What are the conditions such that $a+b+c$ divides $1-abc$, where $(a, b, c)$ are nonzero integers ?
2026-04-14 07:34:25.1776152065
Conditions under which $a+b+c$ divides $1-abc$
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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
$$\begin{align} &(4,4,-7)\\ &(4,4,-9)\\ &(4,4,-5)\\ &(4,4,-11)\\ &(4,4,35)\\ &(4,4,-51)\\ &(4,4,121)\\ &(4,4,-137) \end{align}$$