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

$$
f(f(x) f(y))=f(x) y
$$
We check that $f(x)=0$ is a solution to the functional equation. Now suppose there exists a real number $a$ such that $f(a) \neq 0$. Let $b$ be a real number, and set $x=0$ and $y=\frac{b}{f(a)}$, we then have:

$$
f\left(f(a) f\left(\frac{b}{f(a)}\right)=b\right.
$$

so $\mathbf{f}$ is surjective. We can therefore replace $f(x)$ with $t$, let $k$ be the antecedent of 1, we then have $f(t)=t k$, it is then sufficient to test all functions of the form $f(x)=k x$, and it follows that the only solution is $f(x)=x$.

## - Cauchy's Equation -

This part has not been covered in class, and I invite anyone interested to refer to the course offered to group $C$ on this subject, on page 175.

## 2 Arithmetic

## 1 Bézout's and Gauss's Theorems

One can directly refer to the arithmetic course given to group A, and whose notes are on page 38.

## 2 Properties of the Fibonacci Sequence

- Exercises -