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# 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

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

Note by Sharky Kesa
8 months, 4 weeks ago

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Call 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 · 8 months, 3 weeks ago

@Sharky Kesa What say? · 8 months, 3 weeks ago

It should say $$x + y \in \{0,1\}$$. · 8 months, 4 weeks ago

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

Hence, no such function $$f$$ exists.

@Sharky Kesa · 8 months, 4 weeks ago

Sorry, I had meant to write for distinct $$x$$ and $$y$$. · 8 months, 3 weeks ago

Then, we get $|f(x)|=1 \quad \forall x \in \mathbb{Q}\\f(x)f(\frac{1}{x})=-1 \quad \forall x \in \mathbb{Q}-\{1,-1\}\\f(x)f(-x)=-1 \quad \forall x \in \mathbb{Q}-\{0\}\\f(x)f(1-x)=-1 \quad \forall x \in \mathbb{Q}-\left\{\frac{1}{2}\right\}\\\implies f(1+x)=f(x) \quad \forall x \in \mathbb{Q}-\left\{0,\frac{-1}{2}\right\}$ · 8 months, 3 weeks ago

This doesn't give us a rule for $$f$$ per se but it's a good starting. · 8 months, 3 weeks ago

Working on figuring $$f(-1/2)$$ and on [0,1]... · 8 months, 3 weeks ago

My hint is to consider continued fractions for all rational values. · 8 months, 3 weeks ago

Well, I don't know much about continued fractions...

Thanks for the hint anyways. · 8 months, 3 weeks ago