Balti Tee 1995 · Ülesanne 10

Solution: Substituting $x=y=\frac{1}{2} z$ in (ii) we get

$$f\left(\frac{1}{z}\right)=2 f\left(\frac{2}{z}\right)$$

for all $z \neq 0$. Substituting $x=y=\frac{1}{z}$ in (iii) yields

$$\frac{2}{z} f\left(\frac{2}{z}\right)=\frac{1}{z^{2}}\left(f\left(\frac{1}{z}\right)\right)^{2}$$

for all $z \neq 0$, and hence

$$2 f\left(\frac{2}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2} .$$

From (1) and (2) we get

$$f\left(\frac{1}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2},$$

or, equivalently,

$$f(x)=x(f(x))^{2}$$

for all $x \neq 0$. If $f(x)=0$ for some $x$, then by (iii) we would have

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

which contradicts the condition (i). Hence $f(x) \neq 0$ for all $x$, and (3) implies $x f(x)=1$ for all $x$, and thus $f(x)=\frac{1}{x}$. It is easily verified that this function satisfies the given conditions.