Problem 917

## Solution

Inserting $y=0$ yields the equation $f(1)=(f(x)+f(0)) /(f(x)-f(0))$ for $x \neq 0$, i.e.

$$f(x)(f(1)-1)=f(0)(f(1)+1)$$

From this follows $f(1)=1$ and therefore also $f(0)=0$, because otherwise $f$ would be constant on $\mathbb{R} \backslash\{0\}$, i.e. not injective. Set $y=-x \neq 0$, then follows

$$f(-x)=-f(x) \quad \forall x \neq 0$$

If you set $y=x z$ with $z \neq 1$ and $x \neq 0$, then on the one hand

$$f\left(\frac{x+x z}{x-x z}\right)=\frac{f(x)+f(x z)}{f(x)-f(x z)}$$

On the other hand

$$f\left(\frac{x+x z}{x-x z}\right)=f\left(\frac{1+z}{1-z}\right)=\frac{1+f(z)}{1-f(z)}$$

A comparison of (3) and (4) now results for all $z \neq 1$ and $x \neq 0$ in the equation

$$f(x z)=f(x) f(z)$$

However, we already know that $f(0)=0$ and $f(1)=1$, so (5) is true for all real $x, z$. With $x=z$ it further follows from (5) that $f(x) \geq 0$ for all $x \geq 0$. Since $f$ is injective and $f(0)=0$, even $f(x)>0$ applies for all $x>0$. If $x>y \geq 0$, then $(x+y) /(x-y)>0$ applies and therefore the left-hand side of the original equation is positive, as is the right-hand side. This also applies to the numerator on the right, i.e. also to the denominator, i.e. $f(x)>f(y)$, and therefore $f$ is strictly monotonically increasing to $\mathbb{R}_{\geq 0}$. Together with (5), this implies that for $x \geq 0$ $f(x)=x^{c}$ with a positive constant $c$. Inserting this into the original equation (where $x>y \geq 0$ ) shows that $c=1$. Together with (2), this now gives the only solution

$$f(x)=x$$