Let $f:\Bbb Z\to \Bbb Z$ be defined by $f(x) = ax+b$, where $a, b$ are integers. Find the necessary and sufficient condition on $a, b$ in order that $f \circ f = id_\Bbb Z$
I can get the answer easily in a similar case where $f(x)=ax$, simply by imposing the condition $f \circ f(x) = x$ and getting $a= \pm 1$; but in this case I can't "get rid" of the $x$, so getting an condition that depends on its value.
$a(ax+b)+b=x=a^2x+ab+b$ for every $x$ in particular if $x=0$ this implies $ab+b=0$ and $a^2=1$. $a=1$ implies $b+b=0, b=0$, $a=-1$ no condition. The solutions are $(1,0), (-1,b)$.
$f(x)=x$ or $f(x)=-x+b$.