TITLE: Functions $f$ such that $f(x)+f(-x)=f(x)f(-x)$
QUESTION [9 upvotes]: I was looking for examples of real valued functions $f$ such that $f(x)+f(-x)=f(x)f(-x)$. Preferably, I'd like them to be continuous, differentiable, etc.
Of course, there are the constant functions $f(x)=0$ and $f(x)=2$. I also showed that $1+b^x$, where $b>0$, is another solution. Are there any other nice ones?

REPLY [4 votes]: Hurkyl's solution is nice, but the change of variables obscures the inherent closed orbit property, which is the crucial part of the problem.
Observe that the functional equation only involves $x$ and $ -x$. In particular, if $h(x) = -x$, then the orbit of $x \neq 0$ is $\{x, -x\}$ and the orbit of $0$ is $\{0\}$. As such, the function is uniquely defined by the non-negative part.
For $x = 0$, we have $2 f(0) = f(0)^2 $, which means $ f(0) = 0$ or $2$.
For $x \neq 0$, we have $ f(-x) = \frac{ f(x) } { f(x) - 1}$, if $ f(x) \neq 1$.
Note that if $ f(x) = 1$, then there is no possible value for $ f(-x)$.
This is a necessary and sufficient condition.