Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

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

for all $x, y \in \mathbb{R}$.

Answer. The only such function is $f(x)=x-1, x \in \mathbb{R}$.
By looking at the variable $y$, one can immediately see that the function $f$ is injective, because from $f(a)=f(b)$ it follows, using the functional equation for $y=a$ and $y=b$, that

$$
a=f(x f(a)+1)-f(f(x) f(a))=f(x f(b)+1)-f(f(x) f(b))=b .
$$

We now set $y=0$ and obtain, due to the injectivity,

$$
x f(0)+1=f(x) f(0) \text {. }
$$

If $f(0)=0$, then $1=0$ would result, which cannot happen. Therefore, we can divide by $f(0)$ and obtain $f(x)=x+c$ for some constant $c$.

By substituting into the original functional equation, we get

$$
\begin{aligned}
& x(y+c)+1+c=y+(x+c)(y+c)+c \\
& \Longleftrightarrow \quad 1=y(1+c)+c^{2}
\end{aligned}
$$

which gives the conditions $c=-1$ (coefficient of $y$) and $c^{2}=1$ (constant term). Thus, the only solution is $f(x)=x-1$.

(Theresia Eisenkölbl)