# Functional equation with three variables

I have a functional equation with three variables. $$f(x,y,z)$$ is a real function with three variables where y is different from z i.e., $$f(x,y,z)$$ defined only for $$y \neq z$$. This function satisfies

1. $$f(x,x,y)=0$$
2. $$f(x,y,x)=1$$
3. $$f(x,y,z)f(z,y,r)=f(x,y,r)$$

What is the general solution for $$f$$?

Or simpler:

Given $$f(x,x,y) = 0$$ and $$f(x,y,x) = 1$$, just set $$x=y$$ to see the contradiction.

There is no such function. If you put $$z=y$$ in 3) you get $$f(x,y,r)=0$$ for all $$x,y,r$$ (provided $$y \neq r$$). But this contradicts 2).

There is no solution.

Note that by 3, For all $$x,y,z$$ you have $$f(x,y,z)=f(x,y,y)f(y,y,z)=f(x,y,y) \cdot 0 =0$$

Edited If $$f(x,y,z)$$ is only defined for $$y \neq z$$ then,a solution is given by $$f(x,y,z)=\frac{x-y}{z-y}$$
• It's not 100% clear from the current phrasing of the question, but I think $f(x,y,z)$ is only defined for $(x,y,z) \in \mathbb{R}^3$ such that $y \neq z$. Hence, $f(x,y,y)$ is undefined. – JimmyK4542 May 5 '19 at 23:25
• Thank you very much. There is a more general solution for this: $f(x,y,z)=\frac{h(x)-h(y)}{h(z)-h(y)}$ or $f(x,y,z)=\frac{h(x)g(y)-g(x)h(y)}{h(z)g(y)-g(z)h(y)}$. I wonder whether there are other solutions. – user409680 May 6 '19 at 0:56