3(k^2+1)-(3k^2-1)$, so $p$ divides $4$. This forces $p=2$, which is impossible if $k$ is even.
The three numbers are not pairwise relatively prime if $k$ is odd, for if $k$ is odd then all three numbers are even.
Remark: Note that the three numbers are pairwise relatively prime if and only if $x$ and $z$ are relatively prime. For example, let us show that if $x$ and $z$ are relatively prime, so are $x$ and $y$. Suppose to the contrary that $x$ and $y$ are not relatively prime. Then some prime $p$ divides $x$ and $y$. It follows that $p$ divides $x^2+y^2$, and therefore $p$ divides $z^3$, and therefore $p$ divides $z$, contradicting the fact that $x$ and $z$ are relatively prime.
