6. Let $\mathbb{R}^{+}$ denote the set of all positive real numbers. Find all functions $f$ : $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ satisfying for all $x, y \in \mathbb{R}^{+}$ the equation

$$
f(x f(y))=f(x y)+x .
$$

(P. Kaňovský)
Solution. If we substitute the value of $f(x)$ for $x$ in the given equation, we get the equation

$$
f(f(x) f(y))=f(f(x) y)+f(x),
$$

from which we can express $f(f(x) y)=f(f(x) f(y))-f(x)$. Another expression for the same term $f(f(x) y)$ is obtained when we interchange the values of $x$ and $y$ in the original equation; we get $f(f(x) y)=f(y x)+y$. By comparing both expressions, we obtain the equation

$$
f(f(x) f(y))=f(y x)+y+f(x)
$$

whose left side does not change if we interchange the values of $x$ and $y$. Therefore, the right side of this equation must have the same property, so it must hold that

$$
f(y x)+y+f(x)=f(x y)+x+f(y), \quad \text { or } \quad y+f(x)=x+f(y) .
$$

A further obvious manipulation yields the equation $f(x)-x=f(y)-y$, which must be satisfied for any $x, y \in \mathbb{R}^{+}$. This means that the function $x \mapsto f(x)-x$ is constant on the set $\mathbb{R}^{+}$, so the sought function $f$ must be of the form $f(x)=x+c$ for some number $c$. By substituting this formula into both sides of the original equation

$$
\begin{gathered}
f(x f(y))=x f(y)+c=x(y+c)+c=x y+c x+c \\
f(x y)+x=(x y+c)+x=x y+x+c
\end{gathered}
$$

we find that it satisfies only $c=1$. The sought function $f$ is thus unique and is given by the formula $f(x)=x+1$.