Exercise 4. Find all functions $\mathrm{f}$ from $\mathbb{R}$ to $\mathbb{R}$ such that for all real numbers $x, y$,

$$
f(x) f(y)=f(x y)+x+y
$$
Solution to Exercise 4 By evaluating at $y=0$ we obtain $f(x) f(0)=f(0)+x$ for all real $x$. For $x=1$, we get that $f(0) \neq 0$, otherwise we would have $1=0$. By evaluating at $x=0$ we get $f(0)^{2}=f(0)$, so $f(0)=1$ and thus $f(x)=1+x$ for all real $x$.

Conversely, let $f$ be the function such that $f(x)=1+x$ for all real $x$ and let $(x, y) \in \mathbb{R}^{2}$, then $f(x) f(y)=(1+x)(1+y)=1+xy+x+y=f(xy)+x+y$ so $f$ does indeed satisfy the equation. The unique solution to the equation is therefore the function $f$ such that $f(x)=1+x$ for all real $x$.

Grader's Comment The exercise is generally well done. However, it is important to distinguish between two parts in a functional equation:

- The analysis, which involves choosing a function $\mathrm{f}$ that satisfies the equation and determining the properties satisfied by $\mathrm{f}$. Here, it is established that if $\mathrm{f}$ is a solution to the equation, then for all real $\mathrm{x}, \mathrm{f}(\mathrm{x})=x+1$.
- The synthesis, which involves verifying that conversely, the function $\mathrm{f}$ that maps $x$ to $x+1$ is indeed a solution to the equation.

Some students forgot the second part and were penalized.