Exercise 3. Find all functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that, for all real numbers $x$ and $y$, we have

$$
f(x f(y))+x=f(x) f(y+1)
$$
Solution to Exercise 3 By taking $x=0$, we obtain $f(0)=f(0) f(y+1)$ for all $y$, so either $f(0)=0$, or $f(y+1)=1$ for all $y$. In the second case, $f$ is constantly equal to 1, but then the equation becomes

$$
1+x=1 \times 1
$$

for all $x$, which is impossible. Therefore, we have $f(0)=0$. By taking $y=0$, we then obtain $x=f(x) f(1)$ for all $x$. In particular, by taking $x=1$, we get $1=f(1)^{2}$, so $f(1)$ is either 1 or -1.

If $f(1)=1$, then $x=f(x)$, so $f$ is the identity function, which is indeed a solution. If $f(1)=-1$, we get $x=-f(x)$ for all $x$, so $f(x)=-x$ for all $x$, and we verify that this function is also a solution to the problem.

## Common Exercises