Functions satisfying $f(x,y)f(y,z)=f(x,w)f(w,z)$

Question

Let $f(x,y):\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be a real valued function.

(I am not very concerned about the domain/codomain actually, so feel free to change it, e.g. to integers/rationals/etc.)

I am interested to find (non-trivial) examples of $f$ that satisfies $$f(x,y)f(y,z)=f(x,w)f(w,z)$$ for all $x,y,z,w$.

So far I thought of the below examples:

1) $f(x,y)=\frac{x}{y}$, we need to modify the domain to $f(x,y):\mathbb{R}\times\mathbb{R}_{>0}\to\mathbb{R}$.

2) $f(x,y)=a^{x-y}$, where $a$ is a some constant.

Are there other such examples?

Thanks a lot.

Answer 1

Paul's comment is almost the complete solution (namely up to constant factor and some exceptions involving zero values):

Expanding on Paul's comment,

we have nice solutions of the form $$\tag1f(x,y)=c\cdot\frac{g(x)}{g(y)} $$ where $c$ is an arbitrary constant and $g$ an arbitrary map without zero.

We have the trivial solution $f(x,y)=0$ (corresponding to $c=0$ and arbitrary $g$). But we also have more compicated "semi-trivial" solutions:

Let $A,B\subseteq \Bbb R$ be subsets with $A\cup B=\Bbb R$. Then $f$ with $$\tag2f(x,y)=\begin{cases}0&\text{if }x\in A\\0&\text{if }y\in B\\\text{arbitrary}&\text{otherwise}\end{cases} $$ is a solution (because the functional equation will always be $0$ on both sides).

Assume that for some $a$, the function $f(a,\cdot)$ is identically zero. Then $f(x,y)f(y,z)=f(x,a)f(a,z)=0$ for all $x,y,z$, i.e., for all $y$ one of the functions $f(\cdot,y)$ and $f(y,\cdot)$ must be zero. By letting $A=\{\,y\mid f(y,\cdot)=0\,\}$, $B=\{\,y\mid f(\cdot,y)=0\,\}$, we see that $f$ is of the form $(2)$.

The same argument applies if for some $a$, $f(\cdot,a)$ is identically zero.

Assume that $f$ is not of the form $(2)$. Pick $a\in\Bbb R$ and let $g(x):=f(x,a)$. As just seen, $f(\cdot ,a)$ is not identically zero; hence there exists $b$ with $g(b)\ne 0$. We wil also use below that $f(b,\cdot)$ is not identically zero. With $z=w=a$, we have $$ f(x,y)g(y)=g(x)g(a).$$ Assume $g(y)=0$ for some $y$. Then $g(x)g(a)=0$ for all $x$, in particular, $g(a)=0$. Then $f(\cdot,b)=0$, contradiction. We conclude that $g(y)\ne0$ for all $y$ and that $$f(x,y)=\frac{g(x)g(a)}{g(y)},$$ whic is of form $(1)$.