Are 0 and -1 the only rational periodic solutions of $z_{n}\equiv z_{n-1}^{2}+c$? Let $c$ be any complex rational number.  Let $z$ be a series of polynomials in $c$ defined by $z_{n}\equiv z_{n-1}^{2}+c$ and $z_{0}\equiv0$ 
The only rational roots of any $z_{n}$ I have been able to find are $c=0$ for all $n$ and $c=-1$ for all even $n$. Are these are the only rational roots and any $z_n$?  Can it be proven that no other rational roots of any $z_n$ exist?
Point of Interest:  If $z_{n}=0$, then $z_{n+k}=z_{k}$ for all $k$, and therefore $z$ has a periodicity equal to $n$ or to some product of the prime factors of $n$.  This makes $c$ a periodic point of the Mandelbrot Set.

Note that for all $n$, the polynomial $z_n\in\Bbb Z[c]$ always is of the form $c^{2^n}+\ldots +c$, i.e., we clearly have $c=0$ as a root and after dividing out this linear factor we have a monic polynomial ending in $\ldots +1$. By the rational root theorem, any rational root must in fact be integer and a divisor of $1$, i.e., the only candidates are $c=1$ and $c=-1$. Of course $z_n$ grows to infinity for $c=1$, so this is never a root; and you already investigated $c=0$ and $c=1$.
