Let $I_A(x)=\begin{cases} 0 & x \in A\\1 & \text{else} \end{cases}$ where $A \subseteq \mathbb{R}.$
A particularly hard functional equation boiled down to determining the sets $A$ satisfying $I_A(x)I_A(x+y)=I_A(y)I_A(x-y).$ Due to analysis of the original functional equation, we know that $I_A(0)=1,$ or $0 \notin A.$
I have determined the following properties:
$I_A(x) = I_A(-x),$ or $x \in A \iff -x \in A$
$x \in A \Rightarrow x/2 \in A$
If $a \in A,$ then $I_A(x)I_A(a-x)=I_A(x)I_A(a+x)=0.$
I've noticed that if A is non-empty, it will contain arbitrarily large and small values. Thus, I've conjectured that the only possible $A$ that satisfy the aforementioned indicational equation are $\emptyset, \, \mathbb{R} \setminus 0.$
The set of all rationals numbers and the set of all dyadic rationals numbers both satisfy the given equation. There are many solutions.