# 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
4 years, 6 months ago

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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

- 4 years, 6 months ago

@Sharky Kesa What say?

- 4 years, 6 months ago

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

Hence, no such function $f$ exists.

- 4 years, 6 months ago

Sorry, I had meant to write for distinct $x$ and $y$.

- 4 years, 6 months ago

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

- 4 years, 6 months ago

This doesn't give us a rule for $f$ per se but it's a good starting.

- 4 years, 6 months ago

Working on figuring $f(-1/2)$ and on [0,1]...

- 4 years, 6 months ago

My hint is to consider continued fractions for all rational values.

- 4 years, 6 months ago

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

Thanks for the hint anyways.

- 4 years, 6 months ago

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

- 4 years, 6 months ago