Find all finite set S such that any $a, b, c \in S$, $ab + bc + ca \in S$ Find all finite set of real numbers S such that: $ab + bc + ca \in S$ with any distinct $a, b, c \in S$

I just can solve the problem when $\exists$ at least $3$ elements $\in S\ge 1$
In another case, I got stuck.

Any set of cardinality $< 3$ is (trivially) a solution.
There  are three-element solutions $\{a, b, c\}$ with $b = \dfrac{a(1-c)}{a+c}$ for any  $a$ and $c$ such that $a \ne -c$ and $a, \dfrac{a(1-c)}{a+c}, c$ are distinct.
There are  four-element solutions in complex numbers, e.g.
$$ \eqalign{a&=\dfrac{1}{3}+\dfrac{\sqrt {13}}{6}+\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr b & =\dfrac{1}{3}+\dfrac{
\sqrt {13}}{6}-\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr c &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}-\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}+\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr
d &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}+\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}-\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr}$$
