I am currently working on this functional equation : find $f$ piecewise-continuous over $[0,b+c]$, and $a,b,c,d>0$ such that
\begin{equation} \left\lbrace \begin{array}{ll} |f(x)|=1& x\in [0,b+c] \\ f(x)=-1& x \in [a,b] \cup [c,d] \\ f(x)=f(x+d)& x \in [0,a] \\ f(x)=-f(x-b)& x \in [b,b+c]$ \end{array}\right. \end{equation}
with $a+d=b+c$, $b-a=d-c$.
Eventually, we assume that $\exists (p,q)\in \mathbb{N}\times\mathbb{N^*}, d=\frac{p}{q}(b+d)-\frac{1}{q}(b-a)$.
The question is the following :
Which relation between $p$ and $q$ allows to define $f$ everywhere in $[0,b+c]$ without any contradiction?
Any tips or partial results will be appreciated.