Functional equations with 3 variables

Question

What are the general solutions of the functional equations?

$$ f(x,y)+f(y,z)=\frac{1}{f(x,z)} $$

$$ f(x,y)f(y,z)f(x,z)=1 $$

Answer 1

Under the false assumption that they need to both hold,

From the first equation, $f$ is never zero. From the second equation, $f(x,y)f(y,z)=\frac{1}{f(x,z)}$; combining with the first we get $$f(x,y)+f(y,z)=f(x,y)f(y,z)$$

Taking $x=y=z$ we get $2f(x,x)=f(x,x)^2$; since $f(x,x)\neq 0$ we have $f(x,x)=2$ for all $x$. Taking $x=y$ we get $2+f(y,z)=2f(y,z)$ so $f(y,z)=2$ for all $y,z$. But this doesn't satisfy the second equation, so there are no solutions.

Answer 2

I am assuming the variables are real.

For the first, setting $x=y=z$ gives $2f(x,x)=\frac 1{f(x,x)}$, so $f(x,x)=\pm\frac {\sqrt 2}2$

As Ivan Loh points out, the sign choice must be the same for all $x$:
$\quad$$y=z$ gives $f(x,z)+f(z,z)=\frac1f{(x,z)}$
$\quad$$x=y$ gives $f(x,x)+f(x,z)=\frac 1{f(x,z)}$
$\quad$so $f(x,x)=f(z,z).$

Assume $f(x,x)=\frac {\sqrt 2}2$:
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)+\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=\frac {\sqrt 2}2, -\sqrt 2$
$\quad$But only the first satisfies the original equation

Now if $f(x,x)=-\frac{\sqrt 2}2$
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=-\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)-\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=-\frac {\sqrt 2}2, \sqrt 2$
$\quad$But only the first solves the original equation

So $f(x,y)$ is the same for all $x,y$ and is $\pm \frac {\sqrt 2}2$

For the second, setting $x=y=z$ gives $f(x,x)=1$
Setting $x=z$ we get $f(x,y)f(y,x)=1$ or $f(x,y)=\frac 1{f(y,x)}$
Setting $x=y$ we have $f(x,z)^2=1$
So $f(x,y)=\pm 1$ for all $x,y$

As Hagen von Eitzen points out, we can divide the reals into two equivalence classes in any way we want (including one of them empty). Let $a,b$ be representatives of the two classes. We have $f(a,a)=f(b,b)=1, f(a,b)=f(b,a)=-1$

Answer 3

There are infinitely many(whose cardinality is $2^\Bbb R$) solutions $f:\Bbb R\to \Bbb R$ of the simple functional equation $$ f(x)^2=1. $$ Let $A$ be an arbitrary subset of $\Bbb R$. Then, $f(x)=1$ for $x\in A$, $f(x)=-1$ for $x\in A^c$ are the solutions of the equation.

Therefore, the above proof(by Millikan) makes no sense.