Conditions under which $a+b+c$ divides $1-abc$ What are the conditions such that $a+b+c$ divides $1-abc$, where $(a, b, c)$ are nonzero integers ?

The triples $(a,b,c)$ with the property that $a+b+c$ divides $1-abc$ can be characterized as triples $(a,b,d-a-b)$, where $d$ is a divisor of $1+ab(a+b)$.  This is because $a+b+c=d$ for such triples, while 
$$1-abc=1-ab(d-a-b)=1+ab(a+b)-abd$$
For example, if $a=b=4$, we have $1+ab(a+b)=129$, which has $8$ divisors:  $d=\pm1,\pm3,\pm43$ and $\pm129$, giving triples
$$\begin{align}
&(4,4,-7)\\
&(4,4,-9)\\
&(4,4,-5)\\
&(4,4,-11)\\
&(4,4,35)\\
&(4,4,-51)\\
&(4,4,121)\\
&(4,4,-137)
\end{align}$$
