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

$$
f(x) \cdot f(y)=f(x-y)
$$
For $x=y=0$, we find $f(0)^{2}=f(0)$, so $f(0)=0$ or $f(0)=1$. Next, for $x=y$, we get $f(x)^{2}=f(0)$. If $f(0)=0$, then $f$ is the zero function. Otherwise, $f(0)=1$, in which case $f(x)=1$ or $f(x)=-1$. But be careful! This is not enough to say that $f$ is constant. Indeed, we must not forget to consider so-called multigraph functions, that is, functions that take the value 1 at some points and -1 at others. We then set $x=0$, and we get $f(y)=f(-y)$, so $f$ is even. Finally, by setting $y=-x$, we arrive at $f(x) f(-x)=f(2 x)$, or $f(2 x)=f(x)^{2}=1$ by the evenness of $f$. Ultimately, since $2 x$ spans $\mathbb{R}$ as $x$ spans $\mathbb{R}$, the only possible solution in this case is the constant function equal to 1. The solutions to the problem are therefore $x \mapsto 0$ and $x \mapsto 1$, which we can verify in reverse that they are valid.