The $c=2$ case also has a fixed point, at $x=2$ (if $x_i=2$ then $x_{i+1}=2$, etc.) but there's a slightly deeper reason than this; that's the other case where it's known that the iteration is _conjugate_ to a bit-shift map. Specifically, taking $x_i=2\cos z_i$ we get $2\cos(z_{i+1})=4\cos^2(z_i)-2$, or $\cos(z_{i+1})=2\cos^2(z_i)-1 = \cos(2z_i)$, so a simple change of variables gets us to the equivalent map $z_{i+1}=2z_i$. Of course, this transformation can't actually be carried out in the integers, but I believe the point is that this makes the topological dynamics of the map 'trivial' enough that it's a questionable choice.