If one iterates the map $z \mapsto z^2 + c$ there is obviously a simple formula for the sequence one gets if $c=0$. Less obviously, there is also a simple formula when $c = -2$ (use the identity $2 \cos(2x) = (2\cos(x))^2 - 2)$. Are there any other values of $c$ for which one can solve this recurrence explicitly? (For all initial values of course: there are many trivial explicit solutions for special initial values, such as fixed points.)


Related links:  

<http://en.wikipedia.org/wiki/Mandelbrot_set> (the points $c$ where 0 remains bounded under iteration of this map: this strongly suggests that there is no simple exact solution for general $c$).  

<http://en.wikipedia.org/wiki/Logistic_map> (gives the explicit solutions above, after a change of variable)


Motivation: I once used the map with $c=-2$ in a lecture to show that one could prove limits exist even without a formula for the exact solution. A first year calculus student pointed out the non-obvious exact solution above, and I don't want to be caught out like this again. 



When does iterating $z \mapsto z^2 + c$ have an exact solution?
No, there are no others. 


Analytically, one can show that if a Julia set contains an analytic arc, it is in fact a straight line or a circle (up to conjugation). For the class $z^2+c$, $0$ and $-2$ are the only values where this occurs.


This does not quite imply that there are no closed-form solutions of the recurrence for any other value of $c$, but such a closed form would naturally describe a fractal. As far as I know, there are no "closed form" functions which **directly** have a fractal as one of its level set.


