Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$,

$$
f(f(x) f(y))=f(x) y
$$
## Solution 1 :

Before trying to solve the functional equation, we can always try to spot easy solutions in advance. The functions $f: x \rightarrow 0$ and $f: x \rightarrow x$ are solutions. We can also see that the function $f: x \rightarrow -x$ is a solution. These functions are likely the only solutions; let's try to prove it.

Among these solutions, two are injective, and the equation involves an output variable, so we are inclined to show injectivity. However, this is impossible because the zero function is a solution to the problem without being injective. To have a chance of showing the injectivity of the function, we will set aside the zero function by only working with functions that admit a real number $l$ such that $f(l) \neq 0$. If $f(a)=f(b)$, then by setting $y=a$ and $y=b$, we get $f(f(x) f(a))=f(x) a$ and $f(f(x) f(b))=f(x) b \Longrightarrow f(x) a=f(x) b$. Thus, for $x=l$, we can divide by $f(l) \neq 0$ and we get $a=b$ so $f$ is injective. Returning to the original equation, and setting $y=1$, we have $f(f(x) f(1))=f(x) \Longrightarrow f(x) f(1)=x$. If $f(1)=0$, then by setting $x=1$ we find $0=1$ which is false, so $f(1) \neq 0$. Thus $f(x)=\frac{x}{f(1)}$ so we can write $f(x)=c x$ for all $x$.

With the original equation, this gives $f(f(x) f(y))=c^{3} x y=f(x) y=c x y$ so $c=0$ or $c= \pm 1$. We recover the 3 solutions identified at the beginning.

## Solution 2 :

In this equation, the left-hand term is symmetric in $x$ and $y$ while the right-hand term is not. In such a situation, it is interesting to use this symmetry, here it gives: $f(f(x) f(y))=f(x) y=x f(y)$. Therefore, for all non-zero real numbers $x, y$, $\frac{f(x)}{x}=\frac{f(y)}{y}$, so the term $\frac{f(x)}{x}$ is constant and we can write $f(x)=c x$ for all non-zero $x$. Moreover, $f(0) \cdot 1=0 \cdot f(1) \Longrightarrow f(0)=0$, so $f(x)=c x$ for all real $x \cdot f(f(x) f(y))=c^{3} x y=f(x) y=c x y$ so $c=0$ or $c= \pm 1$. We verify that the functions $f: x \mapsto 0, f: x \mapsto x$ and $f: x \mapsto -x$ are indeed solutions to the equation.

## Remark 9.

The trick of using the symmetry of one of the terms in the equation inevitably leads to a symmetric equality in $x$ and $y$. The trick in such cases is to group the $x$ variables on one side and the $y$ variables on the other to exploit this symmetry and find a constant expression. In the previous exercise, this gave $\frac{f(x)}{x}$ constant. We leave it to the reader to establish the form of the functions $f$ satisfying $f(x)^{2^{x}}-f(y)^{2}+x^{4}-y^{4}=2\left(x^{2} f(x)-y^{2} f(y)\right)$.