Balti Tee 1997 · Ülesanne 1

Solution:

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.