Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
Answer: $f(x) \equiv 1$ is the only such function.

Since $f$ is not the zero function, there is an $x_{0}$ such that $f\left(x_{0}\right) \neq 0$. From $f\left(x_{0}\right) f(0)=f\left(x_{0}-0\right)=f\left(x_{0}\right)$ we then get $f(0)=1$. Then by $f(x)^{2}=f(x) f(x)=f(x-x)=f(0)$ we have $f(x) \neq 0$ for any real $x$. Finally from $f(x) f\left(\frac{x}{2}\right)=f\left(x-\frac{x}{2}\right)=f\left(\frac{x}{2}\right)$ we get $f(x)=1$ for any real $x$. It is readily verified that this function satisfies the equation.