Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfying $f(x)+f(y) \neq 0$ and

$$
\frac{f(x)-f(x-y)}{f(x)+f(x+y)}+\frac{f(x)-f(x+y)}{f(x)+f(x-y)}=0
$$

for all $x, y \in \mathbf{R}$.
Any function of the form $f(x)=c$, where $c$ is a nonzero constant, clearly satisfies the requirements; we will show these are the only solutions. Bringing the original equation to a common denominator yields $2 f(x)^{2}-f(x-y)^{2}-f(x+y)^{2}=0$ for all $x$, $y$. Define $g(x)=f(x)^{2}$; clearly $g(x) \geq 0$ for all $x$. Our equation states that $2 g(x)=g(x-y)+g(x+y)$. We will show this forces $g$ to be constant.

If $g$ is not constant, there exist some $a, b$ with $g(a)1$ then our equation (with $x=(n-1) a-(n-2) b, y=a-b)$ yields

$$
2 g((n-1) a-(n-2) b)=g(n a-(n-1) b)+g((n-2) a-(n-3) b)
$$

$\Rightarrow 2(n-1) g(a)-2(n-2) g(b)=g(n a-(n-1) b)+(n-2) g(a)-(n-3) g(b)$ (by induction hypothesis)

$$
\Rightarrow g(n a-(n-1) b)=n g(a)-(n-1) g(b)
$$

and the induction step is complete. However, if $n>g(b) /(g(b)-g(a))$ then $g(n a-(n-1) b)=$ $n g(a)-(n-1) g(b)<0$ and we have a contradiction. This shows that $g$ must be constant.

So $g(x)=d$ for some constant $d \geq 0$; then $f(x)= \pm \sqrt{d}$ for all $x$. If both values occur, then the $f(x)+f(y) \neq 0$ constraint is violated; hence we must have $f(x)=\sqrt{d}$ for all $x$ or $f(x)=-\sqrt{d}$ for all $x$ (and $d \neq 0$ ).