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

\[f(x) f(y) = -1\]

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## Comments

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TopNewestCall the required functions as \(f_i \quad i=1,2\).

Then,

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

@Sharky Kesa

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@Sharky Kesa What say?

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Put \(x=y=1\) then, we get \((f(1))^2=-1\).

Hence, no such function \(f\) exists.

@Sharky Kesa

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Sorry, I had meant to write for distinct \(x\) and \(y\).

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Then, we get \[|f(x)|=1 \quad \forall x \in \mathbb{Q}\\f(x)f(\frac{1}{x})=-1 \quad \forall x \in \mathbb{Q}-\{1,-1\}\\f(x)f(-x)=-1 \quad \forall x \in \mathbb{Q}-\{0\}\\f(x)f(1-x)=-1 \quad \forall x \in \mathbb{Q}-\left\{\frac{1}{2}\right\}\\\implies f(1+x)=f(x) \quad \forall x \in \mathbb{Q}-\left\{0,\frac{-1}{2}\right\}\]

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Thanks for the hint anyways.

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It should say \(x + y \in \{0,1\}\).

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