<s>[INST] When is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$? Let $x$, $y$, $z$ be three natural numbers such that $1< x< y< z$. For how many sets of values of $(x,y,z)$, is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$?
I noticed that $(x-1)(y-1)(z-1)=(xyz-1)-z(x+y-1)-xy+x+y$.
But i don't know how to proceed from here. Any clues?
 [/INST] Well, at least there are solutions: $x = 2, y = 4, z = 8$ and $x = 3, y = 5, z = 15$
$(x-1)(y-1)(z-1) = x y z - 1 - p(x, y, z)$ with $p(x, y, z) = x y + y z + x z - x - y - z\,.$
If $(x-1)(y-1)(z-1)$ divides $x y z -1$, it also must divide $p(x, y, z)$.
We can prove easily that $(x-1)(y-1)(z-1) >  \lvert p(x, y, z) \rvert$ for $x, y, z > 6$, so this is not possible and all solutions must have $x \leq 6$. Try to find similar bounds for $y$ in the case $x \leq 6$ and finally for $z$.
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