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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 $f$?

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Or simpler:

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

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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).

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There is no solution.

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

But this contradicts 2.

Edited If $f(x,y,z)$ is only defined for $y \neq z$ then,a solution is given by $$f(x,y,z)=\frac{x-y}{z-y}$$

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  • $\begingroup$ 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. $\endgroup$ – JimmyK4542 May 5 at 23:25
  • $\begingroup$ @JimmyK4542 Edited to include a solution in that case. Looks like a hard equation to solve then, but the OP is only asking for a solution, and the given condition suggest denominator :) $\endgroup$ – N. S. May 5 at 23:31
  • $\begingroup$ 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. $\endgroup$ – user409680 May 6 at 0:56

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