Find the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\forall x \neq 1, f(x)+f\left(\frac{1}{1-x}\right)=x$.
Let $x \notin \{0,1\}$. Then $f(x)+f\left(\frac{1}{1-x}\right)=x$. Moreover, since $x \neq 0$, $\frac{1}{1-x} \neq 1$, we can apply the identity to $\frac{1}{1-x}$ to find $f\left(\frac{1}{1-x}\right)+f\left(1-\frac{1}{x}\right)=\frac{1}{1-x}$. Finally, we apply it again to $1-\frac{1}{x}$ to find $f\left(1-\frac{1}{x}\right)+f(x)=1-\frac{1}{x}$.

We thus have three equations fixing three possible values for the function. By adding the first and the third, and subtracting the second, we find $2 \cdot f(x) = \left(x + 1 - \frac{1}{x}\right) - \frac{1}{1-x}$, i.e., $f(x) = \frac{1}{2}\left(x + 1 - \frac{1}{x} - \frac{1}{1-x}\right)$ for $x \notin \{0,1\}$. The values at 0 and 1, on the other hand, are only fixed by $f(0) + f\left(\frac{1}{1-0}\right) = 0$, i.e., $f(1) = -f(0)$.

More generally, the presence of a homography (a function of the form $x \mapsto \frac{a x + b}{c x + d}$) in an exercise is often a sign of a possible clever transformation: the composition of two homographies remains a homography, and appropriate choices of the parameters $a, b, c, d$ often reveal the intent of the exercise.

## Periodicity

Sometimes, the goal of an exercise is not to find the exact set of solutions to a functional equation, but only to determine a property of such solutions. Most commonly, this property turns out to be periodicity. For example: