Strange Functional Equation

A peculiar functional equations question I found:

Find all functions $f:\mathbb{Q} \rightarrow [-1,1]$ such that if $xy=1$ or $x+y\in\{0,1\}$ for distinct rational $x$ and $y$ , then

$f(x) f(y) = -1$

#Algebra

Easy Math Editor

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Comments

Call the required functions as $f_i \quad i=1,2$ .

Then,

$\begin{aligned} f_i(x)&=f_i(\{x\}) \quad \forall x \not\in \left\{-\frac{1}{2}\right\} \cup \mathbb{Z}\\ f_i(x)&=f_i(1) \quad \forall x \in \mathbb{Z}^+\\ f_i(x)&=f_i(0) \quad \forall x \in \mathbb{Z} - \mathbb{Z}^+\\ f_1(x)&= 1 \quad \forall x \in \left [0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_1(x)&=-1 \quad \forall x \in \left[\frac{1}{2}, 1 \right]\\ f_2(x)&=-1 \quad \forall x \in \left[0,\frac{1}{2} \right) \cup \left\{-\frac{1}{2}\right\} \\ f_2(x)&=1 \quad \forall x \in \left[\frac{1}{2}, 1 \right] \end{aligned}$

@Sharky Kesa