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

$$
f(x+y) f(x-y)=f(x)^{2} f(y)^{2}
$$

We recall that a function $f$ is said to be odd if for all $x \in \mathbb{R}, f(x)=-f(-x)$.
As $f$ is odd, in particular $f(0)=0$.

With $x=y$ we then get

$$
f(2 y) f(0)=f(y)^{2} f(y)^{2}
$$

which means

$$
f(y)^{4}=0
$$

Therefore, for all $y \in \mathbb{R}, f(y)=0$.

We verify that the zero function is indeed a solution: $0 \times 0=0^{2} \times 0^{2}$.

We conclude that the only and unique solution is the zero function.