## Task A-4.1.

Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
## First solution.

Let $f(0)=a$ and $f(1)=b$.

By substituting $x=0$ into the initial equation, we get

$$
b=(a-1) f(y)+2
$$

For $a \neq 1$, it follows that $f(y)=\frac{b-2}{a-1}$ for every $y \in \mathbb{R}$. Substituting this into the initial equation, we see that a constant function is not a solution.

Therefore, it must be that $a=1$, and thus $b=2$.

If we now substitute $y=0$ into the initial equation, we get

$$
b=f(x) \cdot a-a-x+2, \quad \text { i.e. } \quad 2=f(x)-x+1
$$

It follows that $f(x)=x+1$.

By verification, we see that this is indeed a solution.