Functions that satisfy the identity $f\left(\frac{x}{t}\right) f\left(-\frac{y}{r}\right)=f\left(\frac{x-y}{t-r}\right)$

Question

I am looking for function(s) which satisfy the following property:

$$f\left(\frac{x}{t}\right) f\left(-\frac{y}{r}\right)=f\left(\frac{x-y}{t-r}\right)$$

I am not sure if there is any function which satisfies this property.

I thought that using an ansatz like $f\left(\frac{x}{t}\right)=\exp \left[g\left(\frac{x}{t}\right)\right]$ should help. But I can't get any further.

Answer 1

Consider the three equations $$\left\{\begin{align} \frac{x}{t} = & z_1,\\ -\frac{y}{r} = & z_2,\\ \frac{x-y}{t-r} = & z_3. \end{align}\right.$$ They can always be solved, except for some special values of the $z_i$ (e.g. $z_1=z_2=0$). This implies that the only solution is $f$ must be constant and equal either $0$ or $1$.

Answer 2

It's not specified in the question, so I'll suppose that the functions under consideration are functions $\Bbb R \to \Bbb R$, and that the desired property should hold for all real $x, y, r, t$ for which both sides of the equality are defined.

If we set $y = 0$ and $r = t - 1$ we get $$f\left(\frac{x}{t}\right) f(0) = f(x)$$ for all $x, t$. Evaluating at $x = 0$ gives that either $f(0) = 0$ or $f(0) = 1$. The former case immediately implies that $f(x) = 0$, and the latter almost immediately that $f(x) = 1$. So, these are the only two solutions.