by 
and the number of distinct prime divisors of 
.
be a positive integer. Prove that there exist infinitely many positive integers 
and 
does not divide 
for any positive integers 
satisfying 
.
such that for any 
:
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
such that 
for all 
Y by Adventure10, Mango247
Given 
are positive integers pairwise distinct and 
is also a positive integer 
.
Determine under which conditions can![$\sqrt[n]{p},\sqrt[n]{q},\sqrt[n]{r}$](//latex.artofproblemsolving.com/3/1/3/3133b60c921168020d149a9d28ef69b756a704d9.png)
form terms of a same arithmetic progression.
are positive integers pairwise distinct and 
is also a positive integer 
.Determine under which conditions can
![$\sqrt[n]{p},\sqrt[n]{q},\sqrt[n]{r}$](http://latex.artofproblemsolving.com/3/1/3/3133b60c921168020d149a9d28ef69b756a704d9.png)
form terms of a same arithmetic progression.
(with 
)![$\sqrt[n]{p}, \sqrt[n]{q}=\sqrt[n]{p}+d, \sqrt[n]{r} = \sqrt[n]{p}+2d$](http://latex.artofproblemsolving.com/2/4/2/2427d624522f63ee6e1ae140ce0d049c6b8e028c.png)
(for 
)![$\sqrt[n]{r} = \sqrt[n]{p}+2(\sqrt[n]{q}-\sqrt[n]{p}) \Rightarrow \sqrt[n]{\frac{p}{q}} + \sqrt[n]{\frac{r}{q}} = 2$](http://latex.artofproblemsolving.com/a/5/c/a5c4ead44d71f94b2a68e435524fa2b33bc076d7.png)
(i).
(for 
)
(for 
)
